codeforces 768D Jon and Orbs

最新推荐文章于 2023-01-26 00:05:27 发布

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本文介绍了一个基于概率动态规划的问题，旨在计算收集多种类型水晶所需的最小天数，使得每种水晶至少出现一次的概率达到特定阈值。通过构建DP状态转移方程，实现了高效查询不同阈值对应的等待天数。

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 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 1
1

output
1

input
2 2
1
2

output
2
2

题意：有k种水晶，每天生产其中一种，生产每种水晶的概率是均等的，有q次询问，问至少需要几天每一种水晶都存在的概率>p/2000。

思路：听说是概率dp，表示没接触过，对我来说和普通的dp没什么两样。。。。dp[i][j]表示第i天生产了j种水晶的概率，很明显有dp转移方程dp[i][j]=dp[i-1][j-1]*(k-j+1)/k+dp[i-1][j]*j/k，因为p范围是1~1000，所以跑到超过范围我们就不用跑了，用滑动窗口的技巧记录1~1000的答案，然后就能O（1）查询了。下面给代码：

#include<cstdio>  
#include<algorithm>  
#include<cstring>  
#include<iostream>  
#include<cmath>  
#include<queue>  
#include<functional>  
typedef long long LL;
using namespace std;
#define maxn 1005
#define ll l,mid,now<<1  
#define rr mid+1,r,now<<1|1  
#define lson l1,mid,l2,r2,now<<1  
#define rson mid+1,r1,l2,r2,now<<1|1  
#define inf 0x3f3f3f3f  
const int mod = 1e9 + 7;
double dp[10 * maxn][maxn];
int ans[maxn];
int main(){
	int k, q;
	scanf("%d%d", &k, &q);
	dp[0][0] = 1;
	for (int i = 1; ; i++){
		dp[i][0] = 0;
		for (int j = 1; j <= k; j++){
			dp[i][j] = dp[i - 1][j - 1] * ((k - j + 1.0) / k) + dp[i - 1][j] * j / k;
		}
		if (dp[i][k] * 2 >= 1)
			break;
	}
	int pos = 1;
	for (int i = 1; i <= 1000; i++){
		while (dp[pos][k] * 2000 < i)
			pos++;
		ans[i] = pos;
	}
	while (q--){
		int p;
		scanf("%d", &p);
		printf("%d\n", ans[p]);
	}
}