题解 | 牛牛的数学作业

#include <iomanip>
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

int jicha(vector<int>& a)
{
    int n = a.size();
    int result = 0;;
    int max= a[0];
    int min= a[0];
    for(int i =0 ;i<n ; i++)
    {
        if(a[i]>max)
        {
            max = a[i];
        }
        if(a[i]<min)
        {
            min = a[i];
        }
    }
    result = max - min;
    return result;
}

double fangcha(vector<int>& a)
{
    int n = a.size();
    int sum =0;
    double result = 0;
    double sum1 = 0;
    double agang = 0;
    for(int i = 0; i< n ;i ++)
    {
        sum = a[i] + sum;
    }
    agang = sum*1.0/n;
    for(int i = 0; i< n ;i++)
    {
        sum1 = sum1 + pow(a[i]-agang,2);
    }
    result = sum1/n;
    return result ;
}

int main() {
    int T;
    cin >> T;
    int n=0 ;
    int num = 0;
    vector<int> a;
    for(int i = 0 ; i <T ; i++ )
    {
        cin >> n;
        for(int j = 0; j< n ; j++ )
        {
            cin >> num;
            a.push_back(num);
        }
        cout << jicha(a) << " " << fixed << setprecision(3) << fangcha(a) << endl;
        a.clear();
    }

    return 0;
}
// 64 位输出请用 printf("%lld")