ACM: 限时训练题解-Rock-Paper-Scissors-前缀和

1. Rock-Paper-Scissors

 

Rock-Paper-Scissors is a two-player game, where each player chooses one of Rock, Paper, or Scissors. Here are the three cases in which a player gets one  point:

 

­          Choosing Rock wins over a player choosing  scissors.

­          Choosing Scissors wins over a player choosing  Paper.

­          Choosing Paper wins over a player choosing  Rock.

 

In all other cases, the player doesn’t get any  points.

 

Bahosain and his friend Bayashout played N rounds of this game. Unlike Bayashout, Bahosain was too lazy to decide what to play next for each round, so before starting to play, he chose three integers X Y Z such that X+Y+Z = N and X, Y, Z ≥ 0, and then played Rock for the first X rounds, Paper for the next Y rounds, and Scissors for the last Z  rounds.

 

Bayashout got more points in the N rounds and won. Given the moves played by Bayashout in each round, Bahosain wants to know the number of ways in which he could have chosen X, Y and Z such that he wins in the N rounds.

The winner of the N rounds is the player that gets more total points in the    N rounds.

 

Input

 

The first line of input contains T (1 ≤ T ≤   64), where T is the number of test cases.

 

The first line of each test case contains an integer N (1 ≤ N ≤   1000) that represents the number of rounds.

 

The next line contains a string of N uppercase letters, the first letter represents the choice of Bayashout for the first round, the second letter represents his choice for the second round, and so on.

 

Each letter in the string is one of the following: R (Rock), P (Paper), or S   (Scissors).

 

Output

 

For each test case, print a single line with the number of ways in which Bahosain could have won.

 

 

 

Sample Input	Sample Output
4	3
3	1
RPS	1
1	5
R
5
PPRSR
5
RPSPR

 

/*
题意；A　Ｂ猜拳，猜Ｎ次
但是Ａ很懒，不想去想怎么猜拳就随机的连续出ｘ次石头　ｙ次布　ｚ次剪刀 
现在给出Ｂ的猜拳的出拳的顺序， 
然后Ａ以石头Ｒ　布Ｐ　剪刀Ｓ　的顺序出，在出了布之后就不在出石头，出了剪刀后就不再出石头和布 
问Ａ有多少种可能赢Ｂ。 


这题一开始不怎么会做，做到限时赛最后还有３０分钟的时候又回过头来看这个题目
不就是一个简单的前缀和么，一开始都没想到。

ＡＣ代码： 
*/

 

#include"iostream"
#include"algorithm"
#include"cstdio"
#include"cstring"
#include"cmath"
#define MX 1000 + 50
using namespace std;
int r[MX],p[MX],s[MX],len;
char str[MX];

int main() {
	int T;
	scanf("%d",&T);
	while(T--) {
		memset(r,0,sizeof(r));
		memset(p,0,sizeof(p));
		memset(s,0,sizeof(s));
		memset(str,0,sizeof(str));
		scanf("%d%s",&len,str+1);
		for(int i=1; i<=len; i++) {
			if(str[i]=='R') {　　　　//这里用数组rps记录全部出石头布剪刀的得分 【dp-前缀和】 
				p[i]=p[i-1]+1;		//如果B出的石头 则A的布得分 +1 
				r[i]=r[i-1];		//A的石头平局，不变 
				s[i]=s[i-1]-1;		//A的剪刀输扣分 -1 
			}						//下面一样 
			if(str[i]=='S') {
				r[i]=r[i-1]+1;
				s[i]=s[i-1];
				p[i]=p[i-1]-1;
			}
			if(str[i]=='P') {
				s[i]=s[i-1]+1;
				p[i]=p[i-1];
				r[i]=r[i-1]-1;
			}
		}
		int ans=0;
		for(int i=0; i<=len; i++) {		//i-1 为连续出石头的次数 r[i]-r[0]为A出石头的得分 
			for(int j=i; j<=len; j++) { //j-i为连续出布的次数 p[j]-p[i]为A出布的得分 
				if(r[i]-r[0]+p[j]-p[i]+s[len]-s[j]>0) {   //len-j 为连续出布的次数 s[len]-p[j]为A出剪刀的得分 
					ans++;		//计算和 如果比0大就算赢。 
				}
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}