2005.txt

北大ACM题目例程 详细的解答过程 程序实现 算法分析

#include"iostream.h"
 #include"stdio.h"
 #include"string.h"
 #include"algorithm"
 
 int num[15];
 char w[20]={'A','2','3','4','5','6','7','8','9','T','J','Q','K'};
 
 int point(char a)
 {
 	if(a>='1'&&a<='9')return a-'0';
	if(a=='A')return 11;
	return 10;
}
	
 
 
 int value(char a,char b)
 {
 	int p;
	
	p=point(a)+point(b);
	if(a=='A'||b=='A')
	{
		if(p>21)p-=10;
	}
		
	
	if(p>21)return -1;
 	return p;
 }
 
 
 
int main()
{
	int n,i,va,win,lost,tt=0;
	char a,b,c;
	while(1)
	{
		cin>>n;
		if(n==0)break;
		
		if(tt)printf("\n");
		tt++;
		for(i=0;i<13;i++)
		num[i]=n*4;
		
		cin>>a>>b>>c;
		//cout<<a<<b<<c<<endl;
		va=value(b,c);
		//cout<<va<<endl;
		
		for(i=0;i<13;i++)
		{
			if(w[i]==a)num[i]--;
			if(w[i]==b)num[i]--;
			if(w[i]==c)num[i]--;
		}
		
		win=0;lost=0;
		
		for(i=0;i<13;i++)
		{	
			//cout<<w[i]<<' '<<num[i]<<' '<<value(a,w[i])<<endl;
			if(va<=value(a,w[i]))lost+=num[i];
			else win+=num[i];
		}
		
		//cout<<win<<' '<<lost<<endl;
		printf("%.3lf%%\n",100.0*win/(win+lost));
	}
 	
	
	return 0;
}//2005