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。

${Ans_i=Min\left \{ x|f[x][k]>\frac{p_i-\varepsilon }{2000} \right \}}$

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*][maxn],p;
 int main()
 {
 //    yyj("D");
     cin>>n>>T;
     //double N=n;
     f[][]=;
     for (llg i=;i<=n*;i++)
         for (llg j=;j<=n;j++)
             f[i][j]=((double)((double)j/n))*f[i-][j]+(double)((double)(n-j+)/n)*f[i-][j-];
     while (T--)
     {
         scanf("%lf",&p);
         p=p-eps; p/=; //llg ans;
         l=; r=n*;
         while (l<=r)
         {
             mid=(l+r)>>;
             if (f[mid][n]>=p){ans=mid; r=mid-;}else l=mid+;
         }
         printf("%d\n",ans);
     }
     return ;
 }