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BI TP LP TRNH C1. 2. 3. 4. 5. 6. Vit chng trnh gii phng trnh bc nht ax+b=0, trong a v b l hai s bt k nhp t bn phm. x l nghim cn tm. Vit chng trnh gii phng trnh bc hai aX2 + bX +c = 0. Cc h s a, b, c c nhp t bn phm(a0) Bi ton Anca. Vit chng trnh tm nhng hnh ch nht c chiu di gp i chiu rng v din tch bng chu vi. Vi chiu di v chiu rng l cc s nguyn dng nh hn 100. Tm nhng gi tr nguyn x, y, z tho mn cng thc Pitago: x2+y2=z2 vi x,y l cc gi tr nguyn nh hn 100. Vit chng trnh nhp vo mt s nguyn 3 ch s (t 100 n 999), sau in ra cc ch s thuc hng trm, hng chc, hng n v. Mt s Amstrong l mt s c c im sau: s gm n k s, tng cc lu tha bc n ca cc k s bgng chnh s . V d: 153 l mt s c 3 k s, v 13+53+33=1+125+27 = 153. Hy tm cc s Amstrong ln hn 100 v nh hn 1000. Bi ton Gobach: Vit chng trnh chng minh bi ton Gobach: Mt s nguyn t bt k ln hn 5 u c th khai trin thnh tng ca 3 s nguyn t khc. Minh ho cho nhng s nguyn t 0, a>0, a !=1.( dng logax=lnx/lna) 5. Vit chng trnh nhp vo ta ca hai im (x1, y1) v (x2, y2) a) Tnh h s gc ca ng thng i qua hai im theo cng thc: H s gc = (y2 - y1) /(x2 - x1) b) Tnh khong cch gia hai im theo cng thc: Khong cch = 6. Vit chng trnh nhp vo mt k t: a) In ra m Ascii ca k t . b) In ra k t k tip ca n. 7. Vit chng trnh nhp vo cc gi tr in tr R1, R2, R3 ca mt mch in : Tnh tng r trong c hai trng hp mc ni tip (R=AR1+R2+R3) v mc song song theo cng thc ( ) 8. Vit chng trnh nhp vo im ba mn Ton, L, Ha ca mt hc sinh. In ra im trung bnh ca hc sinh vi hai s l thp phn. 9. Vit chng trnh nhp vo ngy, thng, nm. In ra ngy thng nm theo dng dd/mm/yy. (dd: ngy, mm: thng, yy : nm. V d: 20/11/99 ) 10. Vit chng trnh o ngc mt s nguyn dng c ng 3 ch s. II. Bi tp cu trc iu khin 1. Vit chng trnh nhp 3 s t bn phm, tm s ln nht trong 3 s , in kt qu ln mn hnh. 2. Vit chng trnh tnh chu vi, din tch ca tam gic vi yu cu sau khi nhp 3 s a, b, c phi kim tra li xem a, b, c c to thnh mt tam gic khng? Nu c th tnh chu vi v din tch. Nu khng th in ra cu " Khng to thnh tam gic". 3. Vit chng trnh gii phng trnh bc nht ax+b=0 vi a, b nhp t bn phm. 4. Vit chng trnh gii phng trnh bc hai ax2 7

