题解 | 牛牛的数学作业

#include <stdio.h>
#include <math.h>

int main()
{
	int T = 0;
	scanf("%d", &T);
	for (int i = 0; i < T; i++)
	{
		int num = 0;
		long long range = 0;//大整形
		double Var = 0;  //double更精准，且后有小数
		int sum = 0;
		scanf("%d", &num);
		long long Max = -1e18;
		long long Min = 1e18;    //要注意两个不要写反了

		long long arr[10000] = { 0 };
		for (int i = 0; i < num; i++)
		{
			int n = 0;
			scanf("%d", &n);
			arr[i] = n;
			sum += n;
			if (n > Max)
			{
				Max = n;
			}
			if (n < Min)
			{
				Min = n;
			}
		}

		range = Max - Min;
		double average = (double)sum / num;  //因为都要加入方差的计算，且有前一个隐性提升
		for (int i = 0; i < num; i++)
		{
			Var += pow((arr[i] - average),2)/num;
		}
		printf("%lld %.3f\n", range, Var);

	}
	return 0;
}