防溢出求平均算法

输入i个x，求其平均数avg。

算法1：x+= (x-avg)/i;

算法2：sum+=x, avg=sum/i;

#include <iostream>
using namespace std;
int main()
{
	double x,avg = 0;
	for (int i = 1; scanf("%lf", &x) == 1; ++i)
	{
		avg += (x - avg) / i;
	}
	printf("%lf\n",avg);
	return 0;
}

证明

Avg[i] = Avg[i-1] + (X[i] - Avg[i-1])/i //很容易理解
= (X[i] + (i-1)Avg[i-1])/i //提取因子
===> iAvg[i] = X[i] + (i-1)*Avg[i-1] //移项i
===> n * Avg[n] = ∑X[i]{i,1,n} //累加结果
===> Avg[n] = ∑X[i] / n //done