题解 | 牛牛的数学作业

#include <iostream>
using namespace std;

int main() {
   int T;
   cin>>T;
   while(T--){
    int n,min=0,max=0,sum=0;
    cin>>n;
    long long a[n];
    for(int i=0;i<n;i++){
        cin>>a[i];
        sum+=a[i];
        if(i==0){
            min=a[i];
            max=a[i];
        }else{
            if(a[i]<min) min=a[i];
            if(a[i]>max) max=a[i];
        }
    }
    double t=0.0f,avg=(double)sum/n;
    for(int i=0;i<n;i++) t+=(a[i]-avg)*(a[i]-avg);
    printf("%d %.3f\n",max-min,t/n);
   }
}