Favorite Dice（期望dp）

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 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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