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
5 1 1 2 2 3 2 3 4 5 3
Sample Output
1.00000 1.50000 1.66667 2.40741 2.44000
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 ", 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……