SGU 495 — Kids and Prizes

最新推荐文章于 2021-06-18 10:48:14 发布

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最新推荐文章于 2021-06-18 10:48:14 发布

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分类专栏： C语言-ACM-算法-题解

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本文介绍了一个关于概率的编程问题“KidsandPrizes”，该问题要求计算在一个特定的获奖流程中，给定数量的奖品和获奖者时，能够实际发放出去的奖品的数学期望值。文章提供了两种解决方案：一种是直接推导公式计算，另一种是使用概率动态规划的方法求解。

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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

比较简单的概率题目，直接可以推出公式，每个物品不被选取的概率为pow(1-1/n,m),因此被选取的概率为1-pow(1-1/n,m)

因此期望为（1-pow(1-1/n,m)）*n;

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<cmath>
using namespace std;
double n,m;
int main()
{
    while(~scanf("%lf%lf",&n,&m))
    {
        double ans=n*(1-pow(1-1/n,m));
        printf("%0.9lf\n",ans);
    }
    return 0;
}

概率dp也可以做

dp[i]表示算到第i个人时得到物品的期望，dp[i]=dp[i-1]+(n-dp[i-1])/n，即上一状态的期望，加上这个人可以得到的物品的期望。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<cmath>
#define maxn 100005
using namespace std;
double n;
int m;
double dp[maxn];
int main()
{
    while(~scanf("%lf%d",&n,&m))
    {
       dp[0]=0;
       for(int i=1;i<=m;i++)
       {
           dp[i]=dp[i-1]+(n-dp[i-1])/n;
       }
       printf("%0.9lf\n",dp[m]);
    }
    return 0;
}