题解 | 牛牛的数学作业

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

int main() {
    int T, n;
    cin >> T;
    while(T) {
        cin >> n;
        int a[n];
        cin >> a[0];
        float sum = a[0];
        int min = a[0], max = a[0];
        for(int i = 1; i < n; i++) {
            cin >> a[i];
            sum += a[i];
            if (min > a[i]) {
                min = a[i];
            }
            if (max < a[i]) {
                max = a[i];
            }
        }
        cout << max - min <<" ";
        double fc = 0.0;//使用double，防止溢出
        for(int i = 0; i < n; i++) {
            fc += ((double)(a[i] - (sum / n)) * (double)(a[i] - (sum / n)));
        }
        printf("%.3f\n", fc/n);
        //cout << fc / n << endl;
        T--;
    }
}
// 64 位输出请用 printf("%lld")