+bx + c = 0 vi a, b, c nhp t bn phm. 5. Vit chng trnh nhp t bn phm 2 s a, b v mt k t ch. Nu: ch l + th thc hin php tnh a + b v in kt qu ln mn hnh. ch l th thc hin php tnh a - b v in kt qu ln mn hnh. ch l * th thc hin php tnh a * b v in kt qu ln mn hnh. ch l / th thc hin php tnh a / b v in kt qu ln mn hnh. 6. Vit chng trnh nhp vo 2 s l thng v nm ca mt nm. Xt xem thng c bao nhiu ngy? Bit rng: Nu thng l 4, 6, 9, 11 th s ngy l 30. Nu thng l 1, 3, 5, 7, 8, 10, 12 th s ngy l 31. Nu thng l 2 v nm nhun th s ngy 29, ngc li th s ngy l 28. 7. C hai phng thc gi tin tit kim: gi khng k hn li sut 2.4%/thng, mi thng tnh li mt ln, gi c k hn 3 thng li sut 4%/thng, 3 thng tnh li mt ln. Vit chng trnh tnh tng cng s tin c vn ln li sau mt thi gian gi nhp t bn phm. 8. Mt s nguyn dng chia ht cho 3 nu tng cc ch s ca n chia ht cho 3. Vit chng trnh nhp vo mt s c 3 ch s, kim tra s c chia ht cho 3 dng tnh cht trn.( if ) 9. Tr chi "On t t": tr chi c 2 ngi chi mi ngi s dng tay biu th mt trong 3 cng c sau: Ko, Bao v Ba. Nguyn tc: Ko thng bao. Bao thng ba. Ba thng ko. Vit chng trnh m phng tr chi ny cho hai ngi chi v ngi chi vi my. (switch) 10. Vit chng trnh tnh tin in gm cc khon sau: Tin thu bao in k : 1000 ng / thng. nh mc s dng in cho mi h l 50 Kw Phn nh mc tnh gi 450 ng /Kwh Nu phn vt nh mc n vi n l mt s nguyn dng nhp t bn phm. 21. Vit chng trnh in ra s o ngc ca mt s nguyn n, vi n nhp t bn phm. 22. Tnh gi tr trung bnh ca mt dy s thc, kt thc dy vi -1. 23. Vit chng trnh m phng php chia nguyn DIV 2 s nguyn a v b nh sau: chia nguyn a v b ta tnh tr a-b, sau ly hiu tm c li tr cho b... tip tc cho n khi hiu ca n nh hn b. S ln thc hin c cc php tr trn s bng tr ca php chia nguyn. 24. Tm s nguyn dng N nh nht sao cho 1+1/2+ ...+1/N > S, vi S nhp t bn phm. 25. Vit chng trnh tnh P=2*4*6*...*(2n), n nhp t bn phm. 26. Vit chng trnh tm UCLN v BCNN ca hai s a v b theo thut ton sau (K hiu UCLN ca a, b l (a,b) cn BCNN l [a,b]) - Nu a chia ht cho b th (a,b) = b - Nu a = b*q + r th (a,b) = (b,r) - [a,b] = a*b/(b,r) 27. Vit chng trnh nhp vo mt s nguyn dng n, in ra mn hnh cc s nguyn t p =1945 In ra nm m lch tng ng. Bit nm m lch c ghp can v chi trong t vi: Can: Gip, t, Bnh, inh, Mu, K, Canh, Tn, Nhm, Qu Chi: T, Su, Dn,, Mo, Thn, T, Ng, Mi, Thn, Du, Tut, Hi. III. Bi tp Mng 1. Vit chng trnh nhp vo mt dy n s thc a[0], a[1],..., a[n-1], sp xp dy s theo th t t ln n nh. In dy s sau khi sp xp. 2. Vit chng trnh sp xp mt mng theo th t tng dn sau khi loi b cc phn t trng nhau. 3. Vit chng trnh nhp vo mt mng, hy xut ra mn hnh: - Phn t ln nht ca mng. - Phn t nh nht ca mng. - Tnh tng ca cc phn t trong mng . 4. Vit chng trnh nhp vo mt dy cc s theo th t tng, nu nhp sai quy cch th yu cu nhp li. In dy s sau khi nhp xong. Nhp thm mt s mi v chn s vo dy c sao cho dy vn m bo th t tng. In li dy s kim tra. 5. Vit chng trnh nhp vo mt ma trn (mng hai chiu) cc s nguyn, gm m hng, n 9

