CSU1602: Needle Throwing Game(投针问题)

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

裸的投针问题，不知道的童鞋可以自己百度，直接公式。。。

对于这种公式题，知道公式与不知道的差别瞬间就体现出来了

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <algorithm>
using namespace std;
#define ls 2*i
#define rs 2*i+1
#define up(i,x,y) for(i=x;i<=y;i++)
#define down(i,x,y) for(i=x;i>=y;i--)
#define mem(a,x) memset(a,x,sizeof(a))
#define w(a) while(a)
#define LL long long
const double PI = acos(-1.0);
#define Len 200005
#define mod 19999997
const int INF = 0x3f3f3f3f;
#define exp 1e-8

int main()
{
    double L,D;
    double P;
    w(~scanf("%lf%lf",&D,&L))
    {
        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;
}