100 Bai Toan Tin Hoc Pascal Va Bai Giai

8/2/2019 100 Bai Toan Tin Hoc Pascal Va Bai Giai

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100 Ton TinTin hc & Nh trng

H Ni - 2002


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100 Problems & Solutions Page 2

Phn 1: BI

Bi 1/1999 -Tr chi cng nhau qua cu

(Dnh cho hc sinh Tiu hc)Bn ngi cn i qua mt chic cu. Do cu yu nn mi ln i khng qu hai ngi, vv tri ti nn phi cm n mi i c. Bn ngi i nhanh chm khc nhau, qua cuvi thi gian tng ng l 10 pht, 5 pht, 2 pht v 1 pht. V ch c mt chic n nnmi ln qua cu phi c ngi mang n tr v cho nhng ngi k tip. Khi hai ngii cng nhau th qua cu vi thi gian ca ngi i chm hn. V d sau y l mt cchi:- Ngi 10 pht i vi ngi 5 pht qua cu, mt 10 pht.- Ngi 5 pht cm n quay v, mt 5 pht.

- Ngi 5 pht i vi ngi 2 pht qua cu, mt 5 pht.- Ngi 2 pht cm n quay v, mt 2 pht.- Ngi 2 pht i vi ngi 1 pht qua cu, mt 2 pht.Thi gian tng cng l 10+5+5+2+2 = 24 pht.Em hy tm cch i khc vi tng thi gian cng t cng tt, v nu di 19 pht th thttuyt vi! Li gii ghi trong tp vn bn c tn l P1.DOC

Bi 2/1999 - T chc tham quan

(Dnh cho hc sinh THCS)Trong t t chc i tham quan danh lam thng cnh ca thnh ph H Ch Minh, Ban tchc hi thi Tin hc tr t chc cho N on ( nh t s 1 n N) mi on i thm quanmt a im khc nhau. on th i i thm a im cch Khch sn Hong di km

(i=1,2,...., N). Hi thi c M xe taxi nh s t 1 n M (MN) phc v vic a ccon i thm quan. Xe th j c mc tiu th xng l vj n v th tch/km.Yu cu: Hy chn N xe phc v vic a cc on i thm quan, mi xe ch phc vmt on, sao cho tng chi ph xng cn s dng l t nht.

D liu:File vn bn P2.INP:

- Dng u tin cha hai s nguyn dng N, M (N M 200);

- Dng th hai cha cc s nguyn dng d1, d2, ..., dN;- Dng th ba cha cc s nguyn dng v1, v2, ..., vM.- Cc s trn cng mt dng c ghi khc nhau bi du trng.

Kt qu: Ghi ra file vn bn P2.OUT:- Dng u tin cha tng lng xng du cn dng cho vic a cc on i thm quan(khng tnh lt v);- Dng th i trong s N dng tip theo ghi ch s xe phc v on i (i=1, 2, ..., N).

V d:

Tin hc & Nh trng 100 Ton - Tin hc


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P2.INP P2.OUT3 47 5 9

17 13 15 10

2562

34

Bi 3/1999 -Mng t bo(Dnh cho hc sinh THPT)Mng t bo c dng mt li vung hnh ch nht. Ti mi nhp thi gian: mi cali cha tn hiu l 0 hoc 1 v c th truyn tn hiu trong n cho mt s k cnh theomt qui lut cho trc. gc trn bn tri c th nhn tn hiu t bn ngoi a vo.Sau nhp thi gian , tn hiu mt s l 0 nu tt c cc tn hiu truyn n n l 0,cn trong trng hp ngc li tn hiu trong n s l 1. Mt khng nhn c tn hiuno t cc k cnh vi n s gi nguyn tn hiu ang c trong n. Ring i vi trntri, sau khi truyn tn hiu cha trong n i, nu c tn hiu vo th trn tri s chnhntn hiu ny, cn nu khng c tn hiu no th trn tri cng hot ng ging nh cc khc. trng thi u tn hiu trong tt c cc l 0.Yu cu:Cho trc s nhp thi gian T v dy tn hiu vo S l mt dy gm T k hiuS1, ..., ST, trong Si l 0 hoc 1 th hin c tn hiu vo, ngc li Si l X th hin

khng c tn hiu vo ti nhp thi gian th i (1 i T), hy xc nh trng thi ca li

sau nhp thi gian th T.D liu: vo t file vn bn P3.INP:- Dng u tin cha 3 s nguyn M, N, T theo th t l s dng, s ct ca li v s

nhp thi gian (1


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V d:

P3.INP P3.OUT2 2 5101XX

2 42 1

1101

Qu trnh bin i trng thi c din t trong hnh di y:

0 0 1 0 0 1 1 0 1 1 1 10 0 0 0 0 0 0 1 1 0 0 1

Bi 4/1999 -Tr chi bc si

(Dnh cho hc sinh Tiu hc)Trn mt t c mt ng si c 101 vin. Hai em hc sinh Hong v Huy chi tr chinh sau: Mi em n lt i phi bc ra t ng si trn ti thiu l 1 vin v ti a l 4vin. Ngi thua l ngi phi bc vin si cui cng. Gi s Hong l ngi c bctrc, Huy bc sau. Cc em th ngh xem ai l ngi thng cuc, Hong hay Huy? Vngi thng cuc phi suy ngh g v thc hin cc bc i ca mnh ra sao?

Bi 5/1999 - 12 vin bi

(Dnh cho hc sinh THCS)C 12 hn bi ging ht nhau v kch thc, hnh dng v khi lng. Tuy nhin trongchng li c ng mt hn bi km cht lng: hoc nh hn hoc nng hn bnh thng.Dng mt cn bn hai bn, bn hy dng 3 ln cn tm ra c vin bi . Cn ch rrng vin bi l nng hn hay nh hn.Vit chng trnh m phng vic t chc cn cc hn bi trn. D liu v hn bi km chtlng do ngi s dng chng trnh nm gi. Yu cu trnh by chng trnh p vm thut.

Bi 6/1999 - Giao im cc ng thng(Dnh cho hc sinh THPT)Trn mt phng cho trc n ng thng. Hy tnh s giao im ca cc ng thngny. Yu cu tnh cng chnh xc cng tt.Cc ng thng trn mt phng c cho bi 3 s thc A, B, C vi phng trnh Ax +By + C = 0, y cc s A, B khng ng thi bng 0.D liu vo ca bi ton cho trong tp B6.INP c dng sau:- Dng u tin ghi s n- n dng tip theo, mi dng ghi 3 s thc A, B, C cch nhau bi du cch.Kt qu ca bi ton th hin trn mn hnh.



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Bi 7/1999 - Min mt phng chia bi cc ng thng(Dnh cho hc sinh THPT)Xt bi ton tng t nh bi 6/1999 nhng yu cu tnh s min mt phng c chia

bi n ng thng ny:

Trn mt phng cho trc n ng thng. Hy tnh s min mt phng c chia bi ccng thng ny. Yu cu tnh cng chnh xc cng tt.Cc ng thng trn mt phng c cho bi 3 s thc A, B, C vi phng trnh Ax +By + C = 0, y cc s A, B khng ng thi bng 0.D liu vo ca bi ton cho trong tp B7.INP c dng sau:- Dng u tin ghi s n- n dng tip theo, mi dng ghi 3 s thc A, B, C cch nhau bi du cch.Kt qu ca bi ton th hin trn mn hnh.

Bi 8/1999 - Cn to(Dnh cho hc sinh Tiu hc)M i ch v mua cho Nga 27 qu to ging ht nhau v kch thc v khi lng. Tuynhin ngi bn hng ni rng trong s cc qu to trn c ng mt qu c khi lngnh hn. Em hy dng mt chic cn bn hai bn tm ra qu to nh . Yu cu sln cn l nh nht.Cc em hy gip bn Nga tm ra qu to nh i. Nu cc em tm ra qu to sau thn 5 ln cn th l tt lm ri.

Bi 9/1999 - Bc dim(Dnh cho hc sinh Tiu hc)Trn bn c 3 dy que dim, s lng que dim ca cc dy ny ln lt l 3, 5 v 8. Hai

bn Nga v An chi tr chi sau: Mi bn n lt mnh c quyn (v phi) bc mts que dim bt k t mt dy trn. Ngi thng l ngi bc c que dim cui cng.Ai l ngi thng cuc trong tr chi trn? V bn phi bc dim nh th no? Cc

bn hy cng suy ngh vi Nga v An nh.

Bi 10/1999 - Dy s nguyn(Dnh cho hc sinh THCS)Dy cc s t nhin c vit ra thnh mt dy v hn trn ng thng:1234567891011121314..... (1)Hi s v tr th 1000 trong dy trn l s no?Em hy lm bi ny theo hai cch: Cch 1 dng suy lun logic v cch 2 vit chngtrnh tnh ton v so snh hai kt qu vi nhau.



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Tng qut bi ton trn: Chng trnh yu cu nhp s K t bn phm v in ra trn mnhnh kt qu l s nm v tr th K trong dy (1) trn. Yu cu chng trnh chy cngnhanh cng tt.

Bi 11/1999 - Dy s Fibonaci(Dnh cho hc sinh THCS)

Nh cc bn bit dy s Fibonaci l dy 1, 1, 2, 3, 5, 8, .... Dy ny cho bi cng thc qui sau:F1 = 1, F2 =1, Fn = Fn-1 + Fn-2 vi n > 21. Chng minh khng nh sau:Mi s t nhin N u c th biu din duy nht di dng tng ca mt s s trong dys Fibonaci.

N = akFk + ak-1Fk-1 + .... a1F1

Vi biu din nh trn ta ni N c biu din Fibonaci l akak-1...a2a1.2. Cho trc s t nhin N, hy tm biu din Fibonaci ca s N.Input:Tp vn bn P11.INP bao gm nhiu dng. Mi dng ghi mt s t nhin.Output:Tp P11.OUT ghi kt qu ca chng trnh: trn mi dng ghi li biu din Fibonaci cacc s t nhin tng ng trong tp P11.INP.

Bi 12/1999 - N-mino(Dnh cho hc sinh THPT)

N-mino l hnh thu c t N hnh vung 1 1 ghp li (cnh k cnh). Hai n-mino cgi l ng nht nu chng c th t chng kht ln nhau.Bn hy lp chng trnh tnh v v ra tt c cc N-mino trn mn hnh. S n nhp t bn

phm.V d: Vi N=3 ch c hai loi N-mino sau y:

3-mino thng 3-mino hnh thc th

Ch : Gi Mn l s cc n-mino khc nhau th ta c M1=1, M2=1, M3=2, M4=5, M5=12,M6=35,...Yu cu bi gii ng v trnh by p.



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Bi 13/1999 - Phn hoch hnh ch nht(Dnh cho hc sinh THPT)Mt hnh vung c th chia thnh nhiu hnh ch nht c cc cnh song song vi cnhhnh vung (xem Hnh v). Xy dng cu trc d liu v lp chng trnh m t php

chia . Tnh xem c bao nhiu cch chia nh vy.InputD liu nhp vo t tp P13.INP bao gm hai s t nhin ln, m - kch thc hnh ch nht.OutputD liu ra nm trong tp P13.OUT c dng sau:- Dng u tin ghi s K l tng s cc php phn hoch.- Tip theo l K nhm, mi nhm cch nhau bng mt dngtrng.

- Mi nhm d liu bao gm cc cp ta ca cc hnh ch nht nm trong phn hoch.

Bi 14/2000 - Tm s trang sch ca mt quyn sch(Dnh cho hc sinh Tiu hc) nh s cc trang sch ca 1 quyn sch cn tt c 1392 ch s. Hi quyn sch c ttc bao nhiu trang?

Bi 15/2000 - Hi ngh i vin(Dnh cho hc sinh Tiu hc)

Trong mt hi ngh lin chi i c mt s bn nam v n. Bit rng mi bn trai u quenvi N cc bn gi v mi bn gi u quen vi ng N bn trai. Hy lp lun chng trng trong hi ngh s cc bn trai v cc bn gi l nh nhau.

Bi 16/2000 - Chia s(Dnh cho hc sinh THCS)Bn hy chia N2 s 1, 2, 3, ...., N2-1, N2 thnh N nhm sao cho mi nhm c s cc shng nh nhau v c tng cc s ny cng bng nhau.

Bi 17/2000 - S nguyn t tng ng(Dnh cho hc sinh THCS)Hai s t nhin c gi l Nguyn t tng ngnu chng c chung cc c snguyn t. V d cc s 75 v 15 l nguyn t tng ng v cng c cc c nguyn tl 3 v 5. Cho trc hai s t nhin N, M. Hy vit chng trnh kim tra xem cc s nyc l nguyn t tng ng vi nhau hay khng.

Bi 18/2000 - Sn b(Dnh cho hc sinh THCS v THPT)

Trn li vung mt con sn xut pht t nh (0,0) cn phi i n im kt thc ti(N,0) (N l s t nhin cho trc).



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Qui tc i: Mi bc (x1, y1) --> (x2, y2) tho mn iu kin (sn b):

- x2 = x1+1,

- y1 -1


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"Mnh s hi, c phi bn cn h s 1, s 2,..., s 63 khng. Nh vy vi nhiu nht63 cu hi mnh s bit c bn cn h no."Bn Nam ni:

"Cn mnh ch cn n 14 cu, 7 cu bit bn tng my v 7 cu c th bit

chnh xc bn cn h s my ".Cn em, em phi hi nhiu nht my ln bit c bn Lan cn h s bao nhiu?

Bi 21/2000 - Nhng trang sch b ri(Dnh cho hc sinh Tiu hc)Mt cun sch b ri mt mt mng. Trang b ri th nht c s 387, cn trang cui cnggm 3 ch s 3, 8, 7 nhng c vit theo mt th t khc.Hi c bao nhiu trang sch b ri ra?

