hdu 4762 Cut the Cake(高精度&概率)

Cut the Cake

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 586 Accepted Submission(s): 283

Problem Description

MMM got a big big big cake, and invited all her M friends to eat the cake together. Surprisingly one of her friends HZ took some (N) strawberries which MMM likes very much to decorate the cake (of course they also eat strawberries, not just for decoration). HZ is in charge of the decoration, and he thinks that it's not a big deal that he put the strawberries on the cake randomly one by one. After that, MMM would cut the cake into M pieces of sector with equal size and shape (the last one came to the party will have no cake to eat), and choose one piece first. MMM wants to know the probability that she can get all N strawberries, can you help her? As the cake is so big, all strawberries on it could be treat as points.

Input

First line is the integer T, which means there are T cases.

For each case, two integers M, N indicate the number of her friends and the number of strawberry.

(2 < M, N <= 20, T <= 400)

Output

As the probability could be very small, you should output the probability in the form of a fraction in lowest terms. For each case, output the probability in a single line. Please see the sample for more details.

Sample Input

3 3

3 4

Sample Output

1/3

4/27

Source

2013 ACM/ICPC Asia Regional Changchun Online

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liuyiding

题意：

随机在一块蛋糕上放n个草莓，然后用最好的方法切成m块相同大小形状的扇形，问你n个草莓在同一块蛋糕上面

的概率。

思路：

高中的组合公式。先考虑一个草莓。它放在一块蛋糕上的概率为(1/m)。剩下的草莓和它放一起的概率为(1/m)^(n-1)。将但将第一个草莓有n种可能。所以一块蛋糕上有全部草莓的概率为n/(m^n)。但是蛋糕他可以先选。所以。他得到全部草莓的概率应该是每一块蛋糕有全草莓概率的和。所以ans=n/(m^(n-1))。n和m的范围。需采用高精度做。

而Java高精度很方便所以就用Java了。第一份Java代码。嘿嘿。

详细见代码：

import java.util.Scanner;  
import java.math.*;  
public class Main {  
    static Scanner cin=new Scanner(System.in);  
    public static void main(String args[])  
    {  
        int t;  
        BigInteger n,m,gc;  
        t=cin.nextInt();  
        while(t--!=0)  
        {  
            m=cin.nextBigInteger();  
            n=cin.nextBigInteger();  
            m=m.pow(n.intValue()-1);  
            gc=m.易做图(n);  
            System.out.println(n.divide(gc)+"/"+m.divide(gc));  
        }  
    }  
}

补充：软件开发 , C++ ,