C语言求同构数

1、C语言求同构数：求指定区间内的同构数#i nclude<stdio.h>#i nclude<stdlib.h>#i nclude<stri ng.h>char *right(char *ms,i nt len);void mai n()long a;int len;char as10;char ms20;prin tf("求1,10000中 的同构数：n");for(a=1;a<=10000;a+)ultoale n=strle n( as);ultoa(a*a,ms,10);if(strcmp(as,right(ms,le n)=

2、0)(a,as,10);printf("%s 其平方为 %sn",as,ms);char *right(char *ms,i nt len)int i,j;for(i=0;i<(sig ned)strle n( ms);ms+);for(j=0;j<le n;j+,ms-);return ms;c语言输出从m到n的孪生素数对#in elude <stdio.h>#in elude <math.h>int prime(int n) int i;if(n<2)return 0;for(i=2;i<=sqrt (n );i+)if(

3、n %i=0)return 0;return 1;void mai n() int i,m,n;prin tf("m:");sea nf("%d",&m);prin tf(" n:");sea nf("%d",&n);for(i=m;i<=n-2;i+)if(prime(i) && prime(i+2)prin tf("%d,%dn",i,i+2);_lc语言折半查找法#i nclude <stdio.h>#in clude <stri n

4、g.h>void mai n()int cz(i nt a,i nt x,i nt y,i nt c);in t gh10 = 1,2, 3, 4, 5, 6, 7, 8, 9, 10; int h,w;scan f("%d", &h);w = cz(h, 0, 9, gh);if (w != -1)prin tf("%d", ghw);elseprintf("没找到 n"); getch();int cz(i nt a,i nt x,i nt y,i nt c)int m;if (x > y) return -1

5、;m = (x + y) / 2;if (a = cm)return m;if (a < cm)return (cz(a, x, m-1, c);elsereturn (cz(a, m+1, y, c);c语言程序计算1500之间的全部“同构数”之和。#in clude<stdio.h>void mai n()int n, he=O,a=O; double pingfang;for(n=4;n <10 ;n+)pingfang=n*n;if(p in gfa ng-n)%10=0)prin tf("%5d", n);he+=n;a+;if(a%5=0

6、)pingfang=n*n;if(p in gfa ng-n)/100=0)prin tf("%5d", n); he+=n;a+;if(a%5=0)prin tf("n");for(n=100; n<=500; n+)pingfang=n*n;if(p in gfa ng-n )/1000=0) prin tf("%5d", n); he+=n;a+;if(a%5=0)prin tf("n");printf("1到500之间所有的同构数之和为：5dn",he);用C语言验证歌德巴赫猜想(

7、任何充分大的偶数都可以由两个素数之和表示)#i nclude<stdio.h>int f(i nt n)int i;for(i=2;i <n ;i+)if(n %i=0)return 0;return 1;int main() int nu m,i,a,b,co un t=1;printf("请输入一个不小于6的偶数:n");sca nf("%d",&nu m);for(i=1;i<=num /2;i+)a=i;b=nu m-i;if(f(a)=1 &&f(b)=1)if(cou nt%5=0)prin t

8、f("n");coun t+;printf("这个数可以为%d,%d组成.n",a,b);return 0;找出一个数组中的最大值和最小值#i nclude<stdio.h>void mai n()int a88=0;int l,r,i,j,mi n=65535,max=-65536;prin tf("i nput line nu mber:");sca nf("%d",&l);prin tf("i nput colu mn nu mber:");sca nf("%d",&r);printf("input the array:'n")；for(i=0;i<l;i+)for(j=0;j<r;j+)sca nf("%d",&aij);if(aij>max) max=aij;01100)6 二+rL 土-u 至-=5p&HA-p&HX-p&H£l/L)£&