Bi 22/2000 - m ng i(Dnh cho hc sinh THCS)Cho hnh sau:a) Bn hy m tt c cc ng i t A n B. Mi ng i ch c i qua mi nhnhiu nht l 1 ln.

b) Bn hy tm tt c cc ng i t A n D, sao cho ng i qua mi cnh ngmt ln.c) Bn hy tm tt c cc ng i qua tt cc cc cnh ca hnh, mi cnh ng mt ln,sao cho:

- im bt u v im kt thc trng nhau.- im bt u v im kt thc khng trng nhau

Bi 23/2000 - Quay Rubic(Dnh cho hc sinh THPT)

Rubic l mt khi lp phng gm 3 3 3 = 27 khi lp phng con. Mi mt rubic

gm 3 3 = 9 mt ca mt lp 9 khi lp phng con. trng thi ban u, mi mt rubicc t mt mu. Cc mt khc nhau c t cc mu khc nhau. Gi s ta ang nhnvo mt mt trc ca rubic. C th k hiu mu cc mt nh sau: F: mu mt trc l

mt ta ang nhn; U: mu mt trn; R: mu mt phi; B: mu mt sau; L: mu mt bntri; D: mu mt di.

Mt lp gm 3 3 khi lp phng con c th quay 90 nhiu ln, trc quay i qua tm

v vung gc vi mt ang xt. Kt qu sau khi quay l khi lp phng 3 3 3 vi ccmu mt b i khc.Mt xu vng quay lin tip rubic c th m t bng xu cc ch ci ca U, R, F, D, B, L,trong mi ch ci l k hiu mt vng quay c s: quay mt tng ng 90 theochiu kim ng h. Hy vit chng trnh gii 3 bi ton di y:1. Cho 2 xu INPUT khc nhau, kim tra xem liu nu p dng vi trng thi u c cho

cng mt kt qu hay khng?



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2. Cho mt xu vo, hy xc nh s ln cn p dng xu vo cho trng thi u rubic li nhn c trng thi u .

Bi 24/2000 - Sp xp dy s

(Dnh cho hc sinh Tiu hc)Cho dy s: 3, 1, 7, 9, 5Cho php 3 ln i ch, mi, ln c i ch hai s bt k. Em hy sp xp li dy strn theo th t tng dn.

Bi 25/2000 - Xy dng s(Dnh cho hc sinh THCS)Cho cc s sau: 1, 2, 3, 5, 7Ch dng php ton cng hy dng dy trn to ra s: 43, 52.

V d to s 130 bn c th lm nh sau: 123 + 7 = 130.

Bi 26/2000 - T mu(Dnh cho hc sinh THCS)Cho li vung 4x4, cn phi t mu cc ca li. c php dng 3 mu: Xanh, ,

vng. iu kin t mu l ba bt k lin nhau theo chiu dc v ngang phi khcmu nhau. Hi c bao nhiu cch nh vy, hy lit k tt c cc cch.

Bi 27/2000 - Bn c

(Dnh cho hc sinh THPT)Cho mt bn c vung 8x8, trn cho trc mt s qun c. V d hnh v sau l mt

bn c nh vy:

D liu nhp c ghi trn tp BANCO.TXT bao gm 8 dng, mi dng l mt su nhphn c di bng 8. V tr cc qun c ng vi s 1, cc trng ng vi s 0. V dtp BANCO.TXT ng vi bn c trn:01010100100110011010001100010100

00100000



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010100011001100001000110Hy vit chng trnh tnh s qun c lin tc ln nht nm trn mt ng thng trn

bn c. ng thng y c th l ng thng ng. ng nm ngang hoc ngcho. Kt qu th hin trn mn hnh.Vi v d nu trn, chng trnh phi in trn mn hnh kt qu l 4.Bi 28/2000 - i tin(Dnh cho hc sinh Tiu hc)Gi s bn c nhiu t tin loi 1, 2 v 3 ngn ng. Hi vi cc t tin bn c baonhiu cch i t 10 ngn ng? Hy lit k cc cch i.

Bi 29/2000 - Chn bn(Dnh cho hc sinh THCS)Trong mt tri h ngi ta tnh c chn ra mt nhm 6 hc sinh. Chng minh rng s tmc 3 trong s 6 bn sao cho 3 bn ny hoc quen nhau (i mt) t trc hoccha h quen nhau. Em hy ch ra cch tm 3 bn .

Bi 30/2000 - Phn t yn nga(Dnh cho hc sinh THCS)Cho bng A kch thc MxN. Phn t Aij c gi l phn t yn nga nu n l phn t

nh nht trong hng ca n ng thi l phn t ln nht trong ct ca n. V d trongbng s sau y:

15 3 955 4 676 1 2

th phn t A22 chnh l phn t yn nga.Bn hy lp chng trnh nhp t bn phm mt bng s kch thc MxN v kim traxem n c phn t yn nga hay khng?

Bi 31/2000 - Biu din phn s(Dnh cho hc sinh PTTH)Mt phn s lun lun c th c vit di s thp phn hu hn hoc v hn tunhon. V d:

23/5 = 4.63/8 = 0.3751/3 = 0.(3)45/56 = 0.803(571428)....

Trong cc v d trn th cc ch s t trong du ngoc ch phn tun hon ca s thpphn.



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Nhim v ca bn l vit mt chng trnh nhp t s (N) v nhp mu s (D), sau a ra kt qu l dng thp phn ca phn s N/D.V d chy chng trnh:

Nhap N, D:1 7

1/7 = 0.(142857)_

Bi 32/2000 - Bi ton 8 hu(Dnh cho hc sinh Tiu hc)Trn bn c vua hy sp xp ng 8 qun Hu sao cho khng cn con no c th n ccon no. Hy tm ra nhiu cch sp nht?

Bi 33/2000 - M ho vn bn(Dnh cho hc sinh THCS)

Bi ton sau m t mt thut ton m ho n gin ( tin ta ly v d ting Anh, ccbn c th m rng cho ting Vit):Tp hp cc ch ci ting Anh bao gm 26 ch ci c nh s th t t 0 n 25 nhsau:

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

a b c d e f g h i j k l m n o p q r s t u v w x y Z

Quy tc m ho mt k t nh sau (ly v d k t X):

- Tm s th t tng ng ca k t ta c 23- Tng gi tr s ny ln 5 ta c 28- Tm s d trong php chia s ny cho 26 ta c 2- Tra ngc bng ch ci ta thu c C.

a. S dng quy tc trn m ho cc dng ch sau:PEACEHEAL THE WORLDI LOVE SPRING

b. Hy tm ra quy tc gii m cc dng ch sau:

N FR F XYZIJSYNSKTVRFYNHXMFSTN SFYNTSFQ ZSNBJVXNYD

Bi 34/2000 - M ho v gii m(Dnh cho hc sinh THCS)Theo quy tc m ho bi trn (33/2000), hy vit chng trnh cho php:

- Nhp mt xu k t v in ra xu k t c m ha- Nhp mt xu k t c m ho v in ra su k t c gii m.

V d khi chy chng trnh:Nhap xau ky tu:



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PEACE Xau ky tu tren duoc ma hoa la:UJFHJ

Nhap xau ky tu can giai ma:

FRXau ky tu tren duoc giai ma la:AM_

Bi 35/2000 - Cc phn s c sp xp(Dnh cho hc sinh THPT)Xt tp F(N) tt c cc s hu t trong on [0,1] vi mu s khng vt qu N.V d tp F(5):

0/1 1/5 1/4 1/3 2/5 1/2 3/5 2/3 3/4 4/5 1/1

Hy vit chng trnh cho php nhp s nguyn N nm trong khong t 1 n 100 vxut ra theo th t tng dn cc phn s trong tp F(N) cng s lng cc phn s .V d khi chy chng trnh:

Nhap so N: 50/1 1/5 1/4 1/3 2/5 1/2 3/5 2/3 3/4 4/5 1/1Tat ca co 11 phan so_

Bi 36/2000 - Anh chng h tin(Dnh cho hc sinh Tiu hc)

Mt chng h tin ra hiu may qun o. Ngi ch hiu bit tnh khch nn ni vi anhta: Ti tnh tin cng theo 2 cch: cch th nht l ly ng 11700 ng. Cch th hai lly theo tin cc: chic cc th nht ti ly 1 ng, chic cc th 2 ti ly 2 ng gp ichic th nht, chic cc th 3 ti ly 4 ng gp i ln chic cc th 2 v c tip tcnh th cho n ht. o ca anh c 18 chic cc. Nu anh thy cch th nht l t th anhc th tr ti theo cch th hai.Sau mt hi suy ngh chng h tin quyt nh chn theo cch th hai. Hi anh ta phi tr

bao nhiu tin v anh ta c b h hay khng?

Bi 37/2000 - S siu nguyn t(Dnh cho hc sinh THCS)S siu nguyn t l s nguyn t m khi b mt s tu cc ch s bn phi ca n th

phn cn li vn to thnh mt s nguyn t.V d 7331 l mt s siu nguyn t c 4 ch s v 733, 73, 7 cng l cc s nguyn t.

Nhim v ca bn l vit chng trnh nhp d liu vo l mt s nguyn N (0< N


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Tat ca co 16 so_

Bi 38/2000 - Tam gic s(Dnh cho hc sinh THPT)

Hnh sau m t mt tam gic s c s hng N=5:

73 8

8 1 02 7 4 4

4 5 2 6 5

i t nh (s 7) n y tam gic bng mt ng gp khc, mi bc ch c i t s hng trn xung mt trong hai s ng k bn phi hay bn tri hng di, v cng

cc s trn ng i li ta c mt tng.V d: ng i 7 8 1 4 6 c tng l S=26, ng i 7 3 1 7 5 c tng l S=23Trong hnh trn, tng Smax=30 theo ng i 7 3 8 7 5 l tng ln nht trong tt c cctng.

Nhim v ca bn v vit chng trnh nhn d liu vo l mt tam gic s cha trongtext file INPUT.TXT v a ra kt qu l gi tr ca tng Smax trn mn hnh.File INPUT.TXT c dng nh sau:Dng th 1: c duy nht 1 s N l s hng ca tam gic s (0


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1.Trt hnh vung pha trn trng.2.Trt hnh vung bn phi trng.3.Trt hnh vung bn phi trng.4.Trt hnh vung pha di trng.

5.Trt hnh vung pha di trng.6.Trt hnh vung bn tri trng.

T R G S JX D O K IM V L NW P A B EU Q H C F

Cu hnh ban u ca ch

Bn hy vit mt chng trnh ca bn cha cu hnh ban u ca ch cng cc nci v ra ch kt qu.Inputu vo ca chng trnh ca bn cha cu hnh ban u ca mt ch v mt dy ccnc i trong ch .

Nm dng u tin m t cu hnh ban u ca ch, mi dng tng ng vi mt hngca ch v cha ng 5 k t tng ng vi 5 hnh vung ca ch trn hng .

trng c din t bng mt du cch.Cc dng tip theo sau l dy cc nc i. Dy cc nc i c ghi bng dy cc chA,B,R v L th hin hnh vung no c trt vo trng. A th hin hnh vung

pha trn trng c trt vo trng, tng ng: B-pha di, R-bn phi, L-bn tri.C th c nhng nc i khng hp cch, ngay c khi n c biu th bng nhng chci trn. Nu xut hin mt nc i khng hp cch th ch coi nh khng c cu hnhkt qu. Dy cc nc i c th chim mt s dng, nhng n s c xem l kt thcngay khi gp mt s 0.Out put

Nu ch khng c cu hnh kt qu th thng bo 'This puzzle has no finalconfiguration.'; ngc li th hin th cu hnh ch kt qu. nh dng mi dng kt qu

bng cch t mt du cch vo gia hai k t k tip nhau. trng cng c s l nhvy. V d nu trng nm bn trong hng th n c xut hin di dng 3 du cch:mt ngn cch n vi k t bn tri, mt th hin chnh trng , cn mt ngncch n vi k t bn phi.Ch : Input mu u tin tng ng vi ch c minh ho trong v d trn.Sample Input 1TRGSJ

XDOKI


TT RR GG SS JJXX OO KK LL IIMM DD VV BB NNWW PP AA EEUU QQ HH CC FF


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M VLNWPABEUQHCFARRBBL0

Sample Output 1T R G S JX O K L IM D V B NW P A EU Q H C FSample Input 2AB C DEF G H I J

KLMNOPQRSTUVWXAAALLLL0Sample Output 2

A B C DF G H I EK L M N J

P Q R S OT U V W XSample Input 3ABCDEFGHIJKLMNOPQRSTUVWXAAAAABBRRRLL0

Sample Output 3This puzzle has no final configuration.

Bi 40/2000 - My nh v RadioMt con tu c trang b ng-ten nh hng c th xc nh v tr hin thi ca mnhnh cc ln c n hiu a phng. Mi n hiu c t mt v tr bit v phtra mt tn hiu n nht. Mi khi bt c tn hiu, tu lin quay ng-ten ca mnh chon khi t c tn hiu cc i. iu cho php xc nh c phng v tng ica n hiu. Cho bit d liu ca ln c trc (thi gian, phng v tng i, v tr

ca n), mt ln c mi xc nh v tr hin thi ca tu. Bn phi vit mtchng trnh xc nh v tr hin thi ca tu t hai ln c n hiu.



