SGU495Kids and Prizes(数学期望||概率DP||公式)
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 |
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InputThe first and only line of the input file contains the values of N and M ().OutputThe 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。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取到礼物=(n-dp[i-1])/n + dp[i-1];
当然,也可以化简为公式 printf("%.10lf\n",n*1.0*(1-pow((n-1)*1.0/n,m)));
#include<cstdio> #include<cstdlib> #include<iostream> #include<algorithm> #include<cstring> #include<algorithm> using namespace std; double dp[100010]; int main() { int n,m,i; while(~scanf("%d%d",&n,&m)){ dp[1]=1; for(i=2;i<=m;i++) dp[i]=dp[i-1]+(n-dp[i-1])/n; printf("%.9lf\n",dp[m]); }return 0; }
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