[UVA][積分] 11346 - Probability
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%
Problem setters: Aleksej Viktorchik, Leonid Shishlo.
Huge Easy Contest #1
題目描述：
取樣點 P (x, y), -a <= x <= a, -b <= y <= b
問 P 與 (0, 0) 矩形面積大於等於 S 的機率為何？
這個矩形邊長平行 x 軸和 y 軸。
題目解法：
積分，先考慮第一象限可行的 P 面積，得到結果如下：
gif.latex.gif
[UVA][積分] 11346 - Probability
最後將面積 *4 / 總面積 = 機率。
#include <stdio.h>
#include <math.h>
int main() {
double a, b, S;
int testcase;
scanf("%d", &testcase);
while(testcase--) {
scanf("%lf %lf %lf", &a, &b, &S);
double ret = 0;
if(S/a >= b) {}
else if(S/b >= a) {}
else
ret = b*(a-S/b)-S*log(a)+S*log(S/b+1e-8);
printf("%.6lf%%\n", ret*100.0/(a*b));
}
return 0;
}