SDUT OJ 2399 Palindrome

题目描述

A string is said to be a palindrome if it reads the same both forwards and backwards, for example "madam" is a palindrome while "acm" is not. 
Now we want to know how many palindrome strings in the given strings.

输入

The first line of the input contains a single integer T, which indicates number of the given strings.
The next T lines contain one string S, whose length is less than 1000 letters.

输出

One integer indicates the number of palindrome strings.

示例输入

5ababaaabbabccccccAbcd

示例输出

2

题意：就是统计回文串的个数；

给出两个代码 一个是对半查找，能省一点儿时间

对半查找：

#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>char s1[1010];using namespace std;int main(){    int n,i,j;    int cnt=0;    int flag;    scanf("%d",&n);    for(i=0;i<n;i++){        flag=0;        memset(s1,0,sizeof(s1));        scanf("%s",s1);        int len=strlen(s1);        for(j=0;j<=(len-1)/2;j++){            if(s1[j]!=s1[len-j-1]){                flag=1;                break;            }        }        if(!flag)            cnt++;    }    printf("%d\n",cnt);    return 0;}

正常做法

#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>char s1[1010],s2[1010];using namespace std;int main(){    int i,j,k,len,n;    int cnt=0;    scanf("%d",&n);    for(i=1; i<=n; i++)    {        memset(s2,0,sizeof(s2));        scanf("%s",s1);        j=0;        len=strlen(s1);        for(k=len-1; k>=0; k--)        {            s2[j]=s1[k];            j++;        }        if(strcmp(s1,s2)==0)            cnt++;    }    printf("%d\n",cnt);    return 0;}