while循环结构例题

1、引子#include "stdio.h"main() int i=1; for(i=1;i<=10000;i+) printf(“%dt”,i);题型1 输入输出多个数据eg1、输出110000之间所有的整数#include "stdio.h"main() int i=1; while(i<=1000) printf(“%dt”,i); i+;拓展：1、换成所有的奇数 2、换成所有的偶数题型2 有限个数连加和连乘eg2.1、求1+2+3+4+100的值#include "stdio.h"main() int i=1,s=0

2、; while(i<=100) s=s+i;i+;printf(“%dn”,s);拓展：1、求1+2+3+4+n的值2、求12+22+32+n2的值 3、求1+1/2+1/3+1/n的值eg2.2、求n!的值#include "stdio.h"main() int i=1,n,p=1; scanf(“%d”,&n); while(i<=n) p=p*i;i+;printf(“%dn”,p);拓展：求1!+2!+3!+n!的值#include "stdio.h"main() int i=1,n,p=1,s; scanf(“%d”,&am

3、p;n); while(i<=n) p=p*i; s=s+p;i+;printf(“%dn”,s);题型3 无限个数连加eg3、求1-1/3+1/5-1/7+的近似值，要求精度要达到10-4#include "stdio.h"#include "math.h"main() float n=1,s=0,f=1,t=1; while(fabs(t)>=1e-4) t=f/(2*n-1); s=s+t;f=-f;n+;printf(“%fn”,s);拓展：求1-1/2+1/4-1/6+的近似值，要求精度要达到10-4题型4 统计eg4.1、输入20

4、个数，统计其中正数、负数和零的个数。#include "stdio.h"main() int i=1,n,p,z; float x; p=n=z=0; while(i<=20) scanf(“%f”,&x);if(x>0)p+; else if(x<0) n+;else z+; i+;printf(“%dt%dt %dn”,p,n,z);拓展：统计各类字符的个数eg4.2 个位为6且能被3整除的五位数有多少？方法1#include "stdio.h"main() long i=10000,c=0; while(i<=999

5、99) if(i%3=0)&& (i%10=6)c+;i+;printf(“%d n”,c);方法2#include "stdio.h"main() long i=10006,c=0; while(i<=99999) if(i%3=0)c+;i=i+10;printf(“%d n”,c);题型5 数列eg5 输出fibo数列的第20位数字#include "stdio.h"main() int f1=1,f2=1, f3,i=3; while(i<=20) f3=f1+f2;f1=f2;f2=f3;i+;printf(“%d

6、n”,f3);拓展：输出fibo数列前20位数字#include "stdio.h"main() int f1=1,f2=1, f3,i=3;printf(“%d t%d t”,f1,f2); while(i<=20) f3=f1+f2;f1=f2;f2=f3; printf(“%d t”,f3);i+;题型6 数据的逆序输出eg6 任意给定一个正整数，个位数字逆序输出。#include "stdio.h"main() long x,t; scanf(“%ld”,&x); while(x!=0) t=x%10;x=x/10;printf(“

7、%d”,t); 题型7 公约数与公倍数eg7 任意输入两个正整数，求其最大公约数和最小公倍数。#include "stdio.h"main() int m,n,a,b,r,t; scanf(“%d%d”,&m, &n); if(m>n) a=m;b=n; else a=n;b=m; while(b!=0) r=a%b;a=b;b=r;printf(“zuida gongyushu shi:%dn”,a);printf(“zuixiao gongbeishu shi:%dn”,m*n/a);题型8 素数问题eg8 从键盘上任意输入一个正整数，判断其是否为

8、素数。#include "stdio.h"main() int x,i=2; scanf(“%d”,&x); while(x%i!=0) i+; if(x=i) printf(“shi!”);else printf(“fou!”);题型9 高次方程的根eg9.1 用二分迭代法求解方程y=2x3-4x2+3x-6=0在(-10,10)之间的根，要求精度10-5#include "stdio.h"#include "math.h"main()float x1=10,x2=-10,x, y ,y1;x=(x1+x2)/2;y=2*x

9、*x*x-4*x*x+3*x-6;while(fabs(y)>1e-5)y1=2*x1*x1*x1-4*x1*x1+3*x1-6; if(y*y1>0) x1=x; else x2=x; x=(x1+x2)/2; y=2*x*x*x-4*x*x+3*x-6; printf("the root is %fn",x);eg9.2 用牛顿迭代法求解方程2x3+ 4x2-7x-6=0在x=1.5附近的根，要求精度10-5#include "stdio.h"#include "math.h"main()float x,x0, y ,

10、y1;x=1.5;while(fabs(x-x0)>1e-5) x0=x;y=2*x0*x0*x0+4*x0*x0-7*x0-6;y1=6*x0*x0+8*x0-7;x=x0-y/y1;printf("the root is %fn",x);牛顿迭代公式：xn+1=xn-f(xn)/f(xn)do-while循环结构举例#include "stdio.h"main() int i=1,s=0; do s=s+i;i+; while(i<=100);printf(“%dn”,s);for循环结构举例f1#include "stdio.

