2017年省赛前最后一水 Problem A: String_abxba的算法-优快云博客

本文介绍了一个简单的算法，用于判断输入的字符串是否为回文字符串。通过对比字符串首尾字符的一致性来实现，并提供了一个C语言实现的例子。

题目描述

Lucy give you a string s（s1s2s3...sn） .
If（s1s2s3...sn）=（snsn-1sn-2...s1） ,you tell Lucy YES,otherwise ,you tell Lucy NO.
For example.
Lucy give you “abba”，you should tell her “YES”
Lucy give you “abbc”，you should tell her “NO”
Lucy give you “abxba”，you should tell her “YES”

输入

The first line of the input gives the number of test cases T(1<T<1000). T test cases follow.
For each test case.
One line contains a string s('a'<=si<='z') .The length of s no more than 1000.

输出

For each test case, output a single line "Case #x: y", where x is the case number, starting from 1. And y is the answer.

样例输入

abb

a
abbc


abxba

样例输出

Case #1: YES


Case #2: NO


Case #3: YES

分析：判断是否是回文字符串

代码如下:

#include <stdio.h>
#include <string.h>
int main (){
    char str[1005];
    int Case=1;
    int t;
    scanf ("%d",&t);
    getchar();
    while (t--){
        scanf ("%s",str);
         
        int x=strlen(str);
        int flag=1;
        for (int i=0,j=x-1;i<=j&&flag;i++,j--){
            if(str[i]!=str[j]){
                printf ("Case #%d: NO\n",Case++);
                flag=0;
            }
        }
        if (flag)
            printf ("Case #%d: YES\n",Case++);
    }
    return 0;
}