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：
1 #include"bits/stdc++.h" 2 3 #define db double 4 #define ll long long 5 #define vl vector<ll> 6 #define ci(x) scanf("%d",&x) 7 #define cd(x) scanf("%lf",&x) 8 #define cl(x) scanf("%lld",&x) 9 #define pi(x) printf("%d ",x) 10 #define pd(x) printf("%f ",x) 11 #define pl(x) printf("%lld ",x) 12 #define rep(i, n) for(int i=0;i<n;i++) 13 using namespace std; 14 const int N = 1e6 + 5; 15 const int mod = 1e9 + 7; 16 const int MOD = 998244353; 17 const db PI = acos(-1.0); 18 const db eps = 1e-10; 19 const ll INF = 0x3fffffffffffffff; 20 int t; 21 db dp[N]; 22 void cal(int n) 23 { 24 dp[0]=0; 25 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 26 printf("%.2f ",dp[n]); 27 } 28 int main() 29 { 30 ci(t); 31 while(t--){ 32 int n; 33 ci(n); 34 cal(n); 35 } 36 return 0; 37 }