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V tr ca cc n hiu v cc con tu c cho trong h to vung gc, trc Ox hngv pha ng, cn Oy hng v pha bc. Hng i ca con tu c o bng , theochiu kim ng h tnh t hng bc. Nh vy, hng bc s l 0 0, hng ng l 900,hng nam l 1800 v hng ty l 2700. Phng v tng i ca n hiu cng c o

bng , tng i vi hng i ca tu v theo chiu kim ng h. ng ten khng thch ra n hiu nm hng no trn phng v. Nh vy, mt phng v 900 c ngha ln hiu c th nm hng 900 hoc 2700.InputDng u tin ca input l mt s nguyn ch s lng cc n hiu (nhiu nht l 30).Mi dng tip theo cho mt n hiu. Mi dng bt u bng tn n (l mt chui k tkhng vt qu 20 k t), sau l v tr ca n cho bng honh v tung . Cctrng ny phn cch bi mt du cch.Dng tip theo ngay sau cc d liu v n hiu l mt s nguyn ch s lng cc kch

bn ng i ca tu. Mi kch bn cha 3 dng gm mt dng cho bit hng i ca tuso vi hng Bc v vn tc vn tc thc ca tu, v hai dng ch hai ln c n hiu.Thi gian c o bng pht, tnh t lc na m trong vng 24 gi. Vn tc o bngn v di (nh cc n v ca h trc to ) trn n v thi gian. Dng th hai cakch bn l ln c th nht gm thi gian (l mt s nguyn), tn n v gc phng vtng i vi hng i ca tu. Ba trng c ngn cch nhau bi mt du cch. Dngth ba ca kch bn l ln c th hai. Thi gian ca ln c ny lun ln hn ln cth nht.Output

Vi mi kch bn, chng trnh ca bn phi ch ra c s th t ca kch bn (Scenario1, Scenario 2,...), v mt thng bo v v tr ca con tu (c lm trn n hai ch sthp phn) ti thi im ca ln c th hai. Nu v tr ca tu khng th xc nh ththng bo: Position cannot be determined.

Mu input v output chnh xc tng ng c cho nh sau:Sample Input4First 2.0 4.0Second 6.0 2.0

Third 6.0 7.0Fourth 10.0 5.020.0 1.01 First 270.02 Fourth 90.0116.5651 2.23614 Third 126.86995 First 319.3987

Sample OutputScenario 1: Position cannot be determined



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Scenario 2: Position is (6.00, 5.00)

Bi 41/2000 - C Othello(Dnh cho hc sinh THCS v THPT)

C Othello l tr chi cho 2 ngi trn mt bn c kch thc 8x8 , dng nhng quntrn mt mt en, mt mt trng. Cc u th s c ln lt i mt qun vo cntrng trn bn c. Khi i mt qun, u th phi lt c t nht mt qun ca u thkia. Cc qun s lt c nu chng nm lin tip trn cng mt ng thng (ngang, dchoc cho) m hai u ca ng l hai qun c mu ca u th ang i. Khi xongmt lt i, tt c cc qun b lt c i sang mu ca u th va i. Trong mtlt i c th lt c nhiu hng.V d: Nu th c hin thi bn c bn tri v lt i l ca u th trng, th anh ta cth i c mt trong cc nc sau: (3,5) (4,6) (5,3) (6,4). Nu anh ta i nc (3,5) th

sau nc i th c s nh bn c bn phi.V bn cBn hy vit mt chng trnh c mt vn c t mt text file c qui cch:8 dng u tin l bn c th, mi dng cha 8 k t, mi k t c th l:'-' th hin mt trng,'B' th hin mt c qun en,'W' th hin mt c qun trng.Dng th 9 cha mt trong hai k t 'B' hoc 'W' ch nc i thuc v u th no.Cc dng tip theo l cc lnh. Mi lnh c th l: lit k tt c cc nc i c th ca

u th hin thi, thc hin mt nc i, hay thi chi vn c . Mi lnh ghi trn mtdng theo qui cch sau:Lit k tt c cc nc i c th ca u th hin thi:Lnh l mt ch 'L' ct u tin ca dng. Chng trnh phi kim tra c bn c v inra tt c cc nc i hp l ca u th hin thi theo dng (x,y) trong x l hng v yl ct ca nc i. Cc nc i ny phi c in theo qui cch:+ Mi nc i trn hng i s c in trc mi nc i trn hng j nu j>i.+ Nu trn hng i c nhiu hn 1 nc i th cc nc i c in theo th t ca ct.Mi nc i hp l phi in trn mt dng. Nu khng c nc i no hp l v u th

hin thi khng th lt bt c mt qun no th phi in ra thng bo 'No legal move'.Thc hin mt nc iLnh l mt ch 'M' ct u tin ca dng, tip theo sau l 2 ch s ct th hai v th

ba ca dng. Cc ch s ch ra hng v ct ca trng trn bn c ni u th hin this t qun ca mnh, tr phi anh ta khng c nc i hp l no. Nu u th hin thikhng c nc i hp l no th anh ta c thay bi u th kia v by gi nc i lca u th mi. Chng trnh phi kim tra khi nc i l hp l. Bn s phi ghinhn s thay i trn bn c, k c vic thm cc qun mi ln vic thay i mu scqun c b lt. Cui mi nc i hy in ra s lng tt c cc qun c mi mu trn bn

c theo qui cch 'Black - xx White - yy, trong xx l s lng cc qun en cn yy l slng cc qun trng. Sau mt nc i, u th hin thi c thay bi u th kia.



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Thi chi vn c Lnh l mt ch 'Q' ct u tin ca dng, dng lnh ny kt thc Input cho vn cang xt. Chng trnh phi in th c cui cng ca vn c theo qui cch c dng input.

Bn phi kim tra tnh chnh xc ca cc lnh. Khng c dng trng bt c nino trong output.

Bi 42/2000 - Mt cht v t duy s hc(Dnh cho hc sinh Tiu hc)Tm s t nhin nh nht khi chia cho 2, 3, 4, 5, 6, 7, 8, 9, 10 cho phn d tng ng l 1,2, 3, 4, 5, 6, 7, 8, 9.

Bi 43/2000 - Kim gi v pht gp nhau bao nhiu ln trong ngy

(Dnh cho hc sinh Tiu hc)ng h qu lc c 2 kim: gi v pht. Tnh xem trong vng 1 ngy m (t 0h - 24h) c

bao nhiu ln 2 kim gp nhau v l nhng lc no.

Bi 44/2000 - To ma trn s(Dnh cho hc sinh THCS)Cho trc s nguyn dng N bt k. Hy vit thut ton v chng trnh to lp bng

NxN phn t nguyn dng theo quy lut c cho trong v d sau:1 2 3 4 5 6

2 4 6 8 10 123 6 9 12 2 44 8 12 2 4 65 10 2 4 6 86 12 4 6 8 10Thc hin chng trnh trn my vi N=12, a ra mn hnh ma trn kt qu (c dngnh trong v d).

Bi 45/2000 - Cc vng trn Olimpic

(Dnh cho hc sinh THPT)C 5 vng trn Olimpic chia mt phng thnh 15 phn (khng k phn v hn) (hnh v).Hy t vo mi phn mt s sao cho tng s cc s trong mi vng trn bng 39.Lp chng trnh gii quyt bi ton trn v cho bit c bao nhiu cch xp nh vy.

Bi 46/2000 - o ch ci(Dnh cho hc sinh THCS v THPT)Bn phi vit chng trnh a ra tt c cc t c th c pht sinh t mt tp cc ch ci.V d: Cho t abc, chng trnh ca bn phi a ra c cc t "abc", "acb", "bac",

"bca", "cab" v "cba" (bng cch kho st tt c cc trng hp khc nhau ca t hp bach ci cho).



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InputD liu vo c cho trong tp input.txt cha mt s t. Dng u tin l mt s t nhincho bit s t c cho di. Mi dng tip theo cha mt t. Trong , mt t c thcha c ch ci thng hoc hoa t A n Z. Cc ch thng v hoa c coi nh l

khc nhau. Mt ch ci no c th xut hin nhiu hn mt ln.OutputVi mi t cho trong file Input.txt, kt qu nhn c ra file Output.txt phi cha ttc cc t khc nhau c sinh t cc ch ci ca t . Cc t c sinh ra t mt t cho phi c a ra theo th t tng dn ca bng ch ci.Sample Input2abcacba

Sample Outputabcacb

bacbcacabcbaaabcaacb

abacabcaacabacba

baacbacabcaacaabcaba

cbaa

Bi 47/2000 - Xo s trn vng trn(Dnh cho hc sinh THCS v PTTH)Cc s t 1 n 2000 c xp theo th t tng dn trn mt ng trn theo chiu kimng h. Bt u t s 1, chuyn ng theo chiu kim ng h, c bc qua mt s lixo i mt s. Cng vic tip din cho n khi trn vng trn cn li ng mt s. Lpchng trnh tnh v in ra s .



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Bi 48/2000 - Nhng chic gy(Dnh cho hc sinh THCS v THPT)George c nhng chic gy vi chiu di nh nhau v cht chng thnh nhng on c

chiu di ngu nhin cho n khi tt c cc phn tr thnh u c chiu di ti a l 50n v. By gi anh ta mun ghp cc on li nh ban u nhng li qun mt n nhth no v chiu di ban u ca chng l bao nhiu. Hy gip George thit k chngtrnh c tnh nh nht c th ca chiu di nhng ci gy ny. Tt c chiu di c

biu din bng n v l nhng s nguyn ln hn 0.InputD liu vo trong file Input.txt cha cc khi mi khi 2 dng. Dng u tin cha s

phn ca chic gy sau khi ct. Dng th 2 l chiu di ca cc phn ny cch nhau bimt du cch. Dng cui cng kt thc file Input l s 0.

OutputKt qu ra trong file Output.txt cha chiu di nh nht c th ca nhng ci gy, michic trong mi khi trn mt dng.Sample Input95 2 1 5 2 1 5 2 141 2 3 40

Sample Output65

Bi 49/2001 - Mt cht nhanh tr(Dnh cho hc sinh Tiu hc)S t nhin A c tnh cht l khi chia A v lp phng ca A cho mt s l bt k thnhn c s d nh nhau. Tm tt c cc s t nhin nh vy.Bi 50/2001 - Bi ton i mu bi

(Dnh cho hc sinh THCS v THPT)Trn bn c N1 hn bi xanh, N2 hn bi v N3 hn bi vng. Lut chi nh sau:

Nu 2 hn bi khc mu nhau chm nhau th chng s cng bin thnh mu th 3 (v d:xanh, vng --> , ).Tm thut ton v lp chng trnh cho bit rng c th bin tt c cc hn bi thnhmt mu c c khng?

Bi 51/2001 - Thay th t(Dnh cho hc sinh THCS v PTTH)

Hai file INPUT1.TXT v INPUT2.TXT c cho nh sau: File INPUT1.TXT cha mton vn bn bt k. File INPUT2.TXT cha khng qu 50 dng, mi dng gm hai t:



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t u l t ch v t sau l t ngun. Hy tm trong file INPUT1.TXT tt c cc t l tch v thay th chng bng cc t ngun tng ng. Kt qu ghi vo file KQ.OUT (s lmt on vn bn tng t nh trong file INPUT1.TXT nhng c thay th t ch

bi t ngun).

Sample INPUT File INPUT1.TXT cha on vn bn sau:

Nam moi sap den roi, ban co zui khong?Chuc cac ban don mot cai Tet that vui ve va hanh phuc.

Chuc ban luon hoc gioi!

File INPUT2.TXT cha cc dng sau:

ban emzui vui

Sample OUTPUT

File KQ.OUT s cha on vn bn sau:Nam moi sap den roi, em co vui khong?Chuc cac em don mot cai Tet that vui ve va hanh phuc.

Chuc em luon hoc gioi!

Bi 52/2001 - Xc nh cc t gic ng h trong ma trn(Dnh cho hc sinh THCS v THPT)Cho ma trn vung A[i,j] (i,j = 1, 2 ... n). Cc phn t ca A c nh s t 1 n n n.Gi S l s lng cc "t gic" c bn nh l: A[i,j]; A[i,j+1]; A[i+1,j]; A[i+1,j+1] sao

cho cc s nh ca n xp theo th t tng dn theo chiu kim ng h (tnh t mtnh no ).1) Lp chng trnh tnh s lng S.2) Lp thut ton xc nh A sao cho s S l:

a. Ln nht.b. Nh nht.

Bi 53/2001 - Lp lch thng k o(Dnh cho hc sinh THCS v THPT)Lch ca cc thng c biu din bng mt ma trn c s ct bng 7 v s hng nh hn

hoc bng 6.1 2 3 4 5

6 7 8 9 10 11 1213 14 15 16 17 18 1920 21 22 23 24 25 26 27 28 29 30

V d: Trong hnh v, lch ny tha mn tnh cht sau: Mi ma trn con 3 3 khng c trng u l ma trn "k o" theo ngha: Tng cc s ca mi ng cho bng tng catrung bnh cng ca tt c cc ct v hng. Hy xy dng tt c cc lch thng c tnh cht

nh trn. Lp chng trnh m t tt c cc kh nng xy ra.



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Bi 54/2001 - Bn hy gch s(Dnh cho hc sinh Tiu hc v THCS)Chng ta vit lin tip 10 s nguyn t u tin theo th t tng to thnh mt s c

nhiu ch s. Trong s ny hy gch i mt na s ch s s cn li l:a. Nh nht

b. Ln nhtTrong tng trng hp phi nu c th thut gii (ti sao li gch nh vy)?

Bi 55/2001 - Bi ton che mt mo(Dnh cho hc sinh THCS v THPT)Trn bn c vung NxN ti mi c th xp hoc mt con mo con, hoc mt qun c.Hai con mo trn bn c s nhn thy nhau nu trn ng thng ni chng theo hng

ngang, hng dc hay ng cho khng c qun c no c.Hy tm cch xp mo v qun c nh trn sao cho s mo ln nht m khng c hai conmo no nhn thy nhau?

Bi 56/2001 - Chia li(Dnh cho hc sinh THPT)Cho li MN (m, n


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0 12 2 5 0 00 9 2 10 0 00 0 0 0 0 0

Sample Output:0 1 1 1 1 10 1 0 1 1 10 0 0 1 1 10 0 0 1 1 10 0 0 0 0 1

Bi 57/2001 - Chn s(Dnh cho hc sinh Tiu hc v THCS)

Cho 2000 s a1, a2,..., a2000 mi s l +1 hoc -1. Hi c th hay khng t 2000 s chnra cc s no tng cc s c chn ra bng tng cc s cn li? Gi s cho 2001s, liu c th c cch chn khng? Nu cch gii tng qut.

