G - Probability

Time Limit: 1 sec
Memory Limit: 16MB

Consider rectangular coordinate system and point L(X,Y) which is randomly chosen among all points in the area A which is defined in the following manner: A = {(x,y) | x is from interval [-a;a]; y is from interval [-b;b]}. What is the probability P that the area of a rectangle that is defined by points (0,0) and (X,Y) will be greater than S?

INPUT:

The number of tests N <= 200 is given on the first line of input. Then N lines with one test case on each line follow. The test consists of 3 real numbers a > 0, b > 0 ir S => 0.

OUTPUT:

For each test case you should output one number P and percentage " %" symbol following that number on a single line. P must be rounded to 6 digits after decimal point.

SAMPLE INPUT:

SAMPLE OUTPUT:

 23.348371%
0.000000%
100.000000%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的概率思路：转换成x*y > s的图形面积，利用求导函数求面积的方式计算，y=s/x的导函数是y=s*lnx#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

int main() {
	int t;
	double a, b, s;
	scanf("%d", &t);
	while (t--) {
		scanf("%lf%lf%lf", &a, &b, &s);
		if (a * b <= s) {
			puts("0.000000%");
			continue;
		}
		if (s == 0) {
			puts("100.000000%");
			continue;
		}
		printf("%lf%%\n", 100 - (s + s * (log(a) - log(s / b))) / a / b * 100);
	}
	return 0;
}

UVA - 11346 Probability （概率）

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