ct. In ma trn ln mn hnh. Nhp mt s nguyn khc vo v xt xem c phn t no ca ma trn trng vi s ny khng ? v tr no ? C bao nhiu phn t ? 6. Vit chng trnh chuyn i v tr t dng thnh ct ca mt ma trn (ma trn chuyn v) vung 4 hng 4 ct. Sau vit cho ma trn tng qut cp m*n. V d: 1 2 3 4 1 2 9 1 2 5 5 8 2 5 4 5 9 4 2 0 3 5 2 8 1 5 8 6 4 8 0 6 7. Vit chng trnh nhp vo mt mng s t nhin. Hy xut ra mn hnh: - Dng 1 : gm cc s l, tng cng c bao nhiu s l. - Dng 2 : gm cc s chn, tng cng c bao nhiu s chn. - Dng 3 : gm cc s nguyn t. - Dng 4 : gm cc s khng phi l s nguyn t. 8. Vit chng trnh tnh tng bnh phng ca cc s m trong mt mng cc s nguyn. 9. Vit chng trnh thc hin vic o mt mng mt chiu. V d : 1 2 3 4 5 7 9 10 o thnh 10 9 7 5 4 3 2 1 . 10. Vit chng trnh nhp vo hai ma trn A v B c cp m, n. In hai ma trn ln mn hnh. Tng hai ma trn A v B l ma trn C c tnh bi cng thc: cij= aij +bij ( i=0,1,2,...m-1; j=0,1,2...n-1) Tnh ma trn tng C v in kt qu ln mn hnh. 11. Vit chng trnh nhp vo hai ma trn A c cp m, k v B c cp k, n. In hai ma trn ln mn hnh. Tch hai ma trn A v B l ma trn C c tnh bi cng thc: cij= ai1*b1j + ai2 *b2j + ai3 *b3j + ... + aik *bkj (i=0,1,2,...m-1;j=0,1,2...n-1) Tnh ma trn tch C v in kt qu ln mn hnh. 12. Xt ma trn A vung cp n, cc phn t a[i, i] ( i= 1 ... n ) c gi l ng cho chnh ca ma trn vung A. Ma trn vung A c gi l ma trn tam gic nu tt c cc phn t di ng cho chnh u bng 0. nh thc ca ma trn tam gic bng tch cc phn t trn ng cho chnh. Ta c th chuyn mt ma trn vung bt k v ma trn tam gic bng thut ton: - Xt ct i (i =0,1...n-2) - Trong ct i xt cc phn t a[k,i] ( k=i+1...n-1) + Nu a[k,i]=0 th tng k ln xt phn t khc + Nu a[k,i] 0 th lm nh sau: Nhn ton b hng k vi - a[i,i]/a[k,i] Ly hng i cng vo hng k sau khi thc hin php nhn trn. i ch hai hng i v k cho nhau Nhn ton b hng k vi -1 sau khi i ch vi hng i Tng k ln xt phn t khc. Vit chng trnh tnh nh thc cp n thng qua cc bc nhp ma trn, in ma trn, a ma trn v dng tam gic, in ma trn tam gic, in kt qu tnh nh thc. 13. Vit chng trnh thc hin vic trn hai dy c th t thnh mt dy c th t. Yu cu khng c trn chung ri mi sp th t. Khi trn phi tn dng c tnh cht sp ca hai dy con. IV. Bi tp Hm 1. Vit hm tm s ln nht trong hai s. p dng tm s ln nht trong ba s a, b, c vi a, b, c nhp t bn phm. 2. Vit hm tm UCLN ca hai s a v b. p dng: nhp vo t v mu s ca mt phn s, kim tra xem phn s ti gin hay cha. 3. Vit hm in n k t c trn mt dng. Vit chng trnh cho nhp 5 s nguyn cho bit s lng hng bn c ca mt hng A 5 ca hng khc nhau. Dng hm trn v biu so snh 5 gi tr , mi tr dng mt k t ring. 4. Vit mt hm tnh tng cc ch s ca mt s nguyn. Vit chng trnh nhp vo mt s nguyn, dng hm trn kim tra xem s c chia ht cho 3 khng. Mt s chia ht cho 3 khi tng cc ch s ca n chia ht cho 3. 10

5. Tam gic Pascal l mt bng s, trong hng th 0 bng 1, mi mt s hng ca hng th n+1 l mt t hp chp k ca n ( Ck n

=

n! k ! ( n k )!

Tam gic Pascal c dng sau: 1 ( hng 0 ) 1 1 ( hng 1 ) 1 2 1 ( hng 2 ) 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 (hng 6) ...................................................... Vit chng trnh in ln mn hnh tan gic Pascal c n hng (n nhp vo khi chy chng trnh) bng cch to hai hm tnh giai tha v tnh t hp. 6. Yu cu nh cu 5 nhng da vo tnh cht sau ca t hp: hnh thnh thut ton l: to mt hm t hp c hai bin n, k mang tnh quy nh sau:

7. Vit chng trnh tnh cc tng sau:

Trong n l mt s nguyn dng v x l mt s bt k c nhp t bn phm khi chy chng trnh. 8. Vit chng trnh in dy Fibonacci nu trong bng phng php dng mt hm Fibonacci F c tnh quy. F0 = F1 = 1 Fn = Fn-1 + Fn-2 9. Bi ton thp H Ni: C mt ci thp gm n tng, tng trn nh hn tng di (hnh v). Hy tm cch chuyn ci thp ny t v tr th nht sang v tr th hai thng qua v tr trung gian th ba. Bit rng ch c chuyn mi ln mt tng v khng c tng ln trn tng nh.

10. Vit chng trnh phn tch mt s nguyn dng ra tha s nguyn t. V. Bi tp Xu 1. Vit chng trnh nhp mt chui k t t bn phm, xut ra mn hnh m Ascii ca tng k t c trong chui. 2. Vit chng trnh nhp mt chui k t t bn phm, xut ra mn hnh chui o ngc ca chui . V d o ca abcd egh l hge dcba. 3. Vit chng trnh nhp mt chui k t v kim tra xem chui c i xng khng. V d : Chui ABCDEDCBA l chui i xng. 4. Nhp vo mt chui bt k, hy m s ln xut hin ca mi loi k t. 5. Vit chng trnh nhp vo mt chui. - In ra mn hnh t bn tri nht v phn cn li ca chui. V d: Nguyn Vn Minh in ra thnh: Nguyn 11

