hdu 2164 石头剪刀布

最新推荐文章于 2019-06-30 10:41:10 发布

原创最新推荐文章于 2019-06-30 10:41:10 发布 · 638 阅读

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本文介绍了一个简单的程序设计案例——石头剪刀布游戏的胜负判断逻辑。该程序接收玩家输入，并根据游戏规则判断每轮游戏的胜者，最终决定哪位玩家赢得整个比赛。

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Rock, Paper, or Scissors?

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1989 Accepted Submission(s): 1264

Problem Description

Rock, Paper, Scissors is a two player game, where each player simultaneously chooses one of the three items after counting to three. The game typically lasts a pre-determined number of rounds. The player who wins the most rounds wins the game. Given the number of rounds the players will compete, it is your job to determine which player wins after those rounds have been played.

The rules for what item wins are as follows:

?Rock always beats Scissors (Rock crushes Scissors)
?Scissors always beat Paper (Scissors cut Paper)
?Paper always beats Rock (Paper covers Rock)

Input

The first value in the input file will be an integer t (0 < t < 1000) representing the number of test cases in the input file. Following this, on a case by case basis, will be an integer n (0 < n < 100) specifying the number of rounds of Rock, Paper, Scissors played. Next will be n lines, each with either a capital R, P, or S, followed by a space, followed by a capital R, P, or S, followed by a newline. The first letter is Player 1抯 choice; the second letter is Player 2抯 choice.

Output

For each test case, report the name of the player (Player 1 or Player 2) that wins the game, followed by a newline. If the game ends up in a tie, print TIE.

Sample Input

3
2
R P
S R
3
P P
R S
S R
1
P R

Sample Output

Player 2
TIE
Player 1

#include<stdio.h>
  int main()
    {
    	int n,m;
    	int s,t;
    	char a[110],b[110];
    	int i,j;
    	 scanf("%d",&n);
      while(n--)
        {
        	scanf("%d",&m);
        	getchar();
        	  for(i=0;i<m;i++)
        	    {
        	      scanf("%c %c",&a[i],&b[i]);
				    getchar();	//在此处输入一对之后，会再次敲入回车，故此每次都要加getchar(); 
        	    }
        	       s=t=0;
        	  for(i=0;i<m;i++)
        	    {
        	    	if(a[i]==b[i])
        	    	   continue;
        	    	else
        	    	  {
        	    	  	if(a[i]=='R'&&b[i]=='S'||a[i]=='P'&&b[i]=='R'||a[i]=='S'&&b[i]=='P')
        	    	  	    s++;
        	    	  	else
        	    	  	    t++;
        	    	  }
        	    }
          if(s==t)   printf("TIE\n");
          if(s>t)    printf("Player 1\n");
          if(s<t)    printf("Player 2\n");
        }
        
        return 0;
    }