NYOJ 308-Substring【模拟】，nyoj308-substring

NYOJ 308-Substring【模拟】，nyoj308-substring

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

ac代码：

 

#include<string.h>
#include<stdio.h>
#include<algorithm>
using namespace std;
char map[1000];
char map1[1000];
char ans[1000];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int i,j,k;
		scanf("%s",map);
		int len=strlen(map);
		for(i=0;i<len;i++)
		{
			map1[i]=map[len-1-i];
		}
		map1[i]='\0';
		int hh=1;
		int ansx=0,ansy=0;
		for(i=0;i<len;i++)
		{
			for(j=i;j<len;j++)
			{
				memset(ans,'\0',sizeof(ans));
				int jj=0;
				for(k=i;k<=j;k++,jj++)
				{
					ans[jj]=map1[k];
				}
				if(strstr(map,ans)!=NULL)
				{
					if((j-i+1)>=hh)
					{
						hh=j-i+1;
						ansx=len-i-1,ansy=len-j-1;
					}
				}
			}
		}
		for(i=ansy;i<=ansx;i++)
		printf("%c",map[i]);
		printf("\n");
	}
	return 0;
}