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Favorite Dice(期望dp)


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


题目大概:

给出一个n个面的筛子,问要使得每个面都朝上一次,需要投掷多少次。

思路:

期望dp一般逆推。

一般都是dp[i]表示已经出了i个面,距离n个面还差 的期望。

第i次投掷可以落在前i个面,概率为i/n,也可以落在另外的(n-i)个面,概率是(n-i)/n。

每次投掷都百分百的增加投掷次数1.

dp[i]=i/n*dp[i]+(n-i)/n*dp[i+1]+1.

化简的dp[i]=dp[i+1]+n/(n-i).

代码:

#include <bits/stdc++.h>

using namespace std;

double dp[1100];

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

 

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