杭电acm2086

可以推导出:A1 = n / (n + 1) * A0 + 1 / (n + 1)* An+1 - 2*n / (n + 1)* C1 - (2*n - 2)/(n + 1)* C2 - ........2 / (n + 1)* Cn-1。

#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
	int n;
	double Ao, An;
	double num[3005];
	while(scanf("%d", &n) != EOF)
	{
		scanf("%lf%lf", &Ao, &An);
		for(int i = 1; i <= n; i++)
			scanf("%lf", &num[i]);
		double ans = double(n) / double(n + 1) * Ao + 1.0 / (n + 1) * An;
		double k = double(n) * 2.0;
		for(int j = 1; j <= n; j++)
		{
			ans -= k / (n + 1) * num[j];
			k -= 2.0;
		}
		printf("%.2lf\n", ans);
	}
	return 0;
}


你可能感兴趣的:(ACM,HDU)