Rock,Paper,Scissors 水NOJ 1090

Rock, Paper, Scissors

时间限制(普通/Java) : 1500 MS/ 10000 MS 运行内存限制 : 65536 KByte
总提交 : 230 测试通过 : 107

题目描述

Rock, Paper, Scissors is a classic hand game for two people. Each participant holds out either a fist (rock), open hand (paper), or two-finger V (scissors). If both players show the same gesture, they try again. They continue until there are two different gestures. The winner is then determined according to the table below:

Rock beats Scissors
Paper beats Rock
Scissors beats Paper
Your task is to take a list of symbols representing the gestures of two players and determine how many games each player wins.
In the following example:

Turn : 1 2 3 4 5
Player 1 : R R S R S
Player 2 : S R S P S
Player 1 wins at Turn 1 (Rock beats Scissors), Player 2 wins at Turn 4 (Paper beats Rock), and all the other turns are ties.

输入

The input contains between 1 and 20 pairs of lines, the first for Player 1 and the second for Player 2. Both player lines contain the same number of symbols from the set {'R', 'P', 'S'}. The number of symbols per line is between 1 and 75, inclusive. A pair of lines each containing the single character 'E' signifies the end of the input.

输出

For each pair of input lines, output a pair of output lines as shown in the sample output, indicating the number of games won by each player.

样例输入

RRSRS
SRSPS
PPP
SSS
SPPSRR
PSPSRS
E
E

样例输出

P1: 1
P2: 1
P1: 0
P2: 3
P1: 2
P2: 1

水题就不废话了~

实现代码：

<span style="font-size:12px;">#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
using namespace std;
char a[76],b[76];
int main()
{
    while(scanf("%s%s",&a,&b)&&!(a[0]=='E'&&b[0]=='E'))
    {
        int len=strlen(a);
        int aa=0,bb=0;
        for(int i=0;i<len;i++)
        {
            if(a[i]=='R')
            {
                if(b[i]=='S')
                {
                    aa++;
                }
                if(b[i]=='P')
                {
                    bb++;
                }
            }
            else if(a[i]=='P')
            {
                if(b[i]=='R')
                {
                    aa++;
                }
                if(b[i]=='S')
                {
                    bb++;
                }
            }
            else if(a[i]=='S')
            {
                if(b[i]=='P')
                {
                    aa++;
                }
                if(b[i]=='R')
                {
                    bb++;
                }
            }
        }
        printf("P1: %d
P2: %d
",aa,bb);
    }
}</span>