hdoj 1544 Palindromes

r囧r小猫 2022-02-16 01:43 99阅读 0赞

# 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;
    }