Codeforces Round #399 D. Jon and Orbs(概率dp,好题)

题目链接

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 1
1

output

1

input

2 2
12

output

2
2

题意：

这个题题意有点晦涩难懂啊==

给出k种物品，每天你可以获得k种物品中的一种，给出q个询问，每个询问是一个数pi，问你最少经过多少天能够使得集齐k种物品的概率大于(pi-eps)/2000，输出最少的天数。

题解：

dp[n][x] 表示过了n天拥有k种物品的概率，那么很容易得到转移方程 



那么，就是要求最小的n使得

 可以把x小于1000的天数都预处理出来，不过这里不知道n的最大值，所以只能用滚动数组来实现了，而且写的时候有些坑，要注意。详细见代码。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#include<stack>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const int inf=0x3fffffff;
const ll mod=1000000007;
const int maxn=1000+100;
double d[maxn];
const double eps=1e-7;
int ans[maxn];
int k;
void init()
{
d[0]=1;//d[0][0]=1;
memset(ans,0,sizeof(ans));
int p=1;
for(int n=1;p<=1000;n++)
{
for(int x=k;x>0;x--)
{
d[x]=(1.0*x*d[x]+1.0*(k-x+1)*d[x-1])/k;
}
while(p<=1000&&2000*d[k]>=1.0*(p-eps)) //写成乘法防止精度误差
ans[p]=n,p++;
d[0]=0;//d
[0]=0(n>0),注意这个地方
}
}
int main()
{
int q;
scanf("%d%d",&k,&q);
init();
while(q--)
{
int p;
scanf("%d",&p);
printf("%d\n",ans[p]);
}
return 0;
}