sgu 495 Kids and Prizes 概率dp
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495. Kids and Prizes
Time limit per test: 0.25 second(s)
Memory limit: 262144 kilobytes
Memory limit: 262144 kilobytes
input: standard
output: 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.
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个礼盒,m个人取礼物,礼物取走了就没有了 ,求m个人取礼物的期望。
思路:直接转移方程dp[i]=dp[i-1]+(n-dp[i-1])/n,dp[i]表示前i个人拿n个礼物的期望,则第i个人拿到礼物的期望就是(n-dp[i-1])/n,再加上前面的人的就ok了。
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cstdlib>#include <cmath>#include <vector>#include <queue>#include <map>using namespace std;const int N=100005;const double eps=1e-5;int n,m;double dp[N];int main(){ scanf("%d%d",&n,&m); dp[1]=1; for(int i=2;i<=m;i++){ dp[i]=dp[i-1]+(n-dp[i-1])/n; } printf("%.10f\n",dp[m]); return 0;}
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