UVA - 11346 Probability （概率）

Description

 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:

3
10 5 20
1 1 1
2 2 0

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(a,b)组成的矩形面积大于s的概率
思路：只需要求第一象限的即可，转换成a*b>s的图形面积，利用求导函数求面积的方式计算，总面积为m=a*b，求所以概率为（m-s-s*log(m/s))/m.

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int main (){
    int num;
    scanf("%d",&num);
    while(num--){
        double a,b,s;
        scanf("%lf%lf%lf",&a,&b,&s);
        if(s > a*b)
            printf("0.000000%%\n");
        else if(s ==0)
            printf("100.000000%%\n");
        else{
            double m = a * b;
            double ans = (m - s - s * log(m/s)) / m;
            printf("%.6lf%%\n",ans*100);
        }
    }
    return 0;
}