Bi 58/2001 - Tng cc s t nhin lin tip(Dnh cho hc sinh THCS v THPT)Cho trc s t nhin n. Lp thut ton cho bit n c th biu din thnh tng ca haihoc nhiu s t nhin lin tip hay khng?Trong trng hp c, hy th hin tt c cc cch c th c.

Bi 59/2001 - m s vung(Dnh cho hc sinh THCS v THPT)Cho mt bng vung gm NxN im nm trn cc mt li vung. Cc im k nhautrn mt hng hay mt ct c th c ni vi nhau bng mt on thng hoc khngc ni. Cc on s to ra cc vung trn bng. V d vi bng sau y th n = 4v c 3 vung:

Trn mi hng c th c nhiu nht n-1 on thng nm ngang v c tt c n hng nhvy. Tng t nh vy c tt c n-1 hng cc on thng nm dc v trn mi hng cth c nhiu nht n on. m t ngi ta dng hai mng nh phn: mt mng ghi cc on nm ngang kchthc n x (n-1), v mt mng ghi cc on nm dc kch thc (n-1) xn. Trong mng, s



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1 dng m t on thng ni gia 2 im, cn s 0 miu t gia hai im khng con thng ni. Trong v d trn th ma trn "ngang" l:

1 0 1

1 0 0

1 1 11 1 0

v ma trn "dc" l:

1 1 1 0

1 1 0 1

0 1 1 0

Cho trc ma trn "ngang" v ma trn "dc", d liu nhp t cc tp vn bn c tn lNGANG.INP v DOC.INP. Hy lp trnh m s cc vung trn bng.

Bi 60/2001 - Tm s d ca php chia(Dnh cho hc sinh Tiu hc)Mt s nguyn khi chia cho 1976 v 1977 u d 76. Hi s khi chia cho 39 d baonhiu?

Bi 61/2001 - Thut ton in s vo ma trn(Dnh cho hc sinh THCS v THPT)

lp thut ton in cc phn t ca ma trn NN cc s 0, 1 v -1 sao cho:

a) Tng cc s ca mi hnh vung con 2x2 u bng 0.b) Tng cc s ca ma trn trn l ln nht.

Bi 62/2001 - Chn Xu(Dnh cho hc sinh THCS v THPT)Cho mt xu S = 123456789 hy tm cch chn vo S cc du '+' hoc '-' thu c sM cho trc (nu c th). S M nguyn c nhp t bn phm. Trong file OutputChenxau.Out ghi tt c cc phng n chn (nu c) v ghi "Khong co" nu nh khngth thu c M t cch lm trn.

V d: Nhp M = 8, mt trong cc phng n l: '-1+2-3+4+5-6+7';M = -28, mt trong cc phng n l: '-1+2-34+5';

( ra ca bn:L Nhn Tm - 12 Tin Trng THPT Lam Sn)

Bi 63/2001 - Tm s nh nht(Dnh cho hc sinh Tiu hc)Hy vit ra s nh nht bao gm tt c cc ch s 0, 1, 2, 3, ... 9 m n:

a. Chia ht cho 9b. Chia ht cho 5

c. Chia ht cho 20



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C gii thch cho tng trng hp?

Bi 64/2001 - i ma trn s(Dnh cho hc sinh THCS v THPT)

Cho mng s thc vung A kch thc 2nx2n. Hy lp cc mng mi bng cch i chcc khi vung kch thc nxn ca A theo cc cch sau:

a. b.Bi 65/2001 - Li vung v hn

(Dnh cho hc sinh THCS v THPT)Cho li vung v hn v hai pha (trn v phi). Cc ca li c nh s theo quytc sau:

- tri di - v tr (0,0) - c nh s 0.- Cc cn li c nh s theo nguyn tc lan to t v tr (0,0) v theo quy tc:

ti mt v tr s c in vo l s nguyn khng m nh nht cha c in trn hngv ct cha hin thi. V d, ta c hnh dng ca mt s ca li nh sau:

3 2 1 02 3 0 11 0 3 2

0 1 2 3

Cho trc cp s t nhin M, N - kch thc li. Hy vit chng trnh m t litrn, kt qu c ghi vo file KQ.TXT.

Bi 66/2001 - Bng s 9 x 9

(Dnh cho hc sinh Tiu hcv THCS)Hy xp cc s 1, 2, 3, ..., 81 vo bng 9 x 9 sao cho:a) Trn mi hng cc s c xp theo th t tng dn (t tri qua phi).

b) Tng cc s ct 5 l ln nht.



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Yu cu:+ i vi cc bn hc sinh khi Tiu hc ch cn vit ra bng s tho mn tnh cht trn.+ Cc bn hc sinh khi THCS th phi lp trnh hin th kt qu ra mn hnh.

Bi 67/2001 - V cc php bin i "Nhn 2 tr 1"(Dnh cho hc sinh THCS v THPT)Cho ma trn A kch thc M x N, Aij - l cc s t nhin. Cc php bin i c th l:- Nhn tt c cc s ca mt hng vi 2.- Tr tt c cc s ca mt ct cho 1.Tm thut ton sao cho sau mt s php bin i trn ma trn A tr thnh ton s 0.

Bi 68/2001 - Hnh trn v bng vung(Dnh cho hc sinh THPT)

Mt ng trn ng knh 2n -1 n v c v gia bn c 2n 2n. Vi n = 3 cminh ho nh di y:

Vit chng trnh xc nh s vung ca bng b ct bi hnh trn v s vung nmhon ton trong hnh trn.D liu vo trong file Input.txt bao gm: Mi dng l mt s nguyn dng khng lnhn 150 - l cc gi tr ca n.D liu ra trong file Output.txt: Vi mi gi tr vo n, kt qu ra phi tnh c s vung b ct bi hnh trn v s vung nm hon ton trong hnh trn, mi s trn mt

dng. Mi kt qu tng ng vi mt gi tr n phi cch nhau mt dng.

Sample Input34

Sample Output2012

28



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24

Bi 69/2001 - Bi ca 36(Dnh cho hc sinh Tiu hc)

Tm s t nhin nh nht chia ht cho 36 m trong dng vit thp phn ca n c cha ttc cc ch s t 1 ti 9.

Bi 70/2001 - M ho theo kho(Dnh cho hc sinh THCS v THPT)Cho trc kho l mt hon v ca n s (1, 2, ..., n). Khi m ho mt xu k t ta cth chia xu thnhtng nhm n k t (ring nu nhm cui cng khng n k t th tacoa th thm cc du cch vo sau cho ) ri hon v cc k t trong tng nhm. Sau ,ghp li theo th t cc nhm ta c mt xu m ho.

Chng hn: vi kho 3241 (n=4) th ta c th m ho xu 'english' thnh 'gnlehs i'.Hy vit chng trnh m ho mt xu k t cho trc.

Bi 71/2001 - Thc hin php nhn(Dnh cho hc sinh THCS v THPT)Bn hy lp chng trnh nhp 2 s nguyn dng a v b. Sau thc hin php nhn (ax b) nh cch nhn bng tay thng thng. V d:

Bi 72/2001 - Bin i trn li s(Dnh cho hc sinh THCS v THPT)Trn mt li N x N cc c nh s 1 hoc -1. Li trn c bin i theo quy tc

sau: mt no c thay th bng tch ca cc s trong cc k n (k cnh). Lpchng trnh thc hin sao cho sau mt s bc ton li cn li ch s 1.Bi 73/2001 - Bi ton chui s(Dnh cho hc sinh Tiu hcv THCS)Cho mt chui s c quy lut. Bn c th tm c hai s cui ca dy khng, thay thchng trong du hi chm (?). Bi ton khng d dng lm u, v chng c to ra bimt quy lut rt phc tp. Bn th sc xem?

5 8 11 14 17 23 27 32 35 41 49 52 ? ?



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Bi 74/2001 - Hai hng s k o(Dnh cho hc sinh THCS v THPT)Hy xp 2N s t nhin 1, 2, ..., 2N thnh 2 hng s:

A1, A2 ... An

B1, B2 ... BnTha mn iu kin: tng cc s theo n ct bng nhau, tng cc s theo cc hng bngnhau.

Bi 75/2001 - Tr chi Tch - Tc vung(Dnh cho hc sinh THCS v THPT)Trn mt li k vung c 2 ngi chi nh sau: ngi th nht mi ln chi s nhdu x vo 1 trng. Ngi th hai c nh du 0 vo 1 trng. Ngi th nht mun

t c mc ch l nh c 4 du x to thnh 4 nh ca 1 hnh vung. Ngi thhai c nhim v ngn cn mc ch ca ngi th nht.Lp chng trnh tm thut ton ti u cho ngi th nht (ngi th nht c th lunthng).Ch : Li vung c coi l v hn v c hai pha.

Bi 76/2001 -on thng v hnh ch nht(Dnh cho hc sinh THPT)Hy vit mt chng trnh xc nh xem mt on thng c ct hnh ch nht hay khng?V d:Cho ta im bt u v im kt thc ca ng thng: (4,9) v (11,2);V ta nh tri trn v nh phi di ca hnh ch nht: (1,5) v (7,1);

Hnh1:on thng khng ct hnh ch nht

on thng c gi l ct hnh ch nht nu on thng v hnh ch nht c t nht mtim chung.Ch : mc d tt c d liu vo u l s nguyn, nhng ta ca cc giao im tnh ra

cha chc l s nguyn.



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InputD liu vo trong file Input.Inp kim tra N trng hp (N


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Bi 79/2001 -V mt ma trn s(Dnh cho hc sinh THCS)M t thut ton, lp chng trnh xy dng ma trn A[10,10] tho mn cc tnh cht:+ A[i,j] l cc s nguyn t 0..9 (1


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n s u tin l kt qu ca cu a,Tip l mt dng trng v sau l n s kt qu ca cu b (nu tm c dy a).

Bi 82/2001 -Gp g

(Dnh cho hc sinh THPT)Trn mt li vung kch thc MN (M dng, N ct) ngi ta t k rbt. Rbt thi c t (xi,,yi). Mi ca li c th t mt vt cn hay khng. Ti mi bc,mi rbt ch c th di chuyn theo cc hng ln, xung, tri, phi - vo cc k cnhkhng c vt cn. k rbt s gp nhau nu chng cng ng trong mt . k rbt bt udi chuyn ng thi v mi lt c k rbt u phi thc hin vic di chuyn (ngha lkhng cho php mt rbt dng li mt no trong khi rbt khc thc hin bc dichuyn). Bi ton t ra l tm s bc di chuyn t nht m k rbt phi thc hin cth gp nhau. Ch rng, ty trng thi ca li, k rbt c th khng khi no gp c

nhau.D liu vo cho trong file vn bn MEET.INP, bao gm:+ Dng u tin cha 3 s M,N v k (M,N


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3 s (mi phn ca hnh trn ngoi cha c phn ca hai hnh trn trong) c tng bng30.Cc s bn c s dng l:14, 11, 10, 12, 7, 9, 9, 8, 9, 9, 11, 11, 10, 10, 10, 10, 14, 9, 7, 11, 10, 8, 12, 9.

Bi 84/2001 - Cng mt tch(Dnh cho hc sinh THCS v THPT)Cho n s x1, x2, ..., xn ch nhn mt trong cc gi tr -1, 0, 1. V cho mt s nguyn P.

Hy tnh s lng tt c cc cch gn gi tr khc nhau ca n s trn sao cho: i jx x P= (vi i =1..n, j =1..n, i j). Hai cch gn c gi l khc nhau nu s lng cc s xi = 0l khc nhau.

Input: gm 2 s n, P.Output: s cc cch chn khc nhau.Gii hn: 2


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- Dng u s nguyn k l s lng php bin i t nht cn p dng (k = 0 nu khngbin i c)- Dng th i trong s k dng tip theo ghi hai s nguyn xc nh cn chn thc hin

php bin i.

V d:BIENDOI. INP4 51 0 0 0 01 0 0 0 00 1 0 0 00 1 0 0 0

0 0 0 0 0

0 0 0 0 00 0 1 0 00 0 0 0 0BIENDOI.OUT22 13 2( ra ca bn Nguyn Vn c - Cn Th)

Bi 86/2001 - Dy s t nhin logic(Dnh cho hc sinh Tiu hc)y l mt chui cc s t nhin c sp xp theo mt logic no . Hy tm con s utin v cui cng ca dy s thay th cho du ?

? 12 14 15 16 18 20 21 22 ?

Bi 87/2001 - Ghi s trn bng(Dnh cho hc sinh THCS)Trn bng ghi s 0. Mi ln c tng s vit ln bng thm 1 n v hoc tng gp

i. Hi sau t nht l bao nhiu bc s thu c s nguyn dng N?

Bi 88/2001 - V cc s c bit c 10 ch s(Dnh cho hc sinh THCS v THPT)



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Lp chng trnh tnh (v ch ra) tt c cc s c 10 ch s a0a1a2...a9 tho mn cc tnhcht sau:

a0 bng s ch s 0 ca s trn;a1 bng s ch s 1 ca s trn;

a2 bng s ch s 2 ca s trn;.a9 bng s ch s 9 ca s trn;

Bi 89/2001 - Ch s th N(Dnh cho hc sinh THCS v THPT)Khi vit cc s t nhin tng dn t 1, 2, 3, lin tip nhau, ta nhn c mt dy ccch s thp phn v hn, v d: 1234567891011121314151617181920...Yu cu: Hy tm ch s th N ca dy s v hn trn.

