题解 | 牛牛的数学作业

#include <stdio.h>

int main()
{
    int n = 0,i = 0, j = 0,min = 0, max = 0,num = 0, m = 0;
    scanf("%d",&m);
    for( j = 0;j < m; j++)   
    {    
        long long sum1 = 0, sum2 = 0;
        scanf("%d",&n);
        for(i = 0; i < n; i++)
        {
            scanf("%d",&num);
            if(i == 0)
                max = min = num;
            else
            {
                if(num > max)
                    max = num;
                else if(num < min)
                    min = num;
            }
            sum1 += num;
            sum2 += num*num;
        }
        double med1 = (double)sum1/n;
        double med2 = (double)sum2/n;
        printf("%d %.3lf\n",(max-min),med2-med1*med1);
    }
    return 0;
}