495. Kids and Prizes

http://acm.sgu.ru/problem.php?contest=0&problem=495

学习：当一条路走不通，换一种对象考虑，还有考虑对立面。

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

题意：n个盒子里装有礼物，m个人随机选择礼物，选完之后空盒子放回
问选中的礼物数的期望。

m个人是独立的。
对于每个礼物不被人选中的概率为((n-1)/n)^m
那么不被选中的礼物数的期望就是 n*((n-1)/n)^m
所以答案就是 n-n*((n-1)/n)^m;

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int n,m;
double pow(double x,int y)
{
    int i;
    double sum=1;
    for(i=1;i<=y;i++)
      sum*=x;
    return sum;
}
int main()
{
    double cnt;
     double ans;
    while(~scanf("%d%d",&n,&m))
    {
     cnt=double(n-1)/n;
    ans=n-n*pow(cnt,m);
    printf("%.9lf\n",ans);
    }
    return 0;
}