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
",x)
#define pd(x) printf("%f
",x)
#define pl(x) printf("%lld
",x)
#define rep(i, n) for(int i=0;i<n;i++)
using namespace std;
const int N   = 1e6 + 5;
const int mod = 1e9 + 7;
const int MOD = 998244353;
const db  PI  = acos(-1.0);
const db  eps = 1e-10;
const ll INF = 0x3fffffffffffffff;
int t;
db dp[N];
void cal(int n)
{
    dp[0]=0;
    for(int i=1;i<=1000;i++) dp[i]=dp[i-1]+1.0*n/(n-i+1);//抛出新一面的次数
    printf("%.2f
",dp[n]);
}
int main()
{
    ci(t);
    while(t--){
        int n;
        ci(n);
        cal(n);
    }
    return 0;
}