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; }