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[];
 char s[MX];

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