D liu vo t file Number.inp gm mt s dng, mi dng ghi mt s nguyn dngN (N


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(Dnh cho hc sinh THPT)Xt mt dy gm N s nguyn tu . Gia cc s nguyn ta c th t cc du + hoc- thu c cc biu thc s hc khc nhau. Ta ni dy s l chia ht cho K nu mttrong cc biu thc thu c chia ht cho K. Hy vit chng trnh xc nh tnh chia ht

ca mt dy s cho.D liu vo: Ly t mt file vn bn c tn l DIV.INP c cu trc nh sau:

- Dng u l hai s N v K (2 N 10 000, 2 K 100), cch nhau bi du trng.- Cc dng tip theo l dy N s c tr tuyt i khng qu 10 000 cch nhau bi du

trng hoc du xung dng.D liu ra: Ghi ra file vn bn DIV.OUT s 1 nu dy cho chia ht cho K v s 0 nungc li.V d:DIV.INP DIV.OUT DIV.INP DIV.OUT

4 6 0 4 7 11 2 3 5 1 2 3 5

( ra ca bn Trn nh Trung - Lp 11A Tin - Khi PTCT - H Vinh)

Bi 93/2002 - Tr chi bn bi(Dnh cho hc sinh Tiu hc)Cho bng bn bi sau:

Bn c th bn bi vo t mt trong s cc nh ngoi cng. Khi c bn vo trong,

hn bi ch c th tip tc i vo trong nh gn nht hoc ln theo nhiu nht l mtcnh i vo nh k . Bit rng khi n hnh ch nht trong cng, hn bi khng -c ln trn mt cnh no m phi i thng vo tm.Hy tm ng i sao cho tng s im m n i qua l ln nht v c bao nhiu ngi c c s im .

Bi 94/2002 - Biu din tng cc s Fibonaci(Dnh cho hc sinh THCS)Cho s t nhin N v dy s Fibonaci: 1, 1, 2, 3, 5, 8, ....

Bn hy vit chng trnh kim tra xem N c th biu din thnh tng ca ca cc sFibonaci khc nhau hay khng?



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Bi 95/2002 - Dy con c tng ln nht(Dnh cho hc sinh THPT)Cho dy gm n s nguyn a1, a2, ..., an. Tm dy con gm mt hoc mt s phn t lin

tip ca dy cho vi tng cc phn t trong dy l ln nht.D liu: Vo t file vn bn SUBSEQ.INP- Dng u tin cha s nguyn dng n (n < 106).- Dng th i trong s n dng tip theo cha s ai (|ai| 1000).

Kt qu: Ghi ra file vn bn SUBSEQ.OUT- Dng u tin ghi v tr ca phn t u tin ca dy con tm c.- Dng th hai ghi v tr ca phn t cui cng ca dy con tm c- Dng th ba ghi tng cc phn t ca dy con tm c.V d:

SUBSEQ.INP SUBSEQ.OUT8 12 -14 1 23 -622 -34 13

3 6 40

Bi 96/2002 - S chung ln nht(Dnh cho hc sinh THPT)Cho 2 xu:

X = x1x2..xM. (Vi xi l cc k t s t 0 n 9)Y = y1y2..yN.( Vi yi l cc k t s t 0 n 9)

(M, N


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Bng di gm 9 , ban u c in bi cc ch ci. Bn hy thay cc ch ci bi ccch s t 0 n 8 vo sao cho tt c cc s theo hng ngang, hng dc u l s c 3ch s (ch s hng trm phi khc 0) v tho mn:

1 2 3

a b cd e fg h i

Ngang4 - Bi s nguyn ca 8;5 - Tch ca cc s t nhin lin tip u tin;6 - Tch cc s nguyn t k nhau

Dc1 - Bi nguyn ca 11;2 - Tch ca nhiu tha s 2;3 - Bi s nguyn ca 11.

( ra ca bn o Tun Anh - Lp 10A Trng THPT Nng Khiu Ng S Lin - th xBc Giang)

Bi 98/2002 - S phn nguyn t(Dnh cho hc sinh THCS v THPT)

Mt s n gi l s phn nguyn t nu s c s ca n l nhiu nht trong n s t nhinu tin. Cho s K (K


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bit thi gian c th tip trong ngy l t Ai n Bi (i = 1, 2, ..., N). Gi thit rng,khong thi gian cn thit cho mi cuc gp l H v khong thi gian chun b t mtcuc gp n mt cuc gp k tip l T. Bn hy xy dng gip mt lch chc Tt anhta c th chc Tt c nhiu ngi nht.

File d liu vo trong file CHUCTET.INP gm dng u ghi s N, dng th i trong s Ndng tip theo ghi khong thi gian c th tip khch ca ngi i gm 2 s thc Ai v Bi(cch nhau t nht mt du trng). Dng tip theo ghi gi tr H (s thc) v dng cuicng ghi gi tr T (s thc). Gi thit rng cc gi tr thi gian u c vit di dngthp phn theo n v gi, tnh n 1 s l (th d 10.5 c ngha l mi gi ri) v unm trong khong t 8 n 21 (t 8 gi sng n 9 gi ti). S khch ti a khng qu30.Kt qu ghi ra file CHUCTET.OUT gm dng u ghi K l s ngi c thm, K dngtip theo ghi trnh t i thm, mi dng gm 2 s (ghi cch nhau t nht mt du trng):

s u l s hiu ngi c thm, s tip theo l thi im gp tng ng.Th d:CHUCTET.INP20

10.5 12.6

15.5 16.6

14.0 14.1

17.5 21.0

15.0 16.1

10.5 10.6

19.0 21.0

10.5 13.6

12.5 12.6

11.5 13.6

12.5 15.6

16.0 18.1

13.5 14.6

12.5 17.6

13.0 13.1

18.5 21.0

9.0 13.1

10.5 11.6

10.5 12.6

18.0 21.0

0.5

0.1

CHUCTET.OUT16

17 9.0

1 10.5

18 11.1

19 11.78 12.3



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10 12.911 13.5

13 14.15 15.0

2 15.6

12 16.214 16.84 17.5

7 19.016 19.6

20 20.2

( ra ca bn inh Quang Huy - HKHTN - HQG H Ni )

Bi 100/2002 - Mi khch d tic(Dnh cho hc sinh THPT)

Cng ty trch nhim hu hn Vui v c n cn b nh s t 1 n n. Cn b i c nhgi vui tnh l vi (i = 1, 2, ..., n). Ngoi tr Gim c Cng ty, mi cn b c 1 thtrng trc tip ca mnh.Bn ch cn gip Cng ty mi mt nhm cn b n d d tic Vui v sao cho trong snhng ngi c mi khng ng thi c mt nhn vin v th trng trc tip v ngthi tng nh gi vui tnh ca nhng ngi d tic l ln nht.Gi thit rng mi mt th trng c khng qu 20 cn b trc tip di quyn.

D liu: Vo t file vn bn GUEST.INP

- Dng u tin ghi s cn b ca Cng ty: n (1 < n < 1001);- Dng th i trong s n dng tip theo ghi hai s nguyn dng ti, vi; trong ti l shiu ca th trng trc tip v vi l vui tnh ca cn b i (i = 1, 2, ..., n). Quy c ti =0 nu i l s hiu ca Gim c Cng ty.

Kt qu: Ghi ra file vn bn GUEST.OUT- Dng u tin ghi hai s m, v; trong m l tng s cn b c mi cn v l tng vui tnh ca cc cn b c mi d tic;- Dng th i trong s m dng tip theo ghi s hiu ca cn b c mi th i (i = 1, 2, ...,

m).V d:

GUEST.INP GUEST.OUT30 31 62 4

2 713

GUEST.INP GUEST.OUT7 3 63



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0 11 11 122 50

2 13 13 1

345

( ra ca bn Lu Vn Minh)



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Phn II: LI GII

Bi 1/1999 -Tr chi cng nhau qua cu(Dnh cho hc sinh Tiu hc)

p s: 17 pht. Cch i nh sau:Lt 1: 2 + 1 sang, 1 quay v thi gian: 3 phtLt 2: 10 + 5 sang, 2 quay v thi gian: 12 phtLt 3: 2 + 1 sang thi gian: 2 pht

Tng thi gian: 17 pht

Bi 2/1999 - T chc tham quan(Dnh cho hc sinh THCS)Program bai2;uses crt;

const fi = 'P2.inp';fo = 'P2.out';

type _type=array[1..2] of integer;mang=array[1..200] of _type;

var f:text;d,v:mang;m,n:byte;

procedure input;var i:byte;

beginassign(f,fi);reset(f);readln(f,n,m);for i:=1 to n dobeginread(f,d[i,1]);d[i,2]:=i;end;readln(f);

for i:=1 to m dobeginread(f,v[i,1]);v[i,2]:=i;end;close(f);end;

procedure sapxeptang(var m:mang;n:byte);var d:_type;i,j:byte;

beginfor i:=1 to n-1 do



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for j:=i+1 to n doif m[j,1]m[i,1] thenbegind:=m[j];m[j]:=m[i];

m[i]:=d;end;end;

var i:byte;tong:integer;

begininput;sapxeptang(d,n);sapxeptang(v,m);tong:=0;

for i:=1 to n do tong:=tong+v[n-i+1,1]*d[i,1];for i:=1 to n do v[i,1]:=d[n-i+1,2];xapxeptang(v,n);assign(f,fo);rewrite(f);writeln(f,tong);for i:=1 to n do writeln(f,v[i,2]);close(f);end.

Nhn xt: Chng trnh trn s chy chm nu chng ta m rng bi ton (chng hn n


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Pos[2, i] := i;end;

end;procedure QuickSort(t, l, r: Integer);var

x, tg, i, j: Integer;begin

x := Val[t, (l + r) div 2];i := l; j := r;repeatwhile Val[t, i] < x do Inc(i);while Val[t, j] > x do Dec(j);if i j;if i < r then QuickSort(t, i, r);if j > l then QuickSort(t, l, j);

end;procedure WriteOutput;vari: Integer;Sum: LongInt;hf: Text;

beginSum := 0;for i := 1 to n do Inc(Sum, Val[1, n - i + 1] * Val[2, i]);for i := 1 to n do Val[1, Pos[1, n - i + 1]] := Pos[2, i];Assign(hf, Out);Rewrite(hf);Writeln(hf, Sum);for i := 1 to n do Writeln(hf, Val[1, i]);Close(hf);

end;

beginReadInput;QuickSort(1, 1, n);QuickSort(2, 1, m);WriteOutput;

end.

Bi 3/1999 -Mng t bo(Dnh cho hc sinh THPT)Program Bai3/1999;uses crt;const fi = 'P3.inp';



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fo = 'P3.out';

type mang=array[0..201,0..201] of byte;

var m,n,t:byte;

s:string;a:mang;f:text;b,c:^mang;

procedure input;var i,j:byte;

beginassign(f,fi);reset(f);readln(f,m,n,t);

readln(f,s);for i:=1 to m dobeginfor j:=1 to n do read(f,a[i,j]);end;close(f);new(b);new(c);end;

procedure hien;var i,j:byte;

beginfor i:=1 to m dofor j:=1 to n dobegingotoxy(j*2,i);write(b^[i,j]);end;

end;

procedure trans(ch:char);var i,j,d:byte;

beginfillchar(c^,sizeof(mang),0);for i:=1 to m dofor j:=1 to n dobegind:=b^[i,j];case a[i,j] of1:inc(c^[i,j-1],d);

2:inc(c^[i,j+1],d);3:inc(c^[i-1,j],d);



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4:inc(c^[i+1,j],d);5:begin inc(c^[i-1,j],d);inc(c^[i+1,j],d); end;6:begin inc(c^[i,j-1],d);inc(c^[i,j+1],d); end;7:begin inc(c^[i,j-1],d);inc(c^[i-1,j],d); end;8:begin inc(c^[i,j+1],d);inc(c^[i+1,j],d); end;

end;end;if ch'X' then b^[1,1]:=ord(ch)-48;for i:=1 to m dofor j:=1 to n doif (i1) or (j1) then b^[i,j]:=byte(c^[i,j]0);hien;readln;end;

procedure output;

var i,j:byte;beginassign(f,fo);rewrite(f);for i:=1 to m dobeginfor j:=1 to n do write(f,' ',b^[i,j]);writeln(f);end;close(f);end;

var i:byte;beginclrscr;input;fillchar(b^,sizeof(mang),0);fillchar(c^,sizeof(mang),0);for i:=1 to t do trans(s[i]);output;end.

Bi 4/1999 -Tr chi bc si(Dnh cho hc sinh Tiu hc)Huy s l ngi thng cuc. Tht vy s si ban u l 101 l mt s c dng 5k+1,ngha l s nu chia 5 s cn d 1. Hong phi bc trc, do s si ca Hong phi ly lt 1 n 4 do sau lt i u tin, s si cn li s ln hn 96. Huy s bc tip theosao cho s si cn li phi l 96, ngha l s dng 5k+1. Tng t nh vy, Huy lunlun ch ng c sau ln bc ca mnh s si cn li l 5k+1. Ln cui cng s sicn li ch l 1 v Hong bt buc phi bc vin cui cng v ... thua.

Bi ton tng qut: c th cho s vin bi l 5k+1 vin.



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Bi 5/1999 - 12 vin bi(Dnh cho hc sinh THCS)Ta s ch ra rng tn ti 3 ln cn ch ra c vin bi c bit .Gi cc vin bi ny ln lt l 1, 2, ..., 12. Trong khi m t thut ton ta dng k hiu

m t qu hn bi th n

m t mt hn bi bt k

M t mt php cn.Ta gi vin bi c trng lng khc l .

I. Ln cn th nht. Ly ra 8 hn bi bt k v chia lm 2 phn cn:

C 2 trng hp xy ra:1.1. Cn trn cn bng. Suy ra vin bi (khng r nng nh) nm trong 4 vin bi cn li(khng mang ra cn)1.2. Cn trn khng cn bng.1.2.1. Nu (1) nh hn (2) suy ra hoc l nh nm trong (1) hoc l nng nm trong(2).1.2.2. Nu (1) nng hn (2) suy ra hoc l nng nm trong (1) hoc l nh nmtrong (2).D thy cc trng hp 1.2.1. v 1.2.2. l tng t nhau.

