Reverse and Compare(DP)

Reverse and Compare

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Time limit : 2sec / Memory limit : 256MB

Score : 500 points

Problem Statement

You have a string A=A1A2…An consisting of lowercase English letters.

You can choose any two indices i and j such that 1≤i≤j≤n and reverse substring AiAi+1…Aj.

You can perform this operation at most once.

How many different strings can you obtain?

Constraints

• 1≤|A|≤200,000
• A consists of lowercase English letters.

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Input

Input is given from Standard Input in the following format:

A

Output

Print the number of different strings you can obtain by reversing any substring in A at most once.

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Sample Input 1

Copy

aatt

Sample Output 1

Copy

5

You can obtain aatt (don't do anything), atat (reverse A[2..3]), atta (reverse A[2..4]), ttaa (reverse A[1..4]) and taat (reverse A[1..3]).

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Sample Input 2

Copy

xxxxxxxxxx

Sample Output 2

Copy

1

Whatever substring you reverse, you'll always get xxxxxxxxxx.

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Sample Input 3

Copy

abracadabra

Sample Output 3

Copy

44


//很简单的一道DP，但是难想到，想到极其简单
dp[i]为前 i 个数的不同串数的话,dp[i] = dp[i-1] + sum[ str[j]!=str[i] ] (1<=j<i) (前面不等于字符 i 的所有字符的个数)
因为，只要不同，就能reverse[j,i]出一个新串，相同时，reverse[j+1,i-1]是和前相同的，算过了
此题，容易想到回文串什么的，回文串就是个坑，根本不对，跳进去就很难出来了

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define LL long long
#define MX 200500

LL dp[MX];
LL num[26];
char s[MX];

int main()
{
    scanf("%s",s+1);
    int len = strlen(s+1);
    dp[0]=1;
    for (int i=1;i<=len;i++)
    {
        dp[i]=dp[i-1];
        for (int j=0;j<26;j++)
        {
            if (s[i]=='a'+j) continue;
            dp[i] += num[j];
        }
        num[ s[i]-'a' ]++;
    }
    printf("%lld
",dp[len]);
    return 0;
}