题意:给定a,b,s要求在[-a,a]选定x,在[-b,b]选定y,使得(0, 0)和(x, y)组成的矩形面积大于s,求概率
思路:这样其实就是求xy > s的概率,那么画出图形,只要求y = s / x的原函数, y = slnx,带入两点相减就能求出面积,面积比去总面积就是概率
代码:
#include <cstdio>
#include <cstring>
#include <cmath>
int t;
double a, b, s;
int main() {
scanf("%d", &t);
while (t--) {
scanf("%lf%lf%lf", &a, &b, &s);
if (s == 0) printf("100.000000%%\n");
else if (s >= a * b) printf("0.000000%%\n");
else printf("%.6lf%%\n", (a * b - s - s * (log(a) - log(s/b))) / (a * b) * 100);
}
return 0;
}UVA 11346 - Probability(概率),布布扣,bubuko.com
原文地址:http://blog.csdn.net/accelerator_/article/details/37912499