SGU-495 Kids and Prizes（概率DP）

495. Kids and Prizes

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

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

题目大意：有m个人和n个装有奖品的盒子，每个人随机取走一个盒子，若存在奖品则取走，然后放回，否则直接放回，求拿走奖品的个数的期望？

方法一：概率DP  （O(m)）

最终期望是每个人拿到奖品的概率之和，考虑同时只存在一个人拿到盒子

第i个人拿到奖品的概率与前一个人拿到奖品的概率有关，所以设dp[i]表示第i个拿盒子的人拿到奖品的概率

则dp[i]可由两个状态转移而来：

①第i-1个拿盒子的人拿到奖品，概率为dp[i-1]，则第i个拿盒子的人拿到奖品的概率为：dp[i-1]-1/n

②第i-1个拿盒子的人没有拿到奖品，概率为1-dp[i-1]，则第i个拿盒子的人拿到奖品的概率为：dp[i-1]

可得状态转移方程为：dp[i]=(1-dp[i-1])*dp[i-1]+dp[i-1]*(dp[i-1]-1/n);

#include <cstdio>using namespace std;int n,m;double dp[100005],nn,ans;int main() {    while(2==scanf("%d%d",&n,&m)) {        dp[1]=ans=1;        nn=1.0/n;        for(int i=2;i<=m;++i) {            dp[i]=(1-dp[i-1])*dp[i-1]+dp[i-1]*(dp[i-1]-nn);            ans+=dp[i];        }        printf("%.9lf\n",ans);    }    return 0;}

方法二：二项分布  （O(1)）

又看到一种O(1)的算法，但是其成立的前提是m个人没有拿取到某个奖品的是相互独立的（感觉不同时取的话，第一个人必定会取到一个奖品，则后面的人取到该奖品的概率就为0了）

对于每个奖品来说，不被拿走的概率为：(n-1)/n，则不被所有人拿走的期望为：((n-1)/n)^m

则：奖品不被拿走的期望为：n*((n-1)/n)^m，所以奖品被拿走的期望为：n-n*((n-1)/n)^m

#include <cstdio>#include <cmath>using namespace std;int n,m;int main() {    while(2==scanf("%d%d",&n,&m)) {        printf("%.9lf\n",n-n*pow(1.0-1.0/n,m));    }    return 0;}