UV-11346 Probability (Probability)

UV-11346 Probability (Probability)

Description

G-Probability

Time Limit: 1 sec
Memory Limit: 16 MB

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 shoshould 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:

310 5 201 1 12 2 0

Sample output:

23.348371% 0.000000% 100.000000% question: Given a, B, and s, you must select x in [-a, a] AND y in [-B, B] to make (0, 0) the probability that the area of a rectangle composed of (x, y) is greater than the area of s. The thought is to convert it to the area of x * y> s, and calculate the area by using the guide function, y = s/x. The guiding function is 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;}