HDU 1544 Palindromes（回文字符子串）-优快云博客

本文介绍了一种算法，用于统计字符串中所有回文子串的数量。通过从中心位置向两侧扩展的方式，分别处理奇数长度和偶数长度的回文串，最终得出答案。

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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

题意：

问有多少个回文的子串。

题解：

从子串的长度考虑，长度为奇数的子串：它的中心只能在[1,len-2]范围内，分别从中心开始向两边扩展，遇到不回文就直接停止拓展。长度为偶数的子串：中心对称的左数范围在[0,len-2]，分别以这中心的两个数向两边扩展。

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int main()
{
    string s;
    while(cin>>s)
    {
        int ans=0;
        int len=s.length();
        for(int i=1;i<len-1;i++)
        {
            for(int j=1;i-j>=0&&i+j<len;j++)
            {
                if(s[i-j]==s[i+j])
                    ans++;
                else
                    break;
            }
        }
        for(int i=0;i<len-1;i++)
        {
            for(int j=0;i-j>=0&&i+j+1<len;j++)
            {
                if(s[i-j]==s[i+j+1])
                    ans++;
                else
                    break;
            }
        }
        cout<<ans+len<<endl;
    }
    return 0;
}

转载于:https://www.cnblogs.com/orion7/p/7931794.html