SGU495 概率DP
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Kids and Prizes
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)
input:
5 7
4 3
output:
3.951424
2.3125
题意: 有n个奖品,m个人排队来选礼物,对于每个人,他打开的盒子,可能有礼物,也有可能已经被之前的人取走了,然后把盒子放回原处。为最后m个人取走礼物的期望。
思路: 排队取,第1个人取到1个,dp[1]=1;后面的人dp[i]=p取到礼物盒子+dp前面的取到礼物盒子=(n-dp[i-1])/n + dp[i-1];
1 #include"bits/stdc++.h" 2 3 #define db double 4 #define ll long long 5 #define vl vector<ll> 6 #define ci(x) scanf("%d",&x) 7 #define cd(x) scanf("%lf",&x) 8 #define cl(x) scanf("%lld",&x) 9 #define pi(x) printf("%d ",x) 10 #define pd(x) printf("%f ",x) 11 #define pl(x) printf("%lld ",x) 12 #define rep(i, n) for(int i=0;i<n;i++) 13 using namespace std; 14 const int N = 1e6 + 5; 15 const int mod = 1e9 + 7; 16 const int MOD = 998244353; 17 const db PI = acos(-1.0); 18 const db eps = 1e-10; 19 const ll INF = 0x3fffffffffffffff; 20 int t; 21 db dp[N]; 22 int n,m; 23 void cal() 24 { 25 dp[1]=1; 26 for(int i=2;i<=m;i++) dp[i]=dp[i-1]+(n-dp[i-1])/n; 27 printf("%f ",dp[m]); 28 } 29 int main() 30 { 31 32 while(scanf("%d%d",&n,&m)==2){ 33 cal(); 34 } 35 return 0; 36 }
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