Vn Minh - In ra mn hnh t bn phi nht v phn cn li ca chui. V d: Nguyn Vn Minh in ra thnh: Minh Nguyn Vn 6. Vit chng trnh nhp vo mt chui ri xut chui ra mn hnh di dng mi t mt dng. V d: Nguyn Vn Minh In ra : Nguyn Vn Minh 7. Vit chng trnh nhp vo mt chui, in ra chui o ngc ca n theo tng t. V d : chui Nguyn Vn Minh o thnh Minh Vn Nguyn 8. Vit chng trnh i s tin t s thnh ch. 9. Vit chng trnh nhp vo h v tn ca mt ngi, ct b cc khong trng khng cn thit (nu c), tch tn ra khi h v tn, in tn ln mn hnh. Ch n trng hp c h v tn ch c mt t. 10. Vit chng trnh nhp vo h v tn ca mt ngi, ct b cc khong trng bn phi, tri v cc khong trng khng c ngha trong chui. In ra mn hnh ton b h tn ngi di dng ch hoa, ch thng. 11. Vit chng trnh nhp vo mt danh sch h v tn ca n ngi theo kiu ch thng, i cc ch ci u ca h, tn v ch lt ca mi ngi thnh ch hoa. In kt qu ln mn hnh. 12. Vit chng trnh nhp vo mt danh sch h v tn ca n ngi, tch tn tng ngi ra khi h v tn ri sp xp danh sch tn theo th t t in. In danh sch h v tn sau khi sp xp. VI. Bi tp kiu cu trc 1. Hy nh ngha kiu: struct Hoso{ char HoTen[40]; float Diem; char Loai[10]; }; Vit chng trnh nhp vo h tn, im ca n hc sinh. Xp loi vn ha theo cch sau: im Xp loi 9, 10 Gii 7, 8 Kh 5, 6 Trung bnh di 5 Khng t In danh sch ln mn hnh theo dng sau: XEP LOAI VAN HOA HO VA TEN DIEM XEPLOAI Nguyen Van A 7 Kha Ho Thi B 5 Trung binh Dang Kim C 4 Khong dat ........................................................................................................ 2. Xem mt phn s l mt cu trc c hai trng l t s v mu s. Hy vit chng trnh thc hin cc php ton cng, tr, nhn, chia hai phn s. (Cc kt qu phi ti gin ). 3. To mt danh sch cn b cng nhn vin, mi ngi ngi xem nh mt cu trc bao gm cc trng Ho, Ten, Luong, Tuoi, Dchi. Nhp mt s ngi vo danh sch, sp xp tn theo th t t in, in danh sch sp xp theo mu sau: 12

DANH SACH CAN BO CONG NHAN VIEN | STT |HO VA TEN | LUONG | TUOI | DIACHI VII. Bi tp tp 1. Vit chng trnh qun l mt tp tin vn bn theo cc yu cu: a- Nhp t bn phm ni dung mt vn bn sau ghi vo a. b- c t a ni dung vn bn va nhp v in ln mn hnh. c- c t a ni dung vn bn va nhp, in ni dung ln mn hnh v cho php ni thm thng tin vo cui tp tin . 2. Vit chng trnh cho php thng k s ln xut hin ca cc k t l ch (A..Z,a..z) trong mt tp tin vn bn. 3. Vit chng trnh m s t v s dng trong mt tp tin vn bn. 4. Vit chng trnh nhp t bn phm v ghi vo 1 tp tin tn l DMHH.DAT vi mi phn t ca tp tin l 1 cu trc bao gm cc trng: Ma (m hng: char[5]), Ten (Tn hng: char[20]).Kt thc vic nhp bng cch g ENTER vo Ma. Ta s dng tp tin ny gii m hng ha cho tp tin DSHH.DAT s cp trong bi 5. 5. Vit chng trnh cho php nhp t bn phm v ghi vo 1 tp tin tn DSHH.Dat vi mi phn t ca tp tin l mt cu trc bao gm cc trng : mh (m hng: char[5]), sl (s lng : int), dg ( n gi: float), st (S tin: float) theo yu cu: - Mi ln nhp mt cu trc - Trc tin nhp m hng (mh), a mh so snh vi Ma trong tp tin DMHH.DAT c to ra bi bi tp 1, nu mh=ma th in tn hng ngay bn cnh m hng. - Nhp s lng (sl). - Nhp n gi (dg). - Tnh s tin = s lng * n gi. Kt thc vic nhp bng cch nh ENTER vo m hng. Sau khi nhp xong yu cu in ton b danh sch hng ha c s gii m v tn hng theo mu sau:

| STT | MA HANG| TEN HANG | SO LG |DON GIA|SO TIEN | | 1 | a0101 |Duong cat trang | 25 | 10000.00 |250000.00 | | 2 | b0101 |Sua co gai Ha Lan | 10 | 40000.00 |400000.00 |

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