Codeforces 1238D AB-string 【思维题】

传送门：Codeforces 1238D

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D. AB-string time limit per test2 seconds memory limit per test256 megabytes The string t1 t2…tk is good if each letter of this string belongs to at least one palindrome of length greater than 1.

A palindrome is a string that reads the same backward as forward. For example, the strings A, BAB, ABBA, BAABBBAAB are palindromes, but the strings AB, ABBBAA, BBBA are not.

Here are some examples of good strings:

t = AABBB (letters t1, t2 belong to palindrome t1…t2 and letters t3, t4, t5 belong to palindrome t3…t5);
t = ABAA (letters t1, t2, t3 belong to palindrome t1…t3 and letter t4 belongs to palindrome t3…t4);
t = AAAAA (all letters belong to palindrome t1…t5);
You are given a string s of length n, consisting of only letters A and B.

You have to calculate the number of good substrings of string s.

Input
The first line contains one integer n (1≤n≤3⋅105) — the length of the string s.

The second line contains the string s, consisting of letters A and B.

Output
Print one integer — the number of good substrings of string s.

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解题思路：
直接找合法的字符串不好找，所以我们应该换个思路，我们去找不合法的字符串

考虑如何求不合法的串，首先串中连续的相同字符一定是回文串的一部分

然后考虑 AB 交错的情况，发现对于某个 A 它如果左右都有 B 那么一定也是回文串的一部分

对于 B 也是同理

那么只要考虑一段 A 和一段 B 连在一起的情况，发现当 ABBBBB… 的时候，串是不合法的，当然 BAAAAA… ，AAAA…B，BBBB…A 也都是不合法的，其他情况显然都是合法的

AC代码：

#include
#include
#include
#include
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
const int N=3*1e5+10;
ll n;
char s[N];

int main()
{
	scanf("%lld",&n);
	scanf("%s",s);
	int pre=0;
	ll ans=0;
	for(int i=0;i<n;i++)
	{
		if(i&&s[i]!=s[i-1])
		{
			//printf("%d %d\n",i,pre);
			ans+=(i-pre);
			pre=i;
		}
		else if(i&&pre!=0)
			ans++;
	}
	ll sum=(n*(n-1))/2;
	//printf("%lld %d\n",sum,ans);
	printf("%lld\n",sum-ans);
	return 0;
}