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                   ABCABAXYZXCVCX

样例输出

ABAXXCVCX

分析：使用string字符串处理http://blog.csdn.net/zchlww/article/details/45361743

代码一：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int main(){int ncase, i, j, maxlen;scanf("%d", &ncase);while(ncase--){string str, tmp, res; //str原始数据，tmp倒置str，res相同子串cin>>str;tmp = str;reverse(tmp.begin(), tmp.end()); //倒置maxlen = 0;for(i = 0; i < str.size(); i++){for(j = 1; j <= str.size() - i; j++) //j代表截取的长度{if(tmp.find(str.substr(i, j)) != string::npos) //如果截取匹配{if(j > maxlen){maxlen = j; //长度更新res = str.substr(i, j); //最长子串}}}}cout<<res<<endl;}return 0;}

代码二：

#include <iostream>#include <vector>#include <algorithm>using namespace std;int main(){  string s1,s2;  int n;  cin>>n;  while(n--)  {    cin>>s1;    s2=s1;    reverse(s2.begin(),s2.end());//反转字符串    bool flag=false;    for(int i=s1.size();i>0;i--)//可以从最长字符串遍历    {      for(int j=0;j<=s1.size()-i;j++)      {        string t=s1.substr(j,i);        string::size_type pos=s2.find(t);        if(pos!=string::npos)        {          cout<<t<<endl;          flag=true;          break;                 }      }      if(flag)  break;    } } return 0;}