NYOJ 308-Substring【模拟】

雨点打透心脏的1/2处 2022-07-26 08:35 0阅读 0赞

## 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][Link 1]：

#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;
    }

[Link 1]: http://www.xuebuyuan.com/