【概率dp】Divide by Zero 2017 and Codeforces Round #399 (Div. 1 + Div. 2, combined) D. Jon and Orbs

直接暴力dp就行……f(i,j)表示前i天集齐j种类的可能性。不超过10000天就能满足要求。

#include<cstdio>
using namespace std;
#define EPS 1e-10
int K,q,p;
double f[10010][1010];
int main()
{
//	freopen("d.in","r",stdin);
	scanf("%d%d",&K,&q);
	f[0][0]=1.0;
	for(int i=1;i<=10000;++i)
	  for(int j=1;j<=K && j<=i;++j)
	    f[i][j]=f[i-1][j]*((double)j/(double)K)+f[i-1][j-1]*((double)(K-(j-1))/(double)K);
	for(int i=1;i<=q;++i)
	  {
	  	scanf("%d",&p);
	  	for(int j=1;j<=10000;++j)
	  	  if(f[j][K]-((double)p/2000.0)>(-EPS))
	  	    {
	  	      printf("%d\n",j);
	  	      break;
	  	    }
	  }
	return 0;
}