1248 - Dice (III)

1248 - Dice (III)

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Time Limit: 1 second(s)

Memory Limit: 32 MB

Given a dice with n sides, you have to find the expected number of times you have to throw that dice to see all its faces at least once. Assume that the dice is fair, that means when you throw the dice, the probability of occurring any face is equal.

For example, for a fair two sided coin, the result is 3. Because when you first throw the coin, you will definitely see a new face. If you throw the coin again, the chance of getting the opposite side is 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 starts with a line containing an integer n (1 ≤ n ≤ 10⁵).

Output

For each case, print the case number and the expected number of times you have to throw the dice to see all its faces at least once. Errors less than 10^-6 will be ignored.

Sample Input	Output for Sample Input
5 1 2 3 6 100	Case 1: 1 Case 2: 3 Case 3: 5.5 Case 4: 14.7 Case 5: 518.7377517640

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Problem Setter: Jane Alam Jan

http://res.tongyi.com/resources/old_article/student/5608.html

几何分布的期望；

投掷出第一个未出现的点数的概率为p1=n/n = 1。第二个未出现的点数第一次出现的概率为 p2=(n - 1) / n。第i个未出现的点数第一次出现的概率为pi=(n - (i-1)) / n。然后当我们取了第i个点，那么要去取第i+1个点，那么这个点的概率为pi=(n - (i)) / n，那么这时这个点取次数的期望就是满足几何分布的，那么这个点的期望求出来表示，要经过多少此取，才能让第i+1个点出现的期望，总的期望就是这n个点第一次出现的期望之和。

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<stdlib.h>
 5 #include<queue>
 6 #include<string.h>
 7 #include<map>
 8 #include<vector>
 9 using namespace std;
10 typedef long long LL;
11 int main(void)
12 {
13         int n;
14         scanf("%d",&n);
15         int __ca = 0;
16         while(n--)
17         {
18                 __ca++;
19                 int x;
20                 scanf("%d",&x);
21                 int i,j;
22                 double sum = x;
23                 double ac = 0;
24                  for(i = 1;i <= x;i++)
25                 {
26                     ac = ac + 1.0/(double)i;
27                 }//printf("%lf\n",ac);
28                 printf("Case %d: %.10f\n",__ca,ac*sum);
29         }
30         return 0;
31 }