HDU 1544Palindromes（求回文字符串）

Palindromes

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1526    Accepted Submission(s): 722

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
   
   
   
   

 

  题目大意：让求回文字符串的个数。

  分析：由样例可知，要加上字符串的长度，因为每个字符也是回文字符串。

也是够了，看了别人的思路，才会用string来实现

#include<iostream>
#include<string>
using namespace std;
int main()
{
    string str;
    while(cin>>str)
    {
        int sum=0;
        for(int i=1;i<str.size()-1;i++)//以一个字符对称的
        {
            int h=i-1,k=i+1;
            while(h>=0&&k<=str.size())
            {
                if(str[h]==str[k])
                {
                    sum++;
                    h--;
                    k++;
                }
                else
                    break;
            }
        }
        for(int i=0;i<str.size()-1;i++)//以两个字符对称的
        {
            int h=i,k=i+1;
            if(str[h]==str[k])
            {
                while(h>=0&&k<=str.size())
                {
                    if(str[h]==str[k])//多比较了一次，可以完善，将sum++放在前面考虑一次
                    {
                        sum++;
                        h--;
                        k++;
                    }
                    else
                        break;
                }
            }

        }
        cout<<sum+str.size()<<endl;
    }
}