UVA-11346 probability (probability)

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 D Efined in the following manner:a = {(x, y) | x is from interval [-a;a]; y was from interval [-b;b]}. What's the probability P that's the area of a rectangle that's defined by points (0,0) and (x, y) would be greater than S?

INPUT:

The number of tests N <= are given on the first line of input. Then N lines with the 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 Line. P must is rounded to 6 digits after decimal point.

SAMPLE INPUT:

310 5 201 1 12 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 namespacestd;intMain () {intnum; scanf ("%d",&num);  while(num--){        Doublea,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{            Doublem = A *b; DoubleAns = (m-s-S * log (M/s))/m; printf ("%.6lf%%\n", ans* -); }    }    return 0;}

UVA-11346 probability (probability)