Codeforces 768D Jon and Orbs【概率Dp】

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本文介绍了一种使用动态规划解决特定概率收集问题的方法。该问题涉及估算获取一系列不同物品所需的最小天数，使得获得每种物品的概率超过某个阈值。

Jon Snow is on the lookout for some orbs required to defeat the white walkers. There are k different types of orbs and he needs at least one of each. One orb spawns daily at the base of a Weirwood tree north of the wall. The probability of this orb being of any kind is equal. As the north of wall is full of dangers, he wants to know the minimum number of days he should wait before sending a ranger to collect the orbs such that the probability of him getting at least one of each kind of orb is at least $\text{[math]}$ , where ε < 10^- 7.

To better prepare himself, he wants to know the answer for q different values of p_i. Since he is busy designing the battle strategy with Sam, he asks you for your help.

Input

First line consists of two space separated integers k, q (1 ≤ k, q ≤ 1000) — number of different kinds of orbs and number of queries respectively.

Each of the next q lines contain a single integer p_i (1 ≤ p_i ≤ 1000) — i-th query.

Output

Output q lines. On i-th of them output single integer — answer for i-th query.

Examples

Input

1 1
1

Output

Input

2 2
1
2

Output

2
2

题目大意：

一共有K种物品，每天可以生产任意一种，产生每一种物品的概率是相等的，有Q个查询，每个查询表示询问将所有种类物品都生产出来并且概率大于p/2000的最小天数。

思路：

涉及到概率以及最小值的查询问题，考虑Dp。

问题无非就两种状态，一个是天数，一个是生产物品的种类数，那么不妨设定dp【i】【j】表示到第i天，生产了j种物品的概率。

那么就有：Dp【i】【j】=Dp【i-1】【j】*j/k+Dp【i-1】【j-1】*（k-j+1）/k

过程统计维护一下答案即可。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
double dp[10500][1005];
int main()
{
    int k,q;
    while(~scanf("%d%d",&k,&q))
    {
        dp[0][0]=1;
        for(int i=1;i<=10400;i++)
        {
            for(int j=1;j<=k;j++)
            {
                dp[i][j]+=dp[i-1][j]*(j*1.0/k*1.0);
                dp[i][j]+=dp[i-1][j-1]*((k-(j-1))*1.0/k*1.0);
            }
        }
        while(q--)
        {
            int p;
            scanf("%d",&p);
            for(int i=1;i<=10400;i++)
            {
                if(dp[i][k]>(double)p/(double)2000)
                {
                    printf("%d\n",i);
                    break;
                }
            }
        }
    }
}