[Lloj 1248] Dice (III)

G-dice (III) Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%LLD &%llu

Description

Given a dice with n sides and you had to find the expected number of times you had to throw this dice to see all I TS faces at least once. Assume that's the dice is fair, that's means when you throw the dice, the probability of occurring any face is equal.

For example, for a fair, and sided coin, the result is 3. Because When you first throw the coin and you'll definitely see a new face. If you throw the coin again, the chance of getting the opposite side are 0.5, and the chance of getting the same side is 0. 5. So, the result is

1 + (1 + 0.5 * (1 + 0.5 * ...))

= 2 + 0.5 + 0.52 + 0.53 + ...

= 2 + 1 = 3

Input

Input starts with an integer T (≤100), denoting the number of test cases.

Each case is starts with a line containing an integer n (1≤n≤105).

Output

For each case, print the case number and the expected number of the times that you had to throw the dice to see all it faces at L East once. Errors less than 10-6 'll be ignored.

Sample Input

5

1

2

3

6

100

Sample Output

Case 1:1

Case 2:3

Case 3:5.5

Case 4:14.7

Case 5:518.737751764

So set Dp[i] for the case has been thrown out of a different side of the situation to throw n different sides of the expected number, dp[n]=0, the answer is dp[0]
Then \ (dp[i]=dp[i]*\frac{i}{n}+dp[i+1]*\frac{n-i}{n}+1\)

Move term: \ (dp[i]=dp[i+1]+\frac{n}{n-i}\)

#include <iostream>#include<cstring>#include<cstdio>using namespacestd;#defineN 100010intMain () {intT,n,icase=1; DoubleDp[n]; scanf ("%d",&T);  while(t--) {scanf ("%d",&N); Dp[n]=0;  for(inti=n-1; i>=0; i--) dp[i]=dp[i+1]+n*1.0/(ni); printf ("Case %d:%.7f\n", icase++,dp[0]); }    return 0;}

[Lloj 1248] Dice (III)