Description
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:
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(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
#include
#include
#include
#include
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; }