Trong mi trng hp ta c kt lun nm trong s 8 vin hoc nh trong 4 hoc nngtrong 4 cn li.II. Xt trng hp 1.1: Tm c 4 vin bi cha Gi cc hn bi ny l 1, 2, 3, 4Ln cn th hai:

Xt cc trng hp sau:2.1. Cn thng bng. Kt lun: vin bi 4 chnh l .2.2. Trng hp cn tri nh hn phi (du ). Suy ra hoc 3 l nh, hoc 1 hoc 2 l

nng.D thy rng cc trng hp 2.2. v 2.3. l tng t nhau.



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III. Xt trng hp 2.1: vin bi 4 chnh l Ln cn th ba:

Nu cn nghing < th 4 l nh, nu cn nghing > th 4 l nng.IV. Xt trng hp 2.2. Hoc 3 l nng, hoc 1 hoc 2 l nh.Ln cn th ba:

Nu cn thng bng th ta c 1 l hn bi nh.Nu cn nghing > th ta c 3 l hn bi nng.Nu cn nghing < th ta c 2 l hn bi nh.V. Xt trng hp 2.3. Hoc 3 l nh, hoc 1 hoc 2 l nng.

Cch lm tng t trng hp 2.2 m t trong mc IV trn.VI. Xt trng hp 1.2.1.Hoc l nh trong 1, 2, 3, 4 hoc l nng trong 5, 6, 7, 8.Ln cn th hai:

6.1. Trng hp cn thng bng. Suy ra s phi nm trong 4, 7, 8, v do theo githit ca trng hp ny ta c hoc l 4 nh, hoc l nng trong 7, 8. D nhn thy

trng hp ny hon ton tng t nh 2.2. Bc tip theo lm tng t nh m t trongIV.6.2. Trng hp cn nghing , suy ra hoc l 5 nng, hoc l nh 3.VII. Xt trng hp 6.3.Hoc l 5 nng, hoc l 3 nh.Ln cn th ba:

Nu cn thng bng, suy ra 5 l nng.Nu cn nghing < suy ra 3 l nh.Tt c cc trng hp ca bi ton c xem xt.



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Sau y l chng trnh chi tit.

Program bai5;Uses crt;

Constst1=' nang hon.';st2=' nhe hon.';Var i, kq1: integer;kq2: string;ch: char;(* Th tc Kq *)Procedure kq(a: integer; b: string);Beginkq1:=a;kq2:=b;

End;(* Th tc Cn *)Procedure can(lan: integer; t1, t2, t3, t4, p1, p2, p3, p4: string);BeginWriteln('Ln cn th', lan, ' :');Writeln;Writeln(' ', t1, ' ', t2, ' ', t3, ' ', t4, ' ', p1, ' ', p2, ' ', p3, ' ', p4);Writeln;Write(' Bn no nng hn? Tri(t)/Phi(p)/ Hay cn bng(c)');Repeatch:=readkey;ch:=upcase(ch);Until (ch in ['P', 'T', 'C']);Writeln(ch);Writeln(*==========================================*);End;(* Th tc Play *)Procedure play;BeginWriteln('C 12 qu cn: 1 2 3 4 5 6 7 8 9 10 11 12');Writeln('Cho php bn chn ra mt qu cn nng hn hay nh hn nhng qu khc.');

can(1, '1', '2', '3', '4', '5', '6', '7', '8');If (ch='T') then {T}Begincan(2, '1', '2', '5', ' ', '3', '4', '6', ' ');If (ch='T') then {TT}Begincan(3, '1', '6', ' ', ' ', '7', '8', ' ', ' ');If ch='T' then kq(1, st1); {TTT}If ch='P' then kq(6, st2); {TTP}If ch='C' then kq(2, st1); {TTC}End

Else If (ch='P') then {TP}Begin



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can(3, '3', '5', ' ', ' ', '7', '8', ' ', ' ');If ch='T' then kq(3, st1); {TPT}If ch='P' then kq(5, st2); {TPP}If ch='C' then kq(4, st1); {TPC}End

Else If (ch='C') then {TC}Begincan(3, '7', ' ', ' ', ' ', ' ', '8', ' ', ' ');If ch='T' then kq(8, st2); {TCT}If ch='P' then kq(7, st2); {TCP}If ch='C' thenBeginWriteln('Tr li sai!'); kq2:=st2;End;End;End

Else If (ch='P') then {P}Begincan(2, '5', '6', '1', ' ', '7', '8', '2', ' ');If (ch='T') then {PT}Begincan(3, '5', '2', ' ', ' ', '3', '4', ' ', ' ');If ch='T' then kq(5, st1);If ch='P' then kq(2, st2);If ch='C' then kq(6, st1);EndElse If (ch='P') then {PP}Begincan(3, '7', '1', ' ', ' ', '3', '4', ' ', ' ');If ch='T' then kq(7, st1);If ch='P' then kq(1, st2);If ch='C' then kq(8, st1);EndElse If (ch='C') then {PC}Begincan(3, '3', ' ', ' ', ' ', ' ', '4', ' ', '');If ch='T' then kq(4, st2);

If ch='P' then kq(3, st2);If ch='C' thenBeginWriteln('Tr li sai !'); kq2:=st2;End;End;EndElse If (ch='C') then {C}Begincan(2, '9', '10', '11', ' ', '1', '2', '3', ' ');If (ch='T') then

{CT}Begin



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can(3, '9', ' ', ' ', ' ', '10', ' ', ' ', ' ');If (ch='T') then kq(9, st1);If (ch='P') then kq(10, st1);If (ch='C') then kq(11, st1);End

Else If (ch='P') then {CP}Begincan(3, '9', ' ', ' ', ' ', '10', ' ', ' ', ' ');If (ch='T') then kq(10, st2);If (ch='P') then kq(9, st2);If (ch='C') then kq(11, st2); EndElse If (ch='C') then {CC}Begincan(3, '12', ' ', ' ', ' ', '1', ' ', ' ', ' ');If (ch='T') then kq(12, st1);If (ch='P') then kq(12, st2);

If (ch='C') then Writeln('Tr li sai!');kq1:=12;End;End;End;(* Chng trnh chnh*)BeginClrscr;

play;Writeln(' Qu th', kq1, kq2);Writeln(' Nhn Enter kt thc...');Readln;End.

Bi 6/1999 - Giao im cc ng thng(Dnh cho hc sinh THPT)Program Bai6;(* Tinh so giao diem cua n duong thang 0 trung nhau *)Uses Crt;Constfn = 'P6.INP';

fg = 'P6.OUT';max = 100;exp = 0.0001;Vara ,b ,c : array[1..max] of real;n : integer;sgd : integer;Procedure Nhap;Varf: text;i: integer;BeginAssign( f ,fn ); Reset( f );



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Readln( f ,n );For i := 1 to n doReadln( f ,a[i] ,b[i] ,c[i] ); { ax + by = c }Close( f );End;

(*--------------------------------------------------------------------------*)Procedure Chuanbi;Beginsgd := 0;End;(*--------------------------------------------------------------------------*)Function Giaodiem( i ,j : integer;Var x ,y : real ) : boolean;Vard ,dx ,dy : real;Begin

d := a[i] * b[j] - a[j] * b[i];dx := c[i] * b[j] - c[j] * b[i];dy := a[i] * c[j] - a[j] * c[i];If d 0 then

beginx := dx / d;y := dy / d;end;giaodiem := d 0;End;(*--------------------------------------------------------------------------*)Function Giatri( i : integer;x ,y : real ) : real;BeginGiatri := a[i] * x + b[i] * y - c[i];End;(*--------------------------------------------------------------------------*)Function bang( a ,b : real ) : boolean;Begin

bang := abs( a - b )


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ii , gt:integer;x, y : real;Begingt := 0;For ii := 1 to i do

If giaodiem( i ,ii ,x ,y ) thenIf thoaman( i ,ii ,x ,y ) then Inc( gt );catrieng := gt;End;(*--------------------------------------------------------------------------*)Procedure Tinhsl;Vari : integer;BeginFor i := 1 to n doInc( sgd ,catrieng( i ) );

End;(*--------------------------------------------------------------------------*)Procedure GhiKQ;BeginWriteln(So giao diem cua cac duong thang la: ' ,sgd );End;(*--------------------------------------------------------------------------*)BEGINClrScr;

Nhap;Chuanbi;Tinhsl;ghiKQ;END.

Bi 7/1999 - Min mt phng chia bi cc ng thng(Dnh cho hc sinh THPT)Program Bai7;(* Tinh so giao diem cua n duong thang ko trung nhau *)Uses Crt;Constfn = 'P7.INP';fg = 'P7.OUT';max = 100;exp = 0.0001;Vara ,b ,c : array[1..max] of real;n : integer;smien : integer;Procedure Nhap;Varf : text;i : integer;



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BeginAssign( f ,fn ); Reset( f );Readln( f ,n );For i := 1 to n doReadln( f ,a[i] ,b[i] ,c[i] ); { ax + by = c }

Close( f );End;(*--------------------------------------------------------------------------*)Procedure Chuanbi;Beginsmien := 1;End;(*--------------------------------------------------------------------------*)Function Giaodiem( i ,j : integer;Var x ,y : real ) : boolean;Vard ,dx ,dy :real;

Begind := a[i] * b[j] - a[j] * b[i];dx:= c[i] * b[j] - c[j] * b[i];dy := a[i] * c[j] - a[j] * c[i];If d 0 then

beginx := dx / d;y := dy / d;end;Giaodiem := d 0;End;(*--------------------------------------------------------------------------*)Function Giatri( i : integer;x ,y : real ) : real;BeginGiatri := a[i] * x + b[i] * y - c[i];End;(*--------------------------------------------------------------------------*)Function bang( a ,b : real ) : boolean;Begin

bang := abs( a - b )


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Varii , gt : integer;x, y : real;Begingt:= 0;

For ii := 1 to i - 1 doIf Giaodiem( i ,ii ,x ,y ) thenIf Thoaman( ii ,x ,y ) then Inc( gt );cattruoc := gt;End;(*--------------------------------------------------------------------------*)Procedure Tinhslmien;Vari : integer;BeginFor i := 1 to n do

Inc( smien ,cattruoc( i ) + 1 );End;(*--------------------------------------------------------------------------*)Procedure GhiKQ;BeginWriteln(So mien mat phang duoc chia la: ' ,smien );End;(*--------------------------------------------------------------------------*)BEGINClrscr;

Nhap;Chuanbi;Tinhslmien;GhiKQ;END.

Bi 8/1999 - Cn to(Dnh cho hc sinh Tiu hc)S ln cn t nht l 3. Cch cn nh sau:

Ln 1: Chia 27 qu to thnh 3 phn, mi phn 9 qu. t 2 phn ln 2 a cn. Nu cnthng bng th qu to nh nm phn cha cn, nu cn lch th qu to nh nm acn nh hn. Sau ln cn th nht, ta chn ra c 9 qu to trong c qu to nh.

Ln 2: Chia 9 qu to, chn c ra thnh 3 phn, mi phn 3 qu. t 2 phn ln 2 acn. Nu cn thng bng th qu to nh nm phn cha cn, nu cn lch th qu tonh nm a cn nh hn. Sau ln cn th 2, ta chn ra c 3 qu to trong c quto nh.

Ln 3: Ly 2 trong s 3 qu to chn t ln 2 a cn. Nu cn thng bng th qu tonh l qu to cn li, nu cn lch th qu to nh nm a cn nh hn. Sau ba ln cnta chn ra c qu to nh.

Bi 9/1999 - Bc dim



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(Dnh cho hc sinh Tiu hc)Nu s lng que dim ca mi dy l: 3, 5, 8 th hai bn Nga v An bn no bc trcs thng. C nhiu cch ngi bc trc s thng. Gi s:- Dy th nht cso 8 que dim.- Dy th hai c 5 que dim.

- Dy th hai c 3 que dim.Nu Nga l ngi bc trc thng, Nga s lm nh sau:1. Bc ht 8 que dim dy u tin. Nh vy cn 2 dy tng cng 8 que. An s phi bcmt s que mt trong hai dy ny.2. Trong trng hp sau khi An bc s dim ch cn trn mt dy, Nga s bc tt c sdim cn li v s thng. Nu sau khi An bc m s dim vn cn trn hai dy th Ngacng s phi bc sao cho a An vo th bt li: mi dy trong 2 dy cui cng cn ngmt que dim. Nu cha a An c vo th bt li th phi bc sao cho mnh khng

phi th bt li. Chng hn nh:

- An bc 3 que dim dy th 2. Nga s bc 1 que dy cui cng.- An bc 1 que dim tip theo cng dy . Nga cng s bc 1 que dy th 3.- An bc 1 que tip theo. Khi , Nga bc que dim cui cng v thng cuc.Cc bn cng c th th cho cc trng hp khc.

Bi 10/1999 - Dy s nguyn(Dnh cho hc sinh THCS)Dy cho l dy cc s t nhin vit lin nhau:123456789 101112...99 100101102...999 100010011002...9999

10000...9 x 1 = 990 x 2 = 180900 x 3 = 27009000 x 4 = 36000 ...Ta c nhn xt sau:- on th 1 c 9 ch s;- on th 2 c 180 ch s;- on th 3 c 2700 ch s;

- on th 4 c 36000 ch s;- on th 5 c 90000 x 5 = 450000 ch s ...Vi k = 1000 ta c: k = 9 + 180 + 3.270 + 1.Do , ch s th k l ch s u tin ca s 370, tc l ch s 3.Chng trnh:Program Bai10;Uses crt;Var k: longInt;(*--------------------------------------------*)

Function chuso(NN: longInt):char;Var st:string[10];



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dem,M:longInt;Begindem:=0;M:=1;

Repeatstr(M,st);dem := dem+length(st);inc(M);

Until dem >= NN;chuso := st[length(st) - (dem - NN)](*-------------------------------------*)BEGIN

clrscr;;

write('Nhap k:');Readln(k);Writeln('Chu so thu', k,'cua day vo han cac so nguyen khong am');write('123456789101112... la:', chu so(k));Readln;

END.Cch gii khc:var n, Result: LongInt;

procedure ReadInput;

beginWrite('Ban hay nhap so K: '); Readln(n);

end;

procedure Solution;

var

i, Sum, Num, Digits: LongInt;

begin

Sum := 9; Num := 1; Digits := 1;

while Sum < n do

begin

Num := Num * 10; Inc(Digits);

Inc(Sum, Num * 9 * Digits);

end;Dec(Sum, Num * 9 * Digits); Dec(n, Sum);

Num := Num + (n - 1) div Digits;

n := (n - 1) mod Digits + 1;

for i := 1 to Digits - n do Num := Num div 10;

Result := Num mod 10;

end;

procedure WriteOutput;

begin

Writeln('Chu so can tim la: ', Result);

Readln;

end;

begin

ReadInput;

Solution;



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WriteOutput;

end.

