codeforces768D(87/600)

最新推荐文章于 2025-06-11 21:58:25 发布

原创最新推荐文章于 2025-06-11 21:58:25 发布 · 721 阅读

1 ·

CC 4.0 BY-SA版权

ACM练习同时被 3 个专栏收录

312 篇文章

订阅专栏

动态规划

88 篇文章

订阅专栏

概率

15 篇文章

订阅专栏

本文介绍了一个关于收集k种不同类型的球的概率问题，通过动态规划的方法解决了如何计算收集到每种类型至少一个球所需的最小天数，使得收集齐所有类型球的概率达到指定阈值。

D. Jon and Orbs
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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 1
1
output
1
input
2 2
1
2
output
2
2

这题明明就是能直接算出来…
非要dp…
行吧

就是水题

#include<bits/stdc++.h>
using namespace std;
#define int long long 
int k;
double eps=1e-5;
bool jiance(int x,int pp)
{
    int y=x-k+1;
    double gl=1;
    for(int a=1;a<=k;a++,y++)
    {
        gl*=y;
        if(a%2==0)gl/=(k*k);
    }
    if(k%2)gl/=k;
    if(gl>=double(pp)/2000.0)return 1;
    return 0;
}
double dp[10003][1000];
main()
{
    cin>>k;
    int q,w;
    cin>>q;
    dp[0][0]=1;
    for(int a=1;a<=8000;a++)
    {
        for(int b=1;b<=k;b++)
        {
            dp[a][b]=dp[a-1][b-1]*(k-b+1)/k+dp[a-1][b]*b/k;
        }
        if(dp[a][k]>0.5)break;
    }
    for(int a=1;a<=q;a++)
    {
        int w;
        cin>>w;
        for(int b=k;;b++)
        {
            if(dp[b][k]*2000>=w)
            {
                cout<<b<<endl;
                break;
            }
        }
    }
}