SPOJ1026 概率DP

Favorite Dice

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：

dp[i]:抛到i面的期望次数，dp[i]=dp[i-1]+n/(n-i+1)

代码：

 #include"bits/stdc++.h"

 #define db double
 #define ll long long
 #define vl vector<ll>
 #define ci(x) scanf("%d",&x)
 #define cd(x) scanf("%lf",&x)
 #define cl(x) scanf("%lld",&x)
 #define pi(x) printf("%d\n",x)
 #define pd(x) printf("%f\n",x)
 #define pl(x) printf("%lld\n",x)
 #define rep(i, n) for(int i=0;i<n;i++)
 using namespace std;
 const int N   = 1e6 + ;
 const int mod = 1e9 + ;
 const int MOD = ;
 const db  PI  = acos(-1.0);
 const db  eps = 1e-;
 const ll INF = 0x3fffffffffffffff;
 int t;
 db dp[N];
 void cal(int n)
 {
     dp[]=;
     for(int i=;i<=;i++) dp[i]=dp[i-]+1.0*n/(n-i+);//抛出新一面的次数
     printf("%.2f\n",dp[n]);
 }
 int main()
 {
     ci(t);
     while(t--){
         int n;
         ci(n);
         cal(n);
     }
     return ;
 }