nyoj308 Substring 河南第4届省赛，nyoj308substring

nyoj308 Substring 河南第4届省赛，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

来源

第四届河南省程序设计大赛

#include<stdio.h>

#include<string.h>
#include<string>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
using namespace std;
int main()
{
	int N;
	scanf("%d",&N);
	while(N--)
	{
		string s1,s2,s3;
		cin>>s1;
		s2=s1;
		reverse(s2.begin(),s2.end());//倒置字符串s1 
		int maxnum=0;
		int len=s1.size();
		for(int i=0;i<len;i++)
		{
			for(int j=1;j<=len-i;j++)
			{
				string::size_type pos=s2.find(s1.substr(i,j));
				//在s2中寻找到s1的一个子串 
				if(pos!=string::npos&&maxnum<j)
				//返回一个名为 string::npos 的特殊值，说明查找没有匹配。string 类将 npos 定义为保证大于任何有效下标的值。 
				{	
					maxnum=j;
					s3=s1.substr(i,j);//选出合适子串给s3 
				}
			}
		}
		cout<<s3<<endl;
	}
	return 0;
}