SP1026 FAVDICE - Favorite Dice 题解

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Seeing that all the solutions are backwards, but none of them are forward-thinking, let’s talk about it here.

Setting $f_i$Indicates that ii has been rolledThe expected number of times when$i$ different faces are
initialized$f_0=0$
Then, the transfer equation is obviously$f_i=\frac{n-i+1}{n}\cdot f_{i-1}+\frac{n-i}{n}\cdot f_i+1$
化简后得到 $f_i=\frac{f_{i-1}\times (n-i+1)+n}{n-i}$
It can be found that if this is recursive to $When\; n$ , the denominator is$0$ , so this method is not feasible

So, we try to use backwards

Setting $f_i$Indicates that ii has been rolled$After\; i$ different faces, how many more times are expected to reach$n$ faces. Initialize$f_n=0$
转移方程 $f_i=\frac{n-i}{n}\cdot f_{i+1}+\frac{i}{n}\cdot f_i+1$
化简后得 $f_i=f-i+1+\frac{n}{n-i}$
In this way, there will be no denominator of $In$ the case of$0$ , the answer is$f_0$

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int Maxn=1010;
double f[Maxn];
int n,T;
int main()
{
    
    
	scanf("%d",&T);
	while(T--)
	{
    
    
		scanf("%d",&n);
		f[n]=0.0;
		for(int i=n-1;i>=0;--i)
		f[i]=f[i+1]+(n*1.0)/((n-i)*1.0);
		printf("%.2lf\n",f[0]);
	}
	return 0;
}