HDU 2029 Palindromes _easy version（字符串）

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Palindromes _easy version

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34085 Accepted Submission(s): 20700

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

AC代码：

#include<iostream>#include<cstdlib>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<cstdlib>#include<iomanip>#include<algorithm>typedef long long LL;using namespace std;int main(){    int n;char s[100];while(~scanf("%d",&n)){getchar();while(n--){gets(s);int m=strlen(s),flag=1;for(int i=0,j=m-1;i<m;i++,j--)if(s[i]!=s[j]){printf("no\n");flag=0;break;}if(flag)printf("yes\n");}}    return 0;}

AC2：

#include<iostream>#include<cstdlib>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<cstdlib>#include<iomanip>#include<algorithm>typedef long long LL;using namespace std;int main(){    int test;    string s;    cin >> test;    while(test--)    {        cin >> s;        int flag = 0;        for(int i = 0;i <= s.length() / 2;i++)        {            if(s[i] != s[s.length() -1 - i])            {                flag = 1;                break;            }        }        if(flag) cout << "no" << endl;        else cout << "yes" << endl;    }    return 0;}

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