c多线程面试题同步_高级java多线程面试题_多线程面试题(35)

#include<string.h>char *commanstring(char shortstring[], char longstring[])

{

int i, j;char *substring=malloc(256);if(strstr(longstring, shortstring)!=NULL) //如果……，那么返回shortstring

return shortstring; for(i=strlen(shortstring)-1;i>0; i--) //否则，开始循环计算

{

for(j=0; j<=strlen(shortstring)-i; j++){

memcpy(substring, &shortstring[j], i);

substring[i]='/0';

if(strstr(longstring, substring)!=NULL)

return substring;

}

}

return NULL;

}

main()

{

char *str1=malloc(256);

char *str2=malloc(256);

char *comman=NULL;gets(str1);

gets(str2);if(strlen(str1)>strlen(str2)) //将短的字符串放前面

comman=commanstring(str2, str1);

else

comman=commanstring(str1, str2);printf("the longest comman string is: %s/n", comman);

}

11.写一个函数比较两个字符串str1和str2的大小，若相等返回0，若str1大于

str2返回1，若str1小于str2返回－1

int strcmp ( const char * src,const char * dst)

{

int ret = 0 ;

while( ! (ret = *(unsigned char *)src - *(unsigned char *)dst) && *dst)

{

++src;

++dst;

}

if ( ret < 0 )

ret = -1 ;

else if ( ret > 0 )

ret = 1 ;

return( ret );

}3,求1000！的未尾有几个0（用素数相乘的方法来做，如72=2*2*2*3*3）;

求出1->1000里,能被5整除的数的个数n1,能被25整除的数的个数n2,能被125整除的数的个数n3,

能被625整除的数的个数n4.

1000!末尾的零的个数=n1+n2+n3+n4;

#include<stdio.h>

#define NUM 1000int find5(int num){

int ret=0;

while(num%5==0){

num/=5;

ret++;

}

return ret;

}

int main(){

int result=0;

int i;

for(i=5;i<=NUM;i+=5)

{

result+=find5(i);

}

printf(" the total zero number is %d/n",result);

return 0;

}

1. 有双向循环链表结点定义为：

struct node

{ int data;

struct node *front,*next;

};

本文来自电脑杂谈，转载请注明本文网址：
http://www.pc-fly.com/a/jisuanjixue/article-36439-35.html