Kids and Prizes - SGU 495
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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:有n个孩子,m个盒子,每个孩子每次选择一个盒子,如果盒子里有奖品就拿走奖品,并且不动盒子,问拿走的奖品数的期望。
思路:每个盒子不被拿走的概率是((n-1)/n)^m,那么不被拿走的盒子的期望就是n*((n-1)/n)^m。
AC代码如下:
#include<cstdio>#include<cstring>#include<cmath>using namespace std;int n,m;int main(){ int i,j,k; double ans; scanf("%d%d",&n,&m); ans=n-n*pow(1.0*(n-1)/n,m); printf("%.10f\n",ans);}
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