spoj1026Favorite Dice

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

输入样例#1： 复制

2
1
12

输出样例#1： 复制

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[1005];
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        memset(f,0,sizeof(f));
        f[n] = 0;
        for(int i = n - 1;i >= 0;i--)
        {
            f[i] = f[i + 1] + n / (n - (double)i);
        }
        printf("%0.2lf
",f[0]);
     } 
     return 0;
}