3045833_ce.c

北大大牛代码 1240道题的原代码 超级权威

#include <stdio.h>
#include <math.h>

double pi = 4.0*atan(1);

int main()
{
	int cas, i;
	double m, n, t, c;
	double p1, p2, p3, p4, s;

	scanf("%d",&cas);
	for(i = 1; i <= cas; i++)
	{
		scanf("%lf%lf%lf%lf",&n,&m,&t,&c);
		s = n*m*t*t;
		printf("Case %d:\n",i);
		p1 = (t-c)*(t-c)*(n-2.0)*(m-2.0)+(t-c/2.0)*(t-c)*(n-2.0+m-2.0)*2.0+(t-c/2.0)*(t-c/2.0)*4.0;
		p1 /= s;
		p4 = (n-1.0)*(m-1.0)*pi*c*c/4.0;
		p4 /= s;
		p3 = c*c*(m-1.0)*(n-1.0)*(1.0-pi/4.0);
		p3 /= s;
		p2 = 1.0-p1-p3-p4;
		printf("Probability of covering 1 tile   = %.4lf%%\n",p1*100.0);
		printf("Probability of covering 2 tiles  = %.4lf%%\n",p2*100.0);
		printf("Probability of covering 3 tiles  = %.4lf%%\n",p3*100.0);
		printf("Probability of covering 4 tiles  = %.4lf%%\n",p4*100.0);
		printf("\n");
	}
	return 0;
}