Palindromes _easy version

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41117    Accepted Submission(s): 24999

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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#include<iostream>#include<cstdlib>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<cstdlib>#include <set>#include<algorithm>#include <limits.h>#include <ctype.h>#include <map>#include <stack>typedef long long LL;using namespace std;const int N = 105;int data[N][N];int dp[N][N];int f[35];char c[6] = {'A', 'B', 'C', 'D', 'E', 'F'};float obj[5];void fun(int n, int m){    int flag = 0;    if(n < 0){        flag = 1;        n = -n;    }    stack<char> s1;    while(n != 0){        int yushu = n % m;        if(yushu >= 10){            s1.push(c[yushu - 10]);        }else {            s1.push(yushu + '0');        }        n =  n / m;    }    if(flag){        cout << "-";    }    while(!s1.empty()){        cout << s1.top();        s1.pop();    }    cout << endl;}int gcd(int a, int b){    return b == 0 ? a : gcd(b, a % b);}int lcm(int a, int b){    return a  / gcd(a, b) * b;}int main() {    //freopen("out.txt","w",stdout);//    int n, sum;//    while(scanf("%d %d", &n, &sum) != EOF){//        if(n == 0 && sum == 0) break;//        for(int i = 1; i <= n; i++){//            int T = 0;//            for(int j = i; j <= n; j++){//                T += j;//                if(T == sum ){//                    cout << "["<< i << "," << j <<"]"<< endl;//                    break;//                }else if(T < sum){//                    continue;//                }else {//                    break;//                }//            }//        }//        cout << endl;//    }    //freopen("in.txt","r",stdin);    int n;    cin >> n;    getchar();    while(n--){        string str;        getline(cin, str);        int flag = 1;        for(int i = 0; i < str.size() / 2; i++){            if(str[i] != str[str.size() - i - 1]){                flag = 0;                break;            }        }        if(flag){            cout << "yes" << endl;        }else {            cout << "no" << endl;        }    }    return 0;}