国王的魔镜 n 264
递归函数的简单应用
#include<stdio.h>
#include<string.h>
char a[105];
int Match(num) //判断匹配
{
if((num%2)==1)
return num;
else
{
int i;
for(i=0;i<(num/2);i++)
{
if(a[i] != a[num-1-i])
break;
}
if(i < (num/2))
return num;
else
return (Match(num/2));
}
}
int main()
{
int length, n;
scanf("%d", &n);
while(n--)
{
scanf("%s", a);
length = strlen(a);
int minleng = Match(length);
printf("%d\n", minleng);
}
return 0;
}
posted on 2011-12-02 01:23 java课程设计例子 阅读(218) 评论(0) 编辑 收藏 举报