CSU-ACM暑假集训基础组训练赛(5-1) B

文章探讨了一个拥有n面的魔法骰子，如何通过期望投掷次数确保每个面至少出现一次。通过数学计算得出，对于特定的n值，所需投掷次数的期望值为一定数值。理解此概念有助于深入掌握概率论与期望值的计算。

B - Problem B

Time Limit:2000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

Description

BuggyD loves to carry his favorite die around. Perhaps you wonder why it's his favorite? Well, his die is magical and can be transformed into an N-sided unbiased die with the push of a button. Now BuggyD wants to learn more about his die, so he raises a question:

What is the expected number of throws of his die while it has N sides so that each number is rolled at least once?

Input

The first line of the input contains an integer t, the number of test cases. t test cases follow.

Each test case consists of a single line containing a single integer N (1 <= N <= 1000) - the number of sides on BuggyD's die.

Output

For each test case, print one line containing the expected number of times BuggyD needs to throw his N-sided die so that each number appears at least once. The expected number must be accurate to 2 decimal digits.

Sample Input

Input:
2
1
12

Output:
1.00
37.24

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题目大意：
一个有n面的色子抛掷多少次能使所有面都能被抛到过，求期望值
总面数为n，当已经抛到过 i 个不同面时，我们抛出下一个不同面的概率为 (n-i)/n，那么抛的次数为 n/(n-i)，
将所有抛出下个面的次数累加起来就好了

简单的模拟题，前提是读懂题意。代码：

#include <iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
int t,i,n;
scanf("%d",&t);
while(t--)
{
double s=0.0;
scanf("%d",&n);
for(i=0;i<n;i++)
s+=double(n)/double(n-i);
printf("%.2lf\n",s);
}
return 0;
}

总结：好好学习，天天向上。聪哥说：不会只能怪高中老师了~~~