Light OJ 1248-dice (III) probability DP

N-sided dice ask for the expectation that each face is thrown at least once

Set Dp[i] for the expectation that I have been thrown on different faces dp[n] = 0 for dp[0]

Because Dp[i] is also required to throw a different face each time it may throw the face that has been thrown or has not been thrown over the face of the situation 2

So dp[i] = (i/n) *dp[i] + (n-i)/n*dp[i+1] +1 equals 2 sides have dp[i]

Move term dp[i] = dp[i+1]+n/(n-i)

#include <cstdio>
#include <cstring>
#define IMAX 100005
double Dp[imax];
int n;
int main ()
{
	int t,tt;
	scanf ("%d", &t);
	for (tt=1;tt<=t;tt++)
	{
		scanf ("%d", &n);
		dp[n]=0;
		for (int i=n-1;i>=0;i--)
		{
			dp[i]=dp[i+1]+ (double) n/(n-i);
		}
		printf ("Case%d:%.8lf\n", tt,dp[0]);
	}
	return 0;
}