SGN 495

495. Kids and Prizes

Time limit per test: 0.25 second(s)
Memory limit: 262144 kilobytes

input: standard
output: standard

ICPC (International Cardboard Producing Company) is in the business of producing cardboard boxes. Recently the company organized a contest for kids for the best design of a cardboard box and selected M winners. There are N prizes for the winners, each one carefully packed in a cardboard box (made by the ICPC, of course). The awarding process will be as follows:

• All the boxes with prizes will be stored in a separate room.
• The winners will enter the room, one at a time.
• Each winner selects one of the boxes.
• The selected box is opened by a representative of the organizing committee.
• If the box contains a prize, the winner takes it.
• If the box is empty (because the same box has already been selected by one or more previous winners), the winner will instead get a certificate printed on a sheet of excellent cardboard (made by ICPC, of course).
• Whether there is a prize or not, the box is re-sealed and returned to the room.

The management of the company would like to know how many prizes will be given by the above process. It is assumed that each winner picks a box at random and that all boxes are equally likely to be picked. Compute the mathematical expectation of the number of prizes given (the certificates are not counted as prizes, of course).

Input

The first and only line of the input file contains the values of N and M ().

Output

The first and only line of the output file should contain a single real number: the expected number of prizes given out. The answer is accepted as correct if either the absolute or the relative error is less than or equal to 10^-9.

Example(s)

sample input	sample output
5 7	3.951424

sample input	sample output
4 3	2.3125

这道题看上去介绍很繁琐，输出的精度也不低，似乎是道很麻烦的题目，但实际上就是一个式子可以解决问题。如果正向去dp，那么原则上是要向前推才能得到期望值，这样是可以做出来，只是还有更好的方法。既然东西要放进去，我们反过来求不取的礼物的期望不是也可以间接得到礼物的期望么。本质就是n-n*((n-1)/n)^m。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
int N,M;
int main()
{
    while(scanf("%d%d",&N,&M)!=EOF){
        double P,S;
        P=(double)(N-1)/(double)N;
        S=N-N*pow(P,M);//核心表达式
        printf("%.9lf\n",S);
    }
    return 0;
}