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.

Example

Input:
2
1
12

Output:
1.00
37.24

 

题意：

甩一个n面的骰子，问每一面都被甩到的次数期望是多少。

思路：

比较简单，公式：初始化dp[]=0;  dp[i]=i/n*dp[i]+(n-i)/n*dp[i+1]+1;  化简逆推即可。  求的是dp[0];

 

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cstring>
#include<memory>
using namespace std;
double dp[2000];
int main()
{
    int T,i,j,n;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n); dp[n]=0;
        for(i=n-1;i>=0;i--) dp[i]=(dp[i+1]*(n-i)/n+1)*n/(n-i);
        printf("%.2lf\n",dp[0]);
    } return 0;
}

 

版权声明：本文为hua-dong原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接和本声明。
本文链接：https://www.cnblogs.com/hua-dong/p/8042721.html