nyoj 308 Substring，nyoj308substring

nyoj 308 Substring，nyoj308substring

Substring

时间限制：1000 ms  |           内存限制：65535 KB 难度：1

描述

You are given a string input. You are to find the longest substring of input such that the reversal of the substring is also a substring of input. In case of a tie, return the string that occurs earliest in input.

Note well: The substring and its reversal may overlap partially or completely. The entire original string is itself a valid substring . The best we can do is find a one character substring, so we implement the tie-breaker rule of taking the earliest one first.

输入

The first line of input gives a single integer, 1 ≤ N ≤ 10,  the number of test cases. Then follow, for each test case,  a  line  containing between 1 and 50 characters, inclusive. Each character of input will be an uppercase letter ('A'-'Z').

输出

Output for each test case  the longest substring of input such that the reversal of the substring is also a substring of input

样例输入

3                   
ABCABA
XYZ
XCVCX

样例输出

ABA
X
XCVCX

求字符串中最大的子串（反转后同样能在字符串中找到）暴力！！！

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define max 10000+100
using namespace std;
struct record
{
    char s[60];
}num[max];
char str[60],str1[60];
bool cmp(record a,record b)
{
    return strlen(a.s)>strlen(b.s);
}
int main()
{
    int t,i,j,l,k,p,q;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",str);
        l=strlen(str);
        q=0;
        for(i=0;i<l;i++)//开头 
        {
            for(j=i;j<l;j++)//结尾 
            {
                memset(str1,'\0',sizeof(str1));
                for(p=j,k=0;p>=i;p--)
                str1[k++]=str[p];
                if(strstr(str,str1))//找到 
                {
                    strcpy(num[q++].s,str1);//复制 
                }
            }
        }
        sort(num,num+q,cmp); 
        printf("%s\n",num[0].s);
    }
    return 0;
}