题解 | 牛牛的数学作业

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

int main() {
    int T; cin>>T;
    int n; 
    int a[100000]={0};
    double sum;
    double fangcha;
    double p=0.0;
    for (int j=0;j<T;j++){
        cin>>n;
        double t=n*1.0;
        for (int i=0;i<n;i++){
        cin>>a[i];
        }
        int max=a[0];
        int min=a[0];
        sum=0.0;
        for (int x=0;x<n;x++){
            sum+=a[x];
        }
        sum=sum/t;
        p=0;
        for (int x=0;x<n;x++){
        if(max<a[x]){
            max=a[x];
        }
        if(min>a[x]){
            min=a[x];
        }
        p+=(sum-a[x])*(sum-a[x]);
        
        fangcha=p/t;
        }

        cout<<max-min<<" "; 
        printf("%.3f\n",fangcha);
}
    return 0;
}
// 64 位输出请用 printf("%lld")