2029

Palindromes _easy version

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40439    Accepted Submission(s): 24564

Problem Description

“回文串”是一个正读和反读都一样的字符串，比如“level”或者“noon”等等就是回文串。请写一个程序判断读入的字符串是否是“回文”。

 

Input

输入包含多个测试实例，输入数据的第一行是一个正整数n,表示测试实例的个数，后面紧跟着是n个字符串。

 

Output

如果一个字符串是回文串，则输出"yes",否则输出"no".

 

Sample Input

4levelabcdenoonhaha

 

Sample Output

yesnoyesno

 

Author

lcy

 

Source

C语言程序设计练习（五）

 

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View Code

Problem : 2029 ( Palindromes _easy version )     Judge Status : Accepted
RunId : 20067092    Language : C    Author : 837274600
Code Render Status : Rendered By HDOJ C Code Render Version 0.01 Beta

#include<stdio.h>#include<string.h>#include<math.h>int main(){    int n;    while (scanf("%d", &n) != EOF)    {        while (n--)        {            char s[100];             scanf("%s", s);            int  i, target = 0, len = strlen(s);            if (len % 2 == 0)            {                for (i = 0; i < len / 2; i++)                    if (s[i] != s[len - i - 1])                        target = 1;                if (target) printf("no\n");                else printf("yes\n");            }            else            {                for (i = 0; i < (len+1) / 2; i++)                    if (s[i] != s[len - i - 1])                        target = 1;                if (target) printf("no\n");                else printf("yes\n");            }        }    }    return 0;}