# 题目

{% fold 点击显/隐题目 %}

- 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).

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# 题解

可以很容易列出概率dp公式

`dp[i] = dp[i-1] / n * dp[i-1] + (n-dp[i]) / n * (dp[i-1] + 1)`

由于 `i`

最大达到 `100000`

使用递归会爆栈,并且没有必要开数组记录(每个都只计算一次)

也没有必要去推通项公式(不会超时)

直接 `for`

循环一边算出答案即可

当精度要求能达到全部取走的时候就不用往后算了

也可以**从盒子的角度想**：

m个人是独立的

对于每个礼物不被人选中的概率为 `((n-1)/n)^m`

那么不被选中的礼物数的期望就是 `n*((n-1)/n)^m`

所以答案就是 `n-n*((n-1)/n)^m`

# 代码

{% fold 点击显/隐代码 %}```cpp Kids and Prizes https://github.com/OhYee/sourcecode/tree/master/ACM 代码备份

#include

#include

using namespace std;

const double eps = 1e-12;

int main() {

int n, m;

while (scanf("%d%d", &n, &m) != EOF) {

double ans = 0;

for (int i = 0; i < m; i++) {

ans = ((ans * ans) + (n - ans) * (ans + 1)) / n;

if (fabs(ans - n) < eps)

break;

}

printf("%.11f\n", ans);

}

return 0;

}

{% endfold %}