Bi 11/1999 - Dy s Fibonaci(Dnh cho hc sinh THCS)

{$R+}constInp = 'P11.INP';Out = 'P11.OUT';Ind = 46;

varn: LongInt;Fibo: array[1..Ind] of LongInt;

procedure Init;

vari: Integer;

beginFibo[1] := 1; Fibo[2] := 1;for i := 3 to Ind do Fibo[i] := Fibo[i - 1] + Fibo[i - 2];

end;procedure Solution;vari: LongInt;

hfi, hfo: Text;begin

Assign(hfi, Inp);Reset(hfi);Assign(hfo, Out);Rewrite(hfo);while not Eof(hfi) dobeginReadln(hfi, n);

Write(hfo, n, ' = ');i := Ind; while Fibo[i] > n do Dec(i);Write(hfo, Fibo[i]);Dec(n, Fibo[i]);while n > 0 dobeginDec(i);if n >= Fibo[i] thenbegin

Write(hfo, ' + ', Fibo[i]);



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Dec(n, Fibo[i]);end;

end;Writeln(hfo);

end;Close(hfo);Close(hfi);

end;begin

Init;Solution;

end.

Bi 12/1999 - N-mino(Dnh cho hc sinh THPT)Program Bai12;{Tinh va ve ra tat ca Mino}Uses Crt;Const fn = 'NMINO.INP';

fg = 'NMINO.OUT';max = 16;

Type bang = array[0..max+1,0..max+1] of integer;Var n : integer;

lonmin : integer;hinh ,hinh1 ,xet ,dd : bang;hang ,cot: array[1..max] of integer;sl : integer;qi,qj : array[1..max*max] of integer;sh ,sc :integer;hangthieu , cotthieu:integer;slch : longint;f : text;

Procedure Nhap;Var f:text;

BeginAssign(f,fn); Reset(f);Readln(f ,n);Close(f);

End;

Procedure Chuanbi;Beginlonmin:= trunc(sqrt(n));If n sqr(lonmin) then Inc(lonmin);slch := 0;

End;



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Function min2( a ,b : integer ) : integer;BeginIf a < b then min2 := a Else min2 := b;

End;

Procedure Taobien( i ,j : integer );Var ii ,jj : integer;BeginFillChar(dd ,SizeOf(dd),1);FillChar(xet,SizeOf(xet),1);For ii := 1 to i doFor jj := 1 to j dobegin

dd[ii,jj] := 0;xet[ii,jj] := 0;

end;

End;

Procedure Ghinhancauhinh;Var i ,j : integer;BeginInc(slch);Writeln(f,sh ,' ' ,sc);For i := 1 to sh dobegin

For j := 1 to sc do Write(f,(dd[i,j] mod 2):2);Writeln(f)

end;End;

Procedure Quaytrai;Var hinh1 : bang;

i,j : integer;Beginhinh1:= hinh;For i := 1 to sh doFor j := 1 to sc do hinh[i,j] := hinh1[sc-j+1,i];

End;Procedure Lathinh;Var hinh1 : bang;

i ,j : integer;Beginhinh1:= hinh;For i := 1 to sh doFor j := 1 to sc do hinh[i,j] := hinh1[sh-i+1,sc-j+1];

End;

Procedure Daohinh;Var hinh1 : bang;



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i,j : integer;Beginhinh1 := hinh;For i := 1 to sh doFor j := 1 to sc do hinh[i,j] := hinh1[sh-i+1,j];

End;

Function Bethat : boolean;Var ii,jj :integer;BeginBethat := false;For ii := 1 to sh doFor jj := 1 to sc doIf hinh[ii,jj] hinh1[ii,jj] then

beginBethat:= hinh[ii,jj] < hinh1[ii,jj];

exit;end;

End;

Function Behon : boolean;BeginBehon := Bethat;

End;

Function Xethinhvuong : boolean;BeginXethinhvuong := false;Quaytrai;If Behon then exit; Quaytrai;If Behon then exit; Quaytrai;If Behon then exit; Daohinh;If Behon then exit; Quaytrai;If Behon then exit; Quaytrai;If Behon then exit; Quaytrai;If Behon then exit; Xethinhvuong := true;

End;

Function Xetchunhat : boolean;BeginXetchunhat := false;Lathinh;If Behon then exit; Daohinh;If Behon then exit; Lathinh;If Behon then exit; Xetchunhat := true;

End;

Procedure Chuyensang( a : bang;Var b : bang );

Var i,j:integer;Begin



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For i := 1 to sh doFor j := 1 to sc do b[i,j] := a[i,j] mod 2;

End;

Procedure Thughinhancauhinh;

BeginChuyensang(dd ,hinh);hinh1:= hinh;If sh = sc then begin If not Xethinhvuong then exit; end

Else If not Xetchunhat then exit;Ghinhancauhinh;

End;

Procedure Xetthem( i ,j : integer );BeginInc(xet[i,j]);

If xet[i,j] = 1 thenbegin

Inc(sl);qi[sl] := i;qj[sl] := j

end;End;

Procedure Xetbot( i ,j : integer );BeginIf xet[i,j] = 1 then Dec(sl);Dec( xet[i,j] );

End;

Procedure Themdiem( ii : integer );Var i ,j : integer;Begini := qi[ii];j := qj[ii];dd[i,j] := 1;If dd[i,j-1] = 0 then Xetthem(i ,j-1);

If dd[i,j+1] = 0 then Xetthem(i ,j+1);If dd[i-1,j] = 0 then Xetthem(i-1,j);If dd[i+1,j] = 0 then Xetthem(i+1,j);

End;

Procedure Bodiem( ii : integer );Var i , j : integer;Begini := qi[ii];j := qj[ii];dd[i,j] := 0;

If dd[i,j-1] = 0 then Xetbot(i,j-1);If dd[i,j+1] = 0 then Xetbot(i,j+1);



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If dd[i-1,j] = 0 then Xetbot(i-1,j);If dd[i+1,j] = 0 then Xetbot(i+1,j);

End;

Procedure Xethangcot( ii : integer );

Var i ,j :integer;Begini := qi[ii];j := qj[ii];Inc(hang[i]);If hang[i] = 1 then Dec(hangthieu);Inc(cot[j]);If cot[j] = 1 then Dec(cotthieu);

End;

Procedure Xetlaihangcot( ii : integer );

Var i,j : integer;Begini := qi[ii];j := qj[ii];If hang[i] = 1 then Inc(hangthieu);Dec(hang[i]);If cot[j] = 1 then Inc(cotthieu);Dec(cot[j]);

End;

Procedure Duyet( i : integer;last : integer );Var ii :integer;BeginIf i > n then

begin thughinhancauhinh; exit; end;For ii := last + 1 to sl do

beginthemdiem(ii);xethangcot(ii);If hangthieu + cotthieu


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taobien(i ,j);For jj := 1 to j do

beginsl:= 1;qi[1] := 1;

qj[1] := jj;duyet(1,0);dd[1,jj] := 2;

end;End;

Procedure Duyethinhbao;Var i ,j : integer;

minj ,maxj : integer;BeginFor i := lonmin to n do

beginminj := (n-1) div i + 1;maxj := min2(n+1-i,i);For j := minj to maxj do duyetcauhinh(i,j);

end;End;

Procedure Ghicuoi;Var f : file of char;

s : string;i : integer;

Beginstr(slch,s);Assign(f,fg); reset(f);Seek(f,0);For i := 1 to length(s) do Write(f,s[i]);Close(f);

End;

BEGINClrscr;

Assign(f,fg); Rewrite(f);Writeln(f ,' ');Nhap;Chuanbi;duyethinhbao;Close(f);ghicuoi;

END.

Bi 13/1999 - Phn hoch hnh ch nht(Dnh cho hc sinh THPT){Recommend:m,n


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const m=4;n=4;max=m*n;var

a: array[1..m,1..n] of byte;i1,j1,dem,daxep,tg: integer;f: text;

time: longint absolute $0:$46C;save: longint;

{------------------------------------}procedure init;beginfor i1:=1 to m dofor j1:=1 to n do a[i1,j1]:=0;

dem:=0; daxep:=0; tg:=0;end;{------------------------------------}

procedure kq;

beginfor i1:=1 to m dobeginfor j1:=1 to n do write(f,a[i1,j1],' ');writeln(f);

end;end;{------------------------------------}

procedure try(i,j: integer);var i2,j2,flag: integer;

beginif (daxep=max) then begin kq; writeln(f); tg:=tg+1; endelsebeginflag:=j;while (flagif (a[i,flag]0) then flag:=flag-1;for i2:=i to m do for j2:=j to flag dobegindem:=dem+1;for i1:=i to i2 do for j1:=j to j2 do a[i1,j1]:=dem;

daxep:=daxep+(i2-i+1)*(j2-j+1);i1:=i;j1:=j2;while (a[i1,j1]0) do

beginj1:=j1+1;if j1=n+1 then begin j1:=1; i1:=i1+1; end;

end;try(i1,j1);daxep:=daxep-(i2-i+1)*(j2-j+1);for i1:=i to i2 dofor j1:=j to j2 do a[i1,j1]:=0;

dem:=dem-1;end;



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end;end;{------------------------------------}BEGENinit;

assign(f,'kq.dat'); rewrite(f);save:=time;try(1,1);write(f,tg);close(f);write('Time is about:',(time-save)/18.2);readln;END.

Bi 14/2000 - Tm s trang sch ca mt quyn sch

(Dnh cho hc sinh Tiu hc) tin tnh ton, ta s nh s li quyn sch bng cc s 001, 002, 003,..., 009, 010,011, 012, 013,..., 098, 099, 100, 101,... tc l mi s ghi bng ng 3 ch s. Nh vy ta

phi cn thm 9x2=18 ch s cho cc s trc y ch c 1 ch s v 90 ch s cho ccs trc y ch c 2 ch s, tng cng ta phi dng thm 108 ch s. Vi cch nh smi ny, ta phi cn ti 1392+108=1500 ch s. V mi s c ng 3 ch s nn c tt c1500:3=500 s, bt u t 001. Vy quyn sch c 500 trang.

Bi 15/2000 - Hi ngh i vin(Dnh cho hc sinh Tiu hc)

tin tnh ton, c mi mt cp bn trai-bn gi quen nhau ta s ni li bng mt sidy. Nh vy mi bn s b "buc" bi ng N si dy v quen vi N bn khc gii. Gis bn trai l T th tnh c s dy ni l TxN. Gi s bn gi l G th tnh c s dyni l GxN. Nhng v 2 cch tnh cho cng kt qu l s dy ni nn TxN=GxN, suy raT=G. Vy trong hi ngh s cc bn trai v cc bn gi l nh nhau.

Bi 16/2000 - Chia s(Dnh cho hc sinh THCS)Lp mt bng 2NxN . Ln lt ghi N2 s 1, 2, 3,..., N2-1, N2 vo N ct, mi ct N stheo cch sau:

12 N+13 N+2 2N+1... ... ... ... ...N 2N-1 3N-2 ... (N-1)N+1

2N 3N-1 ... N2-(N-2)3N ... N2-(N-3)

... N2-(N-4)...

Trong N hng trn, tng i s trong hng th i l:



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i+[N+(i-1)]+[2N+(i-2)]+...+[(i-1)N+1]= N[1+2+...+(i-1)]+[i+(i-1)+(i-2)+...+1]= Ni(i-1)/2+i(i+1)/2= (Ni2-Ni+i2+i)/2Trong N hng di, tng (N-i) s trong hng th N+i l

(i+1)N+[(i+2)N-1]+[(i+3)N-2]+...+[N 2-(N-i-1)]= N[(i+1)+(i+2)+...+N]-[1+2+...+(N-i-1)]= N(N+i+1)(N-i)/2 - (N-i-1)(N-i)/2= (N2+Ni+i+1)(N-i)/2= (N3+Ni+N-Ni2-i2-i)/2Ct i bng chnh gia theo ng k m v ghp li thnh mt bng vung nh sau:

1 2N 3N-1 ... N2-(N-2)2 N+1 3N ... N2-(N-3)3 N+2 2N+1 ... N2-(N-4)... ... ... ... ...N 2N-1 3N-2 ... (N-1)N+1

Khi tng cc s trong hng th i l(Ni2-Ni+i2+i)/2 + (N3+Ni+N-Ni2-i2-i)/2 = (N3+N)/2 = N(N2+1)/2R rng trong mi hng c N s v tng cc s trong mi hng l nh nhau.

Bi 17/2000 - S nguyn t tng ng(Dnh cho hc sinh THCS)C th vit chng trnh nh sau:Program Nttd;

Var M,N,d,i: integer;{------------------------------------}Function USCLN(m,n: integer): integer;Var r: integer;BeginWhile n0 dobeginr:=m mod n; m:=n; n:=r;

end;USCLN:=m;End;

{------------------------------------}BEGINWrite('Nhap M,N: '); Readln(M,N);d:=USCLN(M,N); i:=2;While d1 dobeginIf d mod i =0 thenbeginWhile d mod i=0 do d:=d div i;While M mod i=0 do M:=M div i;While N mod i=0 do N:=N div i;

end;Inc(i);



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end;If M*N=1 then Write('M va N nguyen to tuong duong.')Else Write('M va N khong nguyen to tuong duong.');Readln;END.

