B1018 锤子剪刀布

思路：一种一种情况做判断，实在是太麻烦

#include<cstdio>
int main()
{
	int n;
	scanf("%d",&n);
	int timesA[3]={0},timesB[3]={0};
	int handA[3]={0},handB[3]={0};//BCJ
	for(int i=0;i<n;i++)
	{ 
	   char c1,c2;
	   scanf("%c%c",&c1,&c2);
	   if(c1=='B'&&c2=='C')
	   {
	   	 timesA[0]++;
	   	 timesB[2]++;
	   	 handA[0]++;
	   }
	   else if(c1=='B'&&c2=='J')
	   {
	   	timesA[2]++;
	   	timesB[0]++;
	   	handB[2]++;
	   }
	    else if(c1=='C'&&c2=='B')
	   {
	   	timesA[2]++;
	   	timesB[0]++;
	   	handB[0]++;
	   }
	    else if(c1=='C'&&c2=='J')
	   {
	   	timesA[0]++;
	   	timesB[2]++;
	   	handA[1]++;
	   }
	   else if(c1=='J'&&c2=='B')
	   {
	   	timesA[0]++;
	   	timesB[2]++;
	   	handA[2]++;
	   }
	    else if(c1=='J'&&c2=='C')
	   {
	   	timesA[2]++;
	   	timesB[0]++;
	   	handB[1]++;
	   }
	   else
	   {
	   	timesA[1]++;
	   	timesB[1]++;
	   }
	   for(int i=0;i<3;i++)
	   {
	   	 printf("%d ",timesA[i]);
	   	 printf("%d ",timesB[i]);
	   	 
	   }
		
	}
}

另：发现锤子剪刀布的关系，此处需完善！！！