spoj1026Favorite Dice

Posted 2021-01-18 rax-

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

 1 #include<cstdio>
 2 #include<cmath>
 3 #include<algorithm>
 4 #include<cstring>
 5 using namespace std;
 6 int t;
 7 double f[1005];
 8 int main()
 9 {
10     scanf("%d",&t);
11     while(t--)
12     {
13         int n;
14         scanf("%d",&n);
15         memset(f,0,sizeof(f));
16         f[n] = 0;
17         for(int i = n - 1;i >= 0;i--)
18         {
19             f[i] = f[i + 1] + n / (n - (double)i);
20         }
21         printf("%0.2lf
",f[0]);
22      } 
23      return 0;
24 }