第四届河南省程序设计大赛 ——Substring

忘是亡心i 2022-02-13 11:25 219阅读 0赞

## 题目描述: ##

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.

## 输入描述: ##

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').

## 输出描述: ##

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: ##

复制

3                   
    ABCABA
    XYZ
    XCVCX

## 样例输出: ##

ABA
    X
    XCVCX

### 这一题最保险的方法就是使用将输入的字符串反转后求两个字符串的最大公共连续子序列 ###

### 如果暴力的话，需要注意这用测试用例——XBCDEFYWFEDCBZ 这个测试用例的答案是BCDEF ###

解题代码：（暴力）

#include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<stack>
    #define INF 9999999;
    using namespace std;
    string s;
    bool lst(string s1){   //判断子串s1在原来的字符串中是否能找到相同项
        for(int i=0;i+s1.size()<=s.size();i++){
        	int bj=1;
    	    for(int j=i;j<i+s1.size();j++){
    		    if(s1[j-i]!=s[j]){
    		    	bj=0;
    		    	break;
    		    }
    		}  
    		if(bj)return true;  
    	}
    	return false;	
    }
    int main(){
    	int n;
    	cin>>n;
    	while(n--){
    		cin>>s; 
    		int bj=1;
    		for(int i=s.size();i>0&&bj;i--){
    		    for(int j=0;j+i-1<s.size();j++){
    		    	string s1;
    				for(int k=i-1;k>=0;k--)s1+=s[j+k]; //将所有可能的子串s1判断
    				if(lst(s1)){
    					for(int i=s1.size()-1;i>=0;i--)cout<<s1[i];  //若成立，则将子串s1反转输出
    					cout<<endl;
    					bj=0;
    					break;
    				}
    		    }
    		    
    	    }
    	}
    	return 0;
    }