rock-paper-scissors

rock-paper-scissors
维护三个前缀和，然后注意顺序，最后做差来确定可行的答案，因为答案比较小，可以考虑这种暴力做法，像这种方案数可以++的题真的不多，如果想不出来特别优秀的想法，不妨简单化思维

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<set>
#include<map>
#include<stack>
#include<cstring>
#define inf 2147483647
#define ls rt<<1
#define rs rt<<1|1
#define lson ls,nl,mid,l,r
#define rson rs,mid+1,nr,l,r
#define N 100010
#define For(i,a,b) for(int i=a;i<=b;i++)
#define p(a) putchar(a)
#define g() getchar()

using namespace std;
int R[1010],S[1010],P[1010];
int n,m;
int T;
char a[1010];
int cnt;
void in(int &x){
    int y=1;
    char c=g();x=0;
    while(c<'0'||c>'9'){
        if(c=='-')y=-1;
        c=g();
    }
    while(c<='9'&&c>='0'){
        x=(x<<1)+(x<<3)+c-'0';c=g();
    }
    x*=y;
}
void o(int x){
    if(x<0){
        p('-');
        x=-x;
    }
    if(x>9)o(x/10);
    p(x%10+'0');
}
int main(){
    in(T);
    while(T--){
        cnt=0;
        in(n);
        cin>>(a+1);
        For(i,1,n){
            if(a[i]=='R'){
                R[i]=R[i-1];
                S[i]=S[i-1]-1;
                P[i]=P[i-1]+1;
            }
            if(a[i]=='S'){
                R[i]=R[i-1]+1;
                S[i]=S[i-1];
                P[i]=P[i-1]-1;
            }
            if(a[i]=='P'){
                R[i]=R[i-1]-1;
                S[i]=S[i-1]+1;
                P[i]=P[i-1];
            }
        }
        For(i,0,n)
            For(j,0,n)
                if(i+j<=n&&R[i]+P[i+j]-P[i]+S[n]-S[j+i]>0)
                    cnt++;
        o(cnt);p('
');
    }
    return 0;
}