Codeforces #339 D. Jon and Orbs（概率dp）

D. Jon and Orbs

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 , where ε < 10 - 7.

To better prepare himself, he wants to know the answer for q different values of pi. 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 pi (1 ≤ pi ≤ 1000) — i-th query.

Output

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

Examples

input

1 11

output

1

input

2 212

output

22

【分析】：说白了就是个dp。

题意：有k个物品，每天可以生产任意一个，概率相等。现在有q次询问，每次询问输入一个整数p，问最少需要多少天使得k中物品全被生产过的概率大于 p/2000

用dp[i][j]代表前i天，已经生产过j种的概率。

第i的天的状态可以由i-1天递推过来，即第i天增加还是不增加一种新物品

状态方程：dp[i][j] = dp[i-1][j-1]*(k-j+1)/k + dp[i-1][j]*j/k;

初始状态dp[0][0]=1

注意：天数是不定的，只是天数越大，dp[i][k]（概率）就越大，当概率>0.5的时候就可以break了

【代码】：

#include<bits/stdc++.h>using namespace std;double dp[10100][1010];int main(){    int k,q,p,i;cin>>k>>q;    dp[0][0]=1;    for(i=1; ;i++)//天数    {        for(int j=1;j<=k;j++)        {            dp[i][j]=dp[i-1][j-1]*(k-j+1)/k+dp[i-1][j]*j/k;        }        if(dp[i][k]>0.5)break;    }    double a[10020];    for(int j=0,x=0;j<=1001;j++)    {        while(x<i&&dp[x][k]<j*1.0/2000)x++;//寻找        a[j]=x;    }    while(q--)    {        cin>>p;        cout<<a[p]<<endl;    }}