Educational Codeforces Round 74 (Rated for Div. 2) D. AB-string（逆向思维）

题目链接

https://codeforces.com/contest/1238/problem/D

题目描述

The string t1t2…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⋅10^5) — 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.

Examples

input

5
AABBB

output

6

input

3
AAA

output

3

input

7
AAABABB

output

15

Note

In the first test case there are six good substrings: s1…s2, s1…s4, s1…s5, s3…s4, s3…s5 and s4…s5.

In the second test case there are three good substrings: s1…s2, s1…s3 and s2…s3.

题解

又是一道逆向思维题，CF好深的套路。。。

这题正面求又求不出来，只能从反面求。

考虑到哪些子串不是所谓的“good string”，统计个数，答案就是子串总数-个数。

事实上，不是“good string”的字符串只有AB...（若干B）以及A...（若干A）B，B...（若干B）A，B...A（若干A）四种情况。

代码

#include <bits/stdc++.h>
#define maxn 300010
#define ll long long
using namespace std;
ll n;
char s[maxn];
ll l[maxn],r[maxn];
int main(){
	ll i;
	scanf("%lld",&n);
	scanf("%s",s+1);
	if(n==1){
		cout<<0;
		return 0;
	}
	s[0]='C';s[n+1]='C';s[n+2]='\0';
	memset(l,0,sizeof(l));
	memset(r,0,sizeof(r));
	for(i=1;i<=n;i++){
		l[i]=1;
		if(s[i]==s[i-1]){
			l[i]=l[i-1]+1;
		}
	}
	for(i=n;i>=1;i--){
		r[i]=1;
		if(s[i]==s[i+1]){
			r[i]=r[i+1]+1;
		}
	}
	ll cnt=0,ans;	
	for(i=2;i<=n;i++){
		if(s[i]!=s[i-1]){
			cnt+=l[i-1];
		}
	}
	for(i=n-1;i>=1;i--){
		if(s[i]!=s[i+1]){
			cnt+=(r[i+1]-1);
		}
	}
	ll len=n;
	ans=len*(len-1)/2-cnt;
	cout<<ans;
	return 0;
}