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第 6 章 循 环 控 制. 6.1 概述 6.2 goto 语句以及用 goto 语句构成循环 6.3 while 语句 6.4 do while 语句 6.5 for 语句 6.6 循环的嵌套 6.7 几种循环的比较 6.8 break 语句和 continue 语句 6.9 程序举例 习题. 6.1 概述. 在许多问题中需要用到循环控制。例如,要输入全校学生成绩;求若干个数之和;迭代求根等。几乎所有实用的程序都包含循环。循环结构是结构化程序设计的基本结构之一,它和顺序结构、选择结构共同作为各种复杂程序的基本构造单元。 - PowerPoint PPT Presentation
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6.1 6.2 gotogoto6.3 while6.4 do while6.5 for 6.6 6.7 6.8 breakcontinue6.96
6.1 (1) gotoif (2) while (3) do\|while (4) for
6.2 gotogotogoto gotogoto label-1;goto 123gotogoto
goto(1) if(2) cbreakcontinue(6.8)gotogoto()
0while6.16.16.2
(1) whilewhilewhilewhilesum=sum+i; (2) i>100ii>100i++i
6.4 do whiledo whiledo while ()() 06.3dowhileNS(6.3(b))
6.3
6.3dowhilen6.46.46.5100n=1
whiledo-whileDo-whilewhile6.36.56.5whiledo-whilewhile6.16.16.5()whiledo-while6.26.3while(0)
6.4(b)N\|S6.4(a) i1006.4(b)i>100i100i>1005\|4(b)i100
6.5 for Cforwhile forfor(123) (1) 1 (2) 2(0)for(3)(0)(5)
(3) 3 (4) (2) (5) for 6.6forfor6.6
for() :for(i=1;i
for(i=1; ;i++) sum=sum+i;12i=1; while(1) {sum=sum+1; i++;}(3) 3
6.96.8
for( ;(c=getchar())!='n';) printf("%c"c);213Entercomputer ()computer ()ccoommppuutteerr
Entercfor(baSIcPascaL)fOR13forfor
6.6 (whiledowhilefor)(1) while( ) { while( ) {} }
(2) do { do { } while( ); } while( );(3) for(;;) { for(; ;) {} }
(4) while( ) { do {} while( ); }(5) for(; ;) { while( ) { } }
(6) do { for (; ;) { } } while( );
6.7 (1) goto(2) whiledowhilewhile(i++i=i+1) for33forwhilefor(3) whiledowhilewhiledowhilefor1
(4) whiledowhileforbreakcontinue(breakcontinue6.8)gotoifbreakcontinue
6.8.1break4.4breakSwitchSwitchbreakfor(r=1;r100) break; printf("%f"area); }6.8breakcontinue
r=1r=10area100forarea>100breakbreakbreak;breakSwitch
6.8.2 continue continue;continuebreakcontinuebreak(1) while(1) { if(2) break
} (2) while(1){if(2) continue;}(1)6.10(2)6.116.106.112
6.106.11
n3continue(printf)n3printf6.5if (n%3!=0) printf("%d"n);continuecontinue
6.9
6.6/41-13+15-17+10-6N\|S(6.12)6.12
#includemain(){int s;float ntPi;t=1Pi=0;n=1.0;S=1;while((fabs(t))>1e-6){Pi=Pi+t; n=n+2; s=-s;
t=S/n;}Pi=Pi*4;printf("Pi=%10.6fn"Pi);}
Pi= 3.141594
6.7fibonacci4012113:f1=1 (n=1)f2=1 (n=2)fn=fn-1+fn-2 (n3)33
6.13
6.13
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040 1346269 2178309 3524578 57022887 9227465 14930352 24157817 39088169 63245986 102334155
6.14
printf%12 ld%12d2332767%ldif4iii12(f1f2)i24
6.8m6.14m2m m2m ik(m)m
2k(m)i1i=k+1ik+12k#include main() { int mik; scanf("%d"&m);
k=sqrt(m+1);/*1m*/ for (i=2;i=k+1) printf("%d is a Prime mubern"m); else printf("%d is not a Prime numbern"m); } 1717 is a Prime number
if (i>=k+1){printf("%d "m);n=n+1;}if(n%10==0) printf("n"); }printf ("n"); } 101 103 107 109 113 127 131 137 139 149151 157 163 167 173 179 181 191 193 197199 n10
6.10aEae4WaXbYcZD6.15china!Glmre!
6.15
#include main() { char c; while((c=getchar())!='n') {if((c>='a' && c='a' && c'Z' && c'z') c=c-26; }
printf("%c"c); } } china! Glmre!4(4)4ZzV(v)6.15aD(ad)
c26ascIIifif(c>'Z'|| c>'z') c=c-26;c>Zc=c-26;c>Z && c'z' && cz
6.106.1 mn6.2 6.3 Sn=a+aa+aaa++aaa naa2+22+222+2222+22222(n=5)n6.4 (1!+2!+3!+4!++20!)6.5 3153153=13+53+33
6.6 61236=1+2+3,610006itS factorS are 1,2,36.7 2/1,3/2,5/3,8/5,13/8,21/13,20
6.8 106.9
x10-5
6.10 ** * ** * * * ** * * * * * ** * * * ** * **