hdoj 1544 Palindromes

Palindromes

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1062    Accepted Submission(s): 460

Problem Description

A regular palindrome is a string of numbers or letters that is the same forward as backward. For example, the string "ABCDEDCBA" is a palindrome because it is the same when the string is read from left to right as when the string is read from right to left.

Now give you a string S, you should count how many palindromes in any consecutive substring of S.

 

Input

There are several test cases in the input. Each case contains a non-empty string which has no more than 5000 characters.

Proceed to the end of file.

 

output

A single line with the number of palindrome substrings for each case. 

 

Sample Input

   
   
   
   

    
    
    
    aba
aa
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    4
3
   
   
   
   

 

#include <iostream>
#include <vector>
#include <string>
using namespace std;
int main()
{
    string s;
    while(cin>>s)
    {
        int p,i,half,left,right,count;
        i=s.size();
        count=0;
        //从左到右钉住最后一个
        for(int p=0;p<i-1;p++)
        {
            half=((i-1)-p)/2;
            if(((i-1)-p)%2==0) //奇数
            {
                left=p+half-1;
                right=p+half+1;
            }
            else  //偶数
            {
                left=p+half;
                right=p+half+1;
            }
            while(p<=left)
            {
                if(s[left]==s[right])
                {
                    count++;
                    left--;
                    right++;
                }
                else break;
            }
        }
        for(int p=i-2;p>=1;p--)
        {
            half=p/2;
            if(p%2==0)
            {
                left=half-1;
                right=half+1;
            }
            else
            {
                left=half;
                right=half+1;
            }
            while(left>=0)
            {
                if(s[left]==s[right])
                {
                    count++;
                    left--;
                    right++;
                }
                else
                    break;
            }
        }
        cout<<count+i<<endl;
    }
    return 0;
}