codeforces 768D Jon and Orbs

题目链接：http://codeforces.com/problemset/problem/768/D

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令$f[i][j]$表示当前产生过了$i$个球，产生过了$j$个不同的球的概率。

${Ans_i=Minleft { x|f[x][k]>frac{p_i-varepsilon }{2000} ight }}$

考虑转移：

${f[i][j]=frac{j}{k}*f[i-1][j]+frac{k-j+1}{k}*f[i-1][j-1]}$

${f[0][0]=1}$

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可能答案并不一定小于$K$所以我把$i$的上界设为了${k*10}$

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<cstdlib>
#include<cmath>
#include<cstring>
using namespace std;
#define maxn 2100
#define llg int
#define UP 10000000
#define eps (double)(1e-7)
#define yyj(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
llg n,T,m,l,r,mid,ans;
double f[maxn*5][maxn],p;
int main()
{
//    yyj("D");
    cin>>n>>T;
    //double N=n;
    f[0][0]=1;
    for (llg i=1;i<=n*10;i++)
        for (llg j=1;j<=n;j++)
            f[i][j]=((double)((double)j/n))*f[i-1][j]+(double)((double)(n-j+1)/n)*f[i-1][j-1];
    while (T--)
    {
        scanf("%lf",&p);
        p=p-eps; p/=2000; //llg ans;
        l=1; r=n*10;
        while (l<=r)
        {
            mid=(l+r)>>1;
            if (f[mid][n]>=p){ans=mid; r=mid-1;}else l=mid+1; 
        }
        printf("%d
",ans);
    }
    return 0;
}