11、h"main() int i=1,s=0; for(i=1;i<=100;i+) s=s+i; printf(“%dn”,s);f2#include "stdio.h"main( )int i,f1,f2,f3;f1=1;f2=1;printf("%d,%d",f1,f2);for(i=3;i<=20;i+) f3=f1+f2; f1=f2; f2=f3; printf(",%d",f3);f3#include "stdio.h"main( )int i;float a,max;scanf(&

12、quot;%f ",&a);max=a;for(i=1;i<=9;i+) scanf("%f ",&a); if(max<a) max=a;printf("%fn",max);f4#include "stdio.h"main( )int i,s=1;for(i=9;i<=1;i-)s=2*(s+1);printf("%dn",s);#include "stdio.h"main()int x,n=0,s=0; while (n<10) scanf(

13、"%d",&x); if (x<0) break; s+=x; n+; printf("s=%dn",s); #include "stdio.h"main( )int x,n=0,s=0;while (n<10) scanf("%d",&x); if (x<0) continue; s+=x; n+; printf("s=%dn",s);#include "stdio.h"main( )int x,n=0,s=0;while (n<10

14、) scanf("%d",&x);n+; if (x<0) continue; s+=x; printf("s=%dn",s);#include "stdio.h"main()int i=2,m; scanf("%d",&m);while(m%i!=0)i+;if(i=m) printf("%d shi sushu!n",m); else printf("%d bu shi sushu!n",m);#include "stdio.h"

15、main()int i,m;scanf("%d",&m);for(i=2;m%i!=0;i+) ;if(i=m) printf("%d shi sushu!n",m);else printf("%d bu shi sushu!n",m);#include "stdio.h"main( ) int i,m; scanf("%d",&m); for (i=2; i<=m; i+) if (m%i=0) break; if(i=m) printf("%d shi sus

16、hu!n",m); else printf("%d bu shi sushu!n",m);#include "stdio.h"#include "math.h"main( ) int i,m,s; scanf("%d",&m); s=sqrt(m); for (i=2; i<=s; i+) if (m%i=0) break; if(i=s+1) printf("%d shi sushu!n",m); else printf("%d bu shi sushu!n&

17、quot;,m);#include "stdio.h"#include "math.h"main() int i,j; for(i=100;i<=200;i+) for(j=2;j<=i;j+) if (i%j=0) break; if(j=i) printf("%-10d",i); #include "stdio.h"#include "math.h"main() int i,j,s; for(i=100;i<=200;i+) s=sqrt(i); for(j=2;j<

18、=s;j+) if (i%j=0) break; if(j=s+1) printf("%-10d",i); #include <include.h>main( ) int i,j,s; for (i=2; i<=10000; i+) s=0; for (j=1; j<i; j+) if (i%j=0) s+=j; if (i=s) printf("%6dn",s); #include "stdio.h"main()int i,j,k;for(i=0;i<=35;i+) for(j=0;j<=35;j

19、+) if(i+j=35)&&(2*i+4*j=94) printf("ni=%-10dj=%-10d",i,j); #include "stdio.h"main() int i,j,k; for(i=0;i<=19;i+) for(j=0;j<=33;j+) for(k=0;k<=100;k+) if(i+j+k=100)&&(5*i+3*j+k/3=300) printf("ni=%-10dj=%-10dk=%-10d",i,j,k);#include "stdio.h&

20、quot;main() int i,j,k;for(i=0;i<=19;i+)for(j=0;j<=33;j+) k=100-i-j; if(15*i+9*j+k=300) printf("ni=%-10dj=%-10dk=%-10d",i,j,k); #include "stdio.h"main() int i,j,k; for(i=0;i<=19;i+) for(j=0;j<=33;j+) k=100-i-j; if(5*i+3*j+k/3=100) printf("ni=%-10dj=%-10dk=%-10d&qu

21、ot;,i,j,k); #include "stdio.h"main() int m,n,k; for (m=1;m<=9;m+) for(n=1;n<=m;n+) printf("%d*%d=%-5d",n,m,n*m); printf("n"); #include "stdio.h"main()int i; for(i=1;i<=5;i+)printf("*n"); #include "stdio.h"main()int i,j; for(i=1;i&l

22、t;=5;i+) for(j=1;j<=5-i;j+) printf(" "); printf("*n"); #include "stdio.h"main( ) int i,j; for (i=1; i<=5; i+) for (j=1; j<=20-i; j+) printf(" "); for (j=1;j<=i;j+) printf("*"); printf("n"); #include "stdio.h"main() int i,j; for (i=1;i<=5;i+) for (j=1;j<=20-i;j+) printf(" "); for (j=1;j<=2*i-1;j+) printf("*"); printf("n"); #includ