【51Nod 1756】【算法马拉松 23】谷歌的恐龙

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1765
设答案为(X)。
则$$X=frac{m}{n} imesfrac{sum a_i}{m}+frac{n-m}{n}left(frac{frac{(n-1)n}{2}-sum a_i}{n-m}+X ight)$$

[X=frac{n(n-1)}{2m} ]

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int n, m;

int main() {
	scanf("%d%d", &n, &m);
	printf("%.6lf
", 1.0 * n * (n - 1) / 2 / m);
	return 0;
}