南邮 OJ 1090 Rock, Paper, Scissors
2015-08-04 10:51
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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 : 241 测试通过 : 115
比赛描述
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
题目来源
ACM Mid-Central Regional 2009
#include<iostream>
#include<string>
using namespace std;
int RPS(const char& c1,const char& c2){
if(c1=='R'&&c2=='S' || c1=='S'&&c2=='P' || c1=='P'&&c2=='R')
return 1;
else if(c2=='R'&&c1=='S' || c2=='S'&&c1=='P' || c2=='P'&&c1=='R')
return -1;
else
return 0;
}
int main(){
string str1,str2;
int i,P1win,P2win;
while(cin>>str1>>str2 && str1[0]!='E'){
P1win = 0;
P2win = 0;
for(i=0;i<(int)str1.size();++i){
if(RPS(str1[i],str2[i])==1)
++P1win;
else if(RPS(str1[i],str2[i])==-1)
++P2win;
}
cout<<"P1: "<<P1win<<endl;
cout<<"P2: "<<P2win<<endl;
}
}