【 AtCoder Beginner Contest 162】D - RGB Triplets

Score : $400$ points

Problem Statement

We have a string $S$.

Constraints

• $1\le N\le 4000$ consisting of R, G, and B.

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Input

Input is given from Standard Input in the following format:

$N$

Output

Print the number of triplets in question.

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Sample Input 1 Copy

Copy

4
RRGB

Sample Output 1 Copy

Copy

1

Only the triplet $(1,3,4)$ satisfies the first condition but not the second, so it does not count.

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Sample Input 2 Copy

Copy

39
RBRBGRBGGBBRRGBBRRRBGGBRBGBRBGBRBBBGBBB

Sample Output 2 Copy

Copy

1800

//题目意思简单说一下 给你字符串SS 求 下标为i<j<k的三个字母不一样并且j-i！=k-j的有多少个
//显然3层for循环会TLE，所以这个可以使用前缀和
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=4010;
int n,pre[maxn][3],suf[maxn][3];
char s[maxn];
int d[maxn];
ll ans;
int main() {
	scanf("%d",&n);
	scanf("%s",s+1);
	for (int i=1;i<=n;i++) {
		if (s[i]=='R') d[i]=0;
		else if (s[i]=='G') d[i]=1;
		else d[i]=2;
	}
	for (int i=1;i<=n;i++) {  //前缀和 
		for (int j=0;j<3;j++) pre[i][j]=pre[i-1][j];
		pre[i][d[i]]++;
	}
	for (int i=n;i>=1;i--) {  //后缀和 
		for (int j=0;j<3;j++) suf[i][j]=suf[i+1][j];
		suf[i][d[i]]++;
	}
	for (int i=1;i<=n;i++) {
		for (int j=0;j<3;j++)
		for (int k=0;k<3;k++)
			if (d[i]!=j&&j!=k&&k!=d[i])  //3个字母不同   i为中间的字母，j为前面的，k为后面的。然后相乘组合就可以了
				ans+=pre[i-1][j]*suf[i+1][k];
	}
	for (int i=1;i<=n;i++)
	for (int j=i+1,k;j<=n;j++) {  //处理j-i==k-j的情况 
		k=j*2-i;
		if (j<k&&k<=n) {
			if (d[i]!=d[j]&&d[j]!=d[k]&&d[k]!=d[i]) ans--;
		}
	}
	printf("%lld\n",ans);
	return 0;
}