【FZU

Yinyangshi is a famous RPG game on mobile phones.

Kim enjoys collecting cards in this game. Suppose there are n kinds of cards. If you want to get a new card, you need to pay W coins to draw a card. Each time you can only draw one card, all the cards appear randomly with same probability 1/n. Kim can get 1 coin each day. Suppose Kim has 0 coin and no cards on day 0. Every W days, Kim can draw a card with W coins. In this problem ,we define W=(n-1)!.

Now Kim wants to know the expected days he can collect all the n kinds of cards.

Input
The first line an integer T(1 ≤ T ≤ 10). There are T test cases.

The next T lines, each line an integer n. (1≤n≤3000)

Output
For each n, output the expected days to collect all the n kinds of cards, rounded to one decimal place.

Sample Input
4
1
2
5
9
Sample Output
1.0
3.0
274.0
1026576.0

借鉴理解的原文
我们假设已经有了a张卡片的话，如果我们想要得到第a+1张，我们抽中他的概率显然就是(n-a)/n，只要我们得到除了这a张卡片的任意一张即可。那么从a张卡片变成a+1张卡片的期望抽奖次数是多少呢？

我们考虑这样一个问题，假设我们抽奖中奖的概率为1/4，那么期望中奖的抽奖次数是多少呢？显然是1/（1/4）=4.

那么回归到当前问题，如果我们已经有了a张卡片，想要得到第a+1张，我们抽中一张新的卡片概率很显然是（n-a）/n，那么期望进行的次数就是n/（n-a）；

将这个问题分成n个独立事件去考虑，那么总期望E=各个单独事件的期望和；

那么Ans=Σn！/i（1<=i<=n）

我们预处理出来n！，这样求一个就可以在o(n)的时间范围。

import java.math.BigInteger;import java.math.BigDecimal;import java.util.Scanner;class Main{    public static void main(String[] arges){        Scanner cin=new Scanner(System.in);        BigInteger[]  k=new BigInteger[3000+10];        k[0] = BigInteger.ONE;        for(int i=1;i<=3000+2;i++){            k[i]=k[i-1].multiply(new BigInteger(""+i));        }        int t ;t=cin.nextInt();        while(t-->0){            BigInteger ans=BigInteger.ZERO;            int n;n=cin.nextInt();            for(int i=1;i<=n;i++){                ans=ans.add(k[n].divide(new BigInteger(""+i)));            }            System.out.println(ans+".0");        }    }}