[C++知识库]M - Marbles Lucky Distribution

M - Marbles Lucky Distribution 补提记录

Juan have N red marbles, M blue marbles, and K bottles. He will put a certain number of marbles on each of the K bottles such that no bottle remains empty and every marble is inside a bottle.

Andres is a Santa Fe fan, so he will pick one bottle at random with an uniform distribution, then he will pick a marble inside of it at random with an uniform distribution, with the hope is a red marble. As Juan is a Millonarios fan, he wants to distribute the marbles in the bottles such that the probability of Andres picking a blue marble is maximized. Juan has a busy life, therefore he needs your help to determine the best arrangement for the marbles and the probability of Andres getting a blue marble.

Input
The input consist of three integers separated by spaces, N M and K (1?≤?N,?M,?K?≤?109) - the number of red marbles, blue marbles and bottles respectively.

Output
Print the probability of Bob getting a blue marble such that the marble arrangement is optimal. Your answer will be considered correct, if its absolute or relative error does not exceed 10?-?6.

Example
Input
50 50 2
Output
0.747474747

题意：有红、蓝俩色弹珠，和一定数量的瓶子，问如何放置使得随机抓取蓝色弹珠的概率最大

解题思路：在损耗蓝色弹珠最少的情况下，使得每个瓶子里抓取蓝色弹珠的概率最大，即在瓶子数少于蓝色弹珠数量的情况下，使其尽可量多的瓶子一定能抓到蓝色瓶子（只放蓝色弹珠）。

注意：这里有个小点要注意，用C++输出double类型的变量时，要使用%f,因为理论上C++的printf函数被没有存%lf。

上代码：

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
    double n, m, k;
    scanf("%lf %lf %lf",&n, &m, &k);
    
    double res;
    if(k > m) res = m / k;
    else 
    {
        res = (k - 1) / k + (m - k + 1) / (m - k + 1 + n) /  k;
    }
    
    printf("%.9f",res);
    
    return 0;
}