题解 | 牛牛的数学作业

#include <bits/stdc++.h>
#include <ostream>
using namespace std;
int qiujicha(int t, int a[]) {
    int max, min;
    max = a[0], min = a[0];
    for (int i = 0; i < t; i++) {
        if (a[i] < min)min = a[i];
        if (a[i] > max)max = a[i];
    }
    return max - min;
}
double qiufangcha(int t, int a[]) {
    double jun = 0.000, fangcha = 0.000;
    for (int i = 0; i < t; i++) {
        jun += a[i];
    }
    jun = jun / t;
    for (int i = 0; i < t; i++) {
        fangcha += (a[i] - jun) * (a[i] - jun);
    }
    fangcha = fangcha / t;
    return fangcha;
}
int T, n;
int main() {
    cin >> T;
    while (T--) {
        int a[100000] = {0};
        cin >> n;
        for (int i = 0; i < n; i++)cin >> a[i];
        cout << qiujicha(n, a) << ' ';
        cout << fixed <<setprecision(3)<<qiufangcha(n, a) << endl;
    }
}