Kids and Prizes
Time Limit:250MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Submit Status Practice SGU
495
Appoint description:
System Crawler (2014-10-25)
Description
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.
Sample Input
sample input |
sample output |
5 7 |
3.951424 |
sample input |
sample output |
4 3 |
2.3125 |
题意：有n件奖品装在n个箱子里，M个人选择其中的一个箱子，每个人取出一个箱子，如果有奖品取出，后将箱子放回，后面的人继续取箱子.问最终取出箱子个数的期望。
另dp[i]表示第i个人拿到奖品的概率则，dp[1]=1;若i-1取到了奖品则i获得奖品的概率为dp[i-1]-1/n,否则i得到奖品的概率为dp[i-1].转移方程为dp[i]=dp[i-1]*(dp[i-1]-1/n)+(1-dp[i-1])*dp[i-1].、
/*************************************************************************
> File Name: t.cpp
> Author: acvcla
> Mail: [email protected]
> Created Time: 2014年10月21日 星期二 21时33分55秒
************************************************************************/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<cstring>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<cstdlib>
#include<ctime>
#include<set>
#include<math.h>
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 10;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define pb push_back
int n,m,f;
double dp[maxn];
int main(int argc, char const *argv[])
{
while(~scanf("%d%d",&n,&m)){
memset(dp,0,sizeof dp);
dp[1]=1;
double ans=1;
for(int i=2;i<=m;i++){
dp[i]=dp[i-1]*(dp[i-1]-1.0/n)+(1-dp[i-1])*dp[i-1];
ans+=dp[i];
}
printf("%.10f\n",ans);
}
return 0;
}