求解概率的坑题

Description

We choose an integer K (K > 0). Then we generate N (N > 0) integers one by one randomly, each of them is in range [0, K - 1], and the appearing probabilities of each interger is the same,they are all 1/K.
Can you tell us the expectation of the number of distinct integers in that N integers?

Input

There is an integer T (1 <= T <= 200) in the first line, means there are T test cases in total.
For each test case, there are two integers K (1 <= K <= 1000), N (1 <= N <= 1000), which have the same meaning as above.

Output

For each test case, you should print the result in one line. You should keep the first 5 digits after the decimal point.

Sample Input

Sample Output

 1 #include <cstdio>
 2 double dp[1005];
 3 int main(){
 4     int n,t;
 5     double k;
 6     scanf("%d", &t);
 7     while (t--) {
 8         scanf("%lf %d", &k, &n);
 9         dp[1] = 1.0;
10         for (int i = 2; i <= n; ++i)
11             dp[i] = dp[i-1] + (k - dp[i-1])/k;
12         printf("%.5f\n", dp[n]);
13     }
14     return 0;
15 }

绝对是被坑了，一直都没有想到通过递推的的方式来求解，结果计算越来越复杂，后来就不能忍了，受不了，以后坚决不能犯这样的低级错误……