Bi 18/2000 - Sn b(Dnh cho hc sinh THCS v THPT)Ta c th thy ngay l con sn phi i N bc (v xi+1 = xi+1), v nu i ln k bc th lidi xung k bc (v yN= y0 = 0). Do , h = N div 2;Chng trnh c th vit nh sau:Program Senbo;Uses Crt, Graph;Var f:Text;

gd, gm, N, W,xo,yo:Integer;Procedure Nhap;

BeginWrite('Nhap so N


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Str(i,sx);str(j,sy);S:='('+sx+','+sy+')');OutTextXY(x,y+5,s);Delay(10000);xo:=x;yo:=y;

End;End;

BeginNhap;Assign(F,'P5.Out');ReWrite(F);Dg:=Detect;InitGraph(Gd,Gm,'');VeLuoi;Bo;

Readln;Close(F);CloseGraph;

End.

Bi 19/2000 - a gic(Dnh cho hc sinh THPT)Ta s chng minh khng nh sau cho n 3:Cc s thc dng a1, a2, a3,..., an lp thnh cc cnh lin tip ca mt a gic ncnh khi v ch khi vi mi k=1, 2,..., n ta c cc bt ng thc sau:

a1 + a2 +... (thiu k)... + an > ak (1)(tng ca n-1 cnh bt k phi ln hn di cnh cn li)Chng minhChng minh c tin hnh qui np theo n. Vi n = 3 th (1) chnh l bt ng thc tamgic quen thuc.Gi s (1) ng n n. Xt (1) cho trng hp n+1.Trc tin ta c nhn xt sau: Cc s a1, a2,..., an, an+1 lp thnh mt a gic n +1 cnhkhi v ch khi tn ti mt s g sao cho a1, a2, a3,..., an-1, g to thnh mt a gic n cnh vg, an, an+1 to thnh mt tam gic.Gi s a1, a2, a3,..., an, an+1 lp thnh mt a gic n +1 cnh. Khi theo nhn xt trn thtn ti a gic n cnh a1, a2, a3,..., an-1, g v tam gic g, an, an+1. Do ta c cc bt ng

thc sau suy t gi thit qui np v bt ng thc tam gic:a1 + a2 + a3 +.... + an-1 > g (2)an + an+1 > g > |an - an+1| (3)Do vy ta ca1 + a2 + a3 +.... + an-1 > |an - an+1| (4)t (4) suy ra ngay cc khng nh sau:a1 + a2 + a3 +.... + an-1 + an > an+1 (5)a1 + a2 + a3 +.... + an-1 + an+1 > an (6)Mt khc t gi thit qui np cho a gic n cnh a1, a2, a3,..., an-1, g, tng t nh (2) ta ccc bt ng thc sau vi k < n:a1 + a2 +... (thiu k)... + an-1 + g > ak

thay th v tri ca (3) ta phi c vi k a1 + a2 +... (thiu k)... + an-1 + an + an+1 > ak (7)



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Cc bt ng thc (5), (6) v (7) chnh l (1). iu kin cn c chng minh.Gi s ngc li, h bt ng thc (1) tho mn, ta ca1 + a2 +... + an-1 + an > an+1 (8)a1 + a2 +... + an-1 + an+1 > an (9)v vi mi k < n ta c:

a1 + a2 +...(thiu k)... + an-1 + an + an+1 > ak (10)T (8) v (9) ta c ngay:a1 + a2 +... + an-1 > |an - an+1| (11)T (10) suy ra vi mi k < n ta c:an + an+1 > ak - a1 - a2 -...(thiu k)... - ak (12)T cc bt ng thc (11) v (12) suy ra tn ti mt s dng g tha mn ng thi cciu kin sau:an + an+1 > g > |an - an+1| (13)a1 + a2 +... + an-1 > g (14)g > ak - a1 - a2 -...(thiu k)... - ak (15)Cc bt ng thc (13), (14) v (15) chnh l iu kin tn ti a gic n cnh a1, a2,

a3,..., an-1, g v tam gic g, an, an+1. iu kin c chng minh.

Chng trnh:Program Dagiac;Uses Crt;Const fn = 'P6.INP';Var i,j,N: integer;

a: array[1..100] of real;s: real;Kq: boolean;

{------------------------------------}Procedure Nhap;Var f: text;BeginAssign(f,fn); Reset(f);Readln(f,N);For i:=1 to N do Read(f,a[i]);Close(f);End;{------------------------------------}BEGIN

Nhap;Kq:=true;For i:=1 to N dobegins:=0;For j:=1 to N do If ji then s:=s+a[j];If s


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(Dnh cho hc sinh Tiu hc)Ta coi nh cc cn h c nh s t 1 n 64 (v ngi nh c 8 tng, mi tng c 8 cnh). Ta c th hi nh sau:- C phi s nh bn ln hn 32?Sau khi Lan tr li, d "ng" hay "khng" ta cng bit chnh xc cn h ca Lan trong

s 32 cn h no. Gi s cu tr li l "khng" ta cng bit chnh xc cn h ca Lan trong s 32 cn h no. Gi s cu tr li l "khng", ta hi tip:- C phi s nh bn ln hn 16?Sau cu hi ny ta bit c 16 cn h trong c cn h Lan ang .Tip tc hi nh vy i vi s ng gia trong cc s cn li. Sau mi cu tr likhong cch gia cc s gim i mt na. C nh vy, ch cn 6 cu hi, ta s bit ccn h Lan .

Bi 21/2000 - Nhng trang sch b ri(Dnh cho hc sinh Tiu hc)

Nu trang b ri u tin nh s 387 th trang cui cng s phi nh s ln hn v phi

l s chn. Do vy trang cui cng phi l 738.Nh vy, c 738 - 378 + 1= 352 trang sch (176 t ) b ri.

Bi 22/2000 - m ng i(Dnh cho hc sinh THCS)a) C tt c 8 ng i t A n B sao cho mi ng i qua mt nh no ch ngmt ln. C th:A BA E BA E F BA E D F BA E F C BA E D C BA E F D C BA E D F C B

b). C tt c 8 ng i t A n D, sao cho ng i qua mi cnh no ch ngmt ln, c th:A B C DA B E DA B F D

A E DA E B F DA E B C DA E F DA E F C Dc). Cc ng i qua tt c cc cnh ca hnh, qua mi cnh ng mt ln (im bt uv im kt thc trng nhau):-

+ Cc ng i qua tt c cc cnh ca hnh, qua mi cnh ng mt ln (im bt u

v im kt thc khng trng nhau):- im bt u l C v im kt thc l D:



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CFBCDFEBAEDCFBCDFEABEDCDFCBFEBAED....Tng t nh th vi im bt u l D v im kt thc l C ta cng tm c cc

ng tho mn tnh cht ny.

Bi 23/2000 - Quay Rubic(Dnh cho hc sinh THPT)Khai trin mt rubic v nh s cc mt nh hnh v sau:Khi ta c th xy dng th tc Quay (mt th i) i mu 8 mt con ca mt ny v12 mt con k vi mt ny. Trn c s gii c 2 bi ton ny. Chng trnh c thvit nh sau:Program Rubic;uses Crt;Type Arr= array[0..5, 0..7] of byte;

const color: Array [0..5] of char=('F', 'U','R', 'B', 'L', 'D');Var

A1, A2, A0, A: Arr;X, X1, X2: String;k: byte;

Procedure Nhap;Var i, j: byte;

BeginClrscr;Writeln ('Bai toan 1. So sanh hai xau:');Writeln ('Nhap xau X1:');Readln (X1);Writeln (' Nhap xau X2:');Readln (X2);Writeln ('Bai toan 2. Tinh so lan xoay:');Write ('Nhap xau X:');Readln (X);For i:= 0 to 5 do

For j:= 0 to 7 do A[i, j]:= i;A:=A0; A1:=A0; A2:=A0;

End;

Procedure Quay (Var A: Arr; k: byte);Const Dir : array[0.. 5, 0.. 3, 0.. 3] of byte = ( ( (1,2,5,4), (6,0,2,4), (5,7,1,3), (4,6,0,2) ),

( (0,4,3,2), (0,0,4,0), (1,1,5,1), (2,2,6,2) ),( (0,1,3,5), (4,4,4,4), (3,3,3,3), (2,2,2,2) ),( (1,4,5,2), (2,0,6,4), (1,7,5,3), (0,6,4,2) ),( (0,5,3,1), (0,0,0,0), (7,7,7,7),(6,6,6,6) ),( (0,2,3,4), (6,6,2,6), (5,5,1,5), (4,4,0,4) ) );

var i,j,tg: byte;Begintg:=A[k,6];

for i:=3 downto 1 do A[k,0] := A[k,2*i-2];A[k,0]:=tg;



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tg:=A[k,7];for i:=3 downto 1 do A[k,2*i] := A[k,2*i -2];A[k,1]:=tg;for i:=1 to 3 do

begin

tg:=A[dir[k,0,3], Dir[k,i,3];for j:=3 downto 1 do A[ dir[k,0,j], Dir[k,i,j] ]:= A[ dir[k,0,j-1], Dir[k,i,j-1] ];A[ [dir[k,0,0], Dir[k,i,0] ]:=tg;end;End;Function Eq(A,B:Arr):Boolean;Var i,j,c:byte;Beginc:=0;for i:=1 to 5 dofor j:=1 to 7 do

If A[i,j] B[i,j] then inc(c);If c=0 then Eq:=true else Eq:=false;End;Procedure QuayXau(x:string; var A: arr);Var i,j:byte;Beginfor i:=1 to length(X) do

beginfor j:= 1 to 5 doIf Color[j] = X[i] then Quay(A,j);end;End;Procedure Bai1;BeginQuayXau(X1,A1);QuayXau(X2,A2);End;Procedure Bai2;Begink:=0;Repeat

QuayXau(X,A);Inc(k);Until Eq(A,A0);End;Procedure Xuat;Var i,j:byte;Beginwriteln;writeln('Ket qua:');writeln('Bai toan 1. So sanh 2 xau:') ;If Eq(A1,A2) then writeln('Hai xau X1 va X2 cho cung mot ket qua.');

writeln('Can ap dung xau X ',k,' lan de Rubic quay ve trang thai ban dau.');Readln;



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End;BeginNhap;Bai1;Bai2;

Xuat;END.

Bi 24/2000 - Sp xp dy s(Dnh cho hc sinh Tiu hc)C th sp xp dy s cho theo cch sau:

Ln th Cch i ch Kt qu0 Dy ban u 3, 1, 7, 9, 51 i ch 1 v 3 1, 3, 7, 9, 5

2 i ch 5 v 7 1, 3, 5, 9, 73 i ch 7 v 9 1, 3, 5, 7, 9

Bi 25/2000 - Xy dng s(Dnh cho hc sinh THCS)C th lm nh sau:

1+35+7 = 4317+35 = 52

Bi 26/2000 - T mu

(Dnh cho hc sinh THCS)K hiu mu Xanh l x, mu l d, mu Vng l v. Ta c 12 cch t mu c lit knh sau:

Bi 27/2000 - Bn c


x d v xd v x dv x d vx d v x

xx dd vv xxvv xx dd vvdd vv xx ddxx dd vv xx

xx dd vv xxdd xx vv ddvv dd xx vvxx vv dd xx

xx dd vv xxvv dd xx vvdd xx vv ddxx vv dd xx

dd vv xx dd

xx dd vv xxvv xx dd vvdd vv xx dd

dd vv xx dd

vv xx dd vvxx dd vv xxdd vv xx dd

dd xx vv dd

xx vv dd xxvv dd xx vvdd xx vv dd

vv xx dd vv

xx dd vv xxdd vv xx ddvv xx dd vv

vv xx dd vvdd vv xx ddxx dd vv xxvv xx dd vv

vv dd xx vvdd xx vv ddxx vv dd xxvv dd xx vv

vv dd xx vvxx vv dd xxdd xx vv ddvv dd xx vv

dd xx vv ddvv dd xx vvxx vv dd xxdd xx vv dd


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(Dnh cho hc sinh THPT)Chng trnh ca bn Nguyn Tin Dng lp 8A2 trng PTTH chuyn Bn Tre, tnhBn Tre.Program Ban_co;Uses Crt;

Var a: array [1..8, 1..8] of 0..1;b, c, d, p: array [0..8,0..8] of integer;

max:integer;Procedure Input;

Var f: text; i, j: integer;st: string[8];

BeginAssign (f, 'banco2.txt');Reset (f);For i:=1 to 8 dobegin

Readln(f,st);For j:=1 to 8 do If st[j]= 0 then a[i,j]:=0 else a[i,j]:=1;

end;Close(f);

End;Procedure Init;Begin

Input;Fillchar(b,sizeof(b),0);c:=b; d:=b; p:=b;

End;Function Get_max(x, y, z, t: integer): integer;

Var k: integer;Begin

k:=x;If k < y then k:=y;

If k < z then k:=z;If k < t then k:=t;Get_max:=k;

End;Procedure Find_max;

Var i, j, k: integer;Begin

max:=0;For i:=1 to 8 doFor j:=1 to 8 doIf a[i, j]= 1 then

beginb[i, j]:=b[i-1,j]+1;c[i, j]:=c[i,j-1]+1;d[i,j]:=d[i-1,j-1]+1;

p[i,j]:=p[i-1,j+1]+1;k:=get_max(b[i,j], c[i,j], d[i,j], p[i,j]);



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If max < k then max:=k;end;

Writeln (max);Readln;

End;

BEGINClrscr;Init;

Find_max;END.

Bi 28/2000 - i tin(Dnh cho hc

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