語言求同構(gòu)數(shù)

1、C語言求同構(gòu)數(shù)：求指定區(qū)間內(nèi)的同構(gòu)數(shù)#include#include#includechar *right(char *ms,int len);void main()long a;int len;char as10;char ms20;printf(求1,10000中的同構(gòu)數(shù)：n);for(a=1;a=10000;a+)ultoalen=strlen(as);ultoa(a*a,ms,10);if(strcmp(as,right(ms,len)=0)(a,as,10);printf(%s 其平方為 %sn,as,ms);char *right(char *ms,int len)int i,j;

2、for(i=0;i(signed)strlen(ms);ms+);for(j=0;jlen;j+,ms-);return ms;c語言 輸出從m到n的孿生素?cái)?shù)對(duì)#include #include int prime(int n) int i; if(n2) return 0; for(i=2;i=sqrt(n);i+) if(n%i=0) return 0; return 1; void main() int i,m,n; printf(m:); scanf(%d,&m); printf(n:); scanf(%d,&n); for(i=m;i=n-2;i+) if(prime(i) & pr

3、ime(i+2) printf(%d,%dn,i,i+2); c語言折半查找法#include #include void main() int cz(int a,int x,int y,int c); int gh10 = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10; int h,w; scanf(%d,&h); w = cz(h, 0, 9, gh); if (w != -1) printf(%d, ghw); else printf(沒找到n); getch();int cz(int a,int x,int y,int c) int m; if (x y) return

4、-1; m = (x + y) / 2; if (a = cm) return m; if (a cm) return (cz(a, x, m-1, c); else return (cz(a, m+1, y, c); c語言程序計(jì)算1500之間的全部“同構(gòu)數(shù)”之和。#includevoid main()int n,he=0,a=0;double pingfang;for(n=4;n10;n+)pingfang=n*n;if(pingfang-n)%10=0) printf(%5d,n);he+=n;a+;if(a%5=0)printf(n);for(n=10;n100;n+)pingfang

5、=n*n;if(pingfang-n)/100=0) printf(%5d,n);he+=n;a+;if(a%5=0)printf(n);for(n=100;n=500;n+)pingfang=n*n;if(pingfang-n)/1000=0)printf(%5d,n);he+=n;a+;if(a%5=0)printf(n);printf(1到500之間所有的同構(gòu)數(shù)之和為：%5dn,he);用C語言驗(yàn)證歌德巴赫猜想(任何充分大的偶數(shù)都可以由兩個(gè)素?cái)?shù)之和表示)。#include int f(int n) int i; for(i=2;in;i+) if(n%i=0) return 0; ret

6、urn 1; int main() int num,i,a,b,count=1; printf(請(qǐng)輸入一個(gè)不小于6的偶數(shù):n); scanf(%d,&num); for(i=1;i=num/2;i+) a=i;b=num-i; if(f(a)=1&f(b)=1) if(count%5=0) printf(n); count+; printf(這個(gè)數(shù)可以為%d,%d 組成.n,a,b); return 0; 找出一個(gè)數(shù)組中的最大值和最小值#include void main() int a88=0;int l,r,i,j,min=65535,max=-65536; printf(input line number:);scanf(%d,&l); printf(input column number:);scanf(%d,&r); printf(input the array:n)；for(i=0;il;i+) for(j=0;jmax) max=aij;if(aijmin ) min=aij; for(i=0;il;i+) for(j=0;jr;j+) if(aij=max) printf(Max=%d,x=%d,y=%