F

首先是要理解概率求期望这种题目是用递归来解的！
大概规律就是：
完成事件A的期望是Ea，一次完成的概率是pa，那么
Ea = pa * Ea + (1 + Ea) * (1 - pa)
如果有好多种可能，比方说完成一个事件A的同时，也有诸如事件B的概率期望pb,Eb，事件C的概率期望pc,Ec等等等等，那么就分别写出来：
Ea = pa * Ea + (1 + Ea) * (~pa) + pb * (1 + Eb) + pc * (1 + Ec) + ...
其中~pa是除了这些已知的事件发生概率以外，的概率，也就是~pa = 1 - pa - pb - pc - ...
所以这个题的期望递推公式就是
E1 = 1/6 * E1 + 5/6 * (1 + E1)
E2 = 1/6 * E2 + 1/6 * (1 + E1) + 4/6 * (1 + E2)
E3 = 1/6 * E3 + 1/6 * (1 + E1) + 1/6 * (1 + E2) + 3/6 * (1 + E3)
E4 = 1/6 * E4 + 1/6 * (1 + E1) + 1/6 * (1 + E2) + 1/6 * (1 + E3) + 2/6 * (1 + E4)
E5 = 1/6 * E5 + 1/6 * (1 + E1) + 1/6 * (1 + E2) + 1/6 * (1 + E3) + 1/6 * (1 + E4) + 1/6 * (1 + E5)
E6 = 1/6 * E6 + 1/6 * (1 + E1) + 1/6 * (1 + E2) + 1/6 * (1 + E3) + 1/6 * (1 + E4) + 1/6 * (1 + E5)
E7 = 1/6 * (1 + E1) + 1/6 * (1 + E2) + 1/6 * (1 + E3) + 1/6 * (1 + E4) + 1/6 * (1 + E5) + 1/6 * (1 + E6)
E8 = 1/6 * (1 + E2) + 1/6 * (1 + E3) + 1/6 * (1 + E4) + 1/6 * (1 + E5) + 1/6 * (1 + E6) + 1/6 * (1 + E7)
规律很明显。。。

/****
	*COPYRIGHT NOTICE 
    *Copyright (c) 2014 
    *All rights reserved
    
	*@author    Shen
	*@name		F
	*@file		G:\My Source Code\比赛与日常练习\0608 - wuyiqi退役赛\F\F.cpp
	*@date		2014/6/8 20:21
	*/

//#pragma GCC optimize ("O2")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define _CRT_SECURE_NO_WARNINGS
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
using namespace std;

/*//STL
#include <map>
#include <vector>
#include <list>
#include <stack>
#include <deque>
#include <queue>
*/

/*//Computational Geometry
#include <complex>
#define x real()
#define y imag()
typedef complex<double> point;
*/

typedef long long int64;

const double p6 = 1.0 / 6.0;
double res[100005];

int t, q;

void init()
{
	int i = 1, j = 7;
	double tmp = 36.0;
	for (; j < 100005; i++, j++)
	{
		res[j] = p6 * (6.0 + tmp);
		tmp = tmp + res[j] - res[i];
	}
}
void solve()
{
	scanf("%d", &q);
	printf("%.2lf\n", res[q]);
}

int main()
{
	res[1] = res[2] = res[3] = res[4] = res[5] = res[6] = 6;
	init();
	scanf("%d", &t);
	while (t--) solve();
	return 0;
}