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Uva 11346 Probability 積分

編輯：關於C++

化成反比函數求積分

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%

Problem setters: Aleksej Viktorchik, Leonid Shishlo.
Huge Easy Contest #1

/* ***********************************************
Author        :CKboss
Created Time  :2015年01月29日 星期四 22時10分37秒
File Name     :UVA11346.cpp
************************************************ */

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

double a,b,s;

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

	int T_T;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%lf%lf%lf",&a,&b,&s);
		if(s==0)
		{
			printf("100.000000%\n");
			continue;
		}
		else if(a*b<=s)
		{
			printf("0.000000%\n");
			continue;
		}
		double area = log(b)-log(s/a);
		double mu = 4*a*b;
		area = s + s*area;
		area*=4;
		printf("%.6lf%\n",100.*(1.-area/mu));
	}
    
    return 0;
}