CSU 1602 Needle Throwing Game (投针问题)

1602: Needle Throwing Game

Time Limit: 5 Sec  Memory Limit: 128 MB
Submit: 48  Solved: 25
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Description

There are many parallel lines on the ground with the distance of D between each adjacent two. Now, throwing a needle randomly on the ground，please calculate the possibility of that the needle can be across one of the lines.

Input

The input consists of multiple test cases. Each test case contains 2 integers D, L on a single line (1 <= D, L <= 100). The input is ended with EOF.

 

Output

For each test case, print an integer of (int)(P*10000) where P is the possibility asked above. For example, when P = 0.25658，you should output 2565.

 

Sample Input

4 2

2 4

Sample Output

3183

8372

HINT

 

Source

 

题意:给定平行线的间距和针的长度，求随机投放针，针与平行线相交的概率是多少。

分析：投针问题，公式题。

当L<D时.公式为:P=2L/Dπ

当L>=D时，P=1-(1/π)[2arcsin(D/L)-(2L/D)+(2/D)sqrt(L^2-D^2)]

#include<cstdio>

#include<iostream>

#include<cstring>

#include<cmath>

#include<stdlib.h>

#include<vector>

#include<queue>

#include<stack>

#include<algorithm>

using namespace std;

const double PI=acos(-1.0);

int main()

{

    double D,L,P;

    while(scanf("%lf %lf",&D,&L)!=EOF)

    {

        if(L<D) P=2*L/(PI*D);

        else P=1+(2.0/PI)*((L*1.0/D)*(1-sqrt((1-(D*D)/(L*L))))-asin(D*1.0/L));

        P=P*10000;

        printf("%d\n",(int)P);

    }

    return 0;

}

View Code