南阳理工：Substring，南阳理工substring

南阳理工：Substring，南阳理工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  

注意：本题意是让求原子符串中某个子串s，如果他的翻转串仍然是原串的子串，并且s是所有子串中最长的，则输出s,而不是求最长对称子串！！！

#include<stdio.h>//求最长对称子串的代码，但是和题意不符合
#include<string.h>
int main()
{
 int N,n,i,j,i0,j0,flag,i2,j2,max;
 char str[60];
 scanf("%d",&N);
 while(N--)
 {
  memset(str,0,sizeof(str));
  scanf("%s",str);
  n=strlen(str);
  i2=j2=0;
  for(i=0,max=0;i<n;i++)
  {
   for(j=i;j<n;j++)
   {
    for(i0=i,j0=j,flag=1;i0<=j0;i0++,j0--)
     if(str[i0]==str[j0]) continue;
     else 
     {
      flag=0;
      break;
     }
    if(flag==1&&j-i>max) 
    {
      max=j-i;   
     i2=i;
     j2=j;
    }
   }
  }
  for(i=i2;i<=j2;i++)
   printf("%c",str[i]);
  printf("\n"); 
 }
 return 0;
}

#include<stdio.h>//正确代码
#include<string.h>
int panduan(char str1[60],char str2[60])
{
 int i=0,j=0,flag=0,n1,n2;
 n1=strlen(str1);
 n2=strlen(str2);
 while(i<n1&&j<n2)
  if(str1[i]==str2[j])
  {
   i++;
   j++;
  }
  else
  {
   i=i-j+1;
   j=0;
  }
 if(j==n2) flag=1;
 return flag;
}

int main()
{
 int N,n,i,j,k,i0,len,max;
 char str[60],s[60],zc[60];
 scanf("%d",&N);
 while(N--)
 {
  memset(str,0,sizeof(str));
  memset(s,0,sizeof(s));
  scanf("%s",str);
  n=strlen(str);
  zc[0]=str[0];
  zc[1]='\0';
  max=1;
  for(i=0;i<n;i++)
   for(j=i;j<n;j++)
   {
    for(i0=j,k=0;i0>=i;i0--)
     s[k++]=str[i0];
    s[k]='\0';
    len=strlen(s);
    if(len>max)
    {
     k=panduan(str,s);
     if(k==1)
     {
      strcpy(zc,s);
      max=len;
     }
    }
   }
  for(i=max-1;i>=0;i--)
   printf("%c",zc[i]);
  printf("\n");
 }
 return 0;
}