题意翻译
一个n面的骰子，求期望掷几次能使得每一面都被掷到。
题目描述
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?
输入输出格式
输入格式：
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.
输出格式：
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.
输入输出样例
2
1
12
1.00
37.24 f [ i ]表示还剩i个面能把骰子的n面全扔一遍
对于扔一次骰子，有(n - i)/n能扔到剩下的面，有扔到之前扔过的面
f [ i ] = f [i + 1] * (( n - i ) / n ) + ( i / n) * f [ i ] + 1;
化简可得到f[i] = f [i + 1] + n/(n - i);(把f[ i ]挪到等式的一侧就可以了)
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作者：anonymity__
来源：CSDN
原文：https://blog.csdn.net/qq_42914224/article/details/83889581
版权声明：本文为博主原创文章，转载请附上博文链接！
代码是我自个的
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
int t;
double f[];
int main()
{
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
memset(f,,sizeof(f));
f[n] = ;
for(int i = n - ;i >= ;i--)
{
f[i] = f[i + ] + n / (n - (double)i);
}
printf("%0.2lf\n",f[]);
}
return ;
}