bzoj 1426 收集邮票

f[i]:当前已拥有i种邮票，还需要买的邮票数的期望值。g[i]:当前已拥有i种邮票，还需要的钱的期望值。每张邮票初始都是1元钱，每买一张邮票，还没购买的邮票每张都涨价1元。 f[i]=1+(n-i)/n*f[i+1]+i/n*f[i]
—>>f[i]=f[i+1]+n/(n-i)g[i]=1+(n-i)/n*(g[i+1]+f[i+1])+i/n*(g[i]+f[i])—>>g[i]=f[i+1]+i/(n-i)*f[i]+g[i+1]+n/(n-i);

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int n;
double f[10005],g[10005];
int main()
{
scanf("%d",&n);
for(int i=n-1;i>=0;i--)
f[i]=f[i+1]+n/(1.0*(n-i));
for(int i=n-1;i>=0;i--)
g[i]=n/(1.0*(n-i))+f[i+1]+g[i+1]+(i/(1.0*(n-i)))*f[i];
printf("%0.2lf\n",g[0]);
return 0;
}