题解 | 牛牛的数学作业

#include <iostream>
#include <vector>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;

int main() {
    int t;
    scanf("%d", &t);

    while (t--) {
        long long n;
        scanf("%lld", &n);

        vector<long long> a(n);
        long long sum1 = 0;
        long long min_val = 1e18;   // 初始最小值设为极大值
        long long max_val = -1;     // 初始最大值设为极小值

        for (int i = 0; i < n; i++) {
            scanf("%lld", &a[i]);
            sum1 += a[i];
            if (a[i] < min_val) min_val = a[i];
            if (a[i] > max_val) max_val = a[i];
        }

        double ave = (double)sum1 / n;
        double sum2 = 0.0;
        for (int i = 0; i < n; i++) {
            double diff = (double)a[i] - ave;
            sum2 += diff * diff;
        }
        double fang = sum2 / n;  // 总体方差（非样本方差）

        printf("%lld %.3f\n", max_val - min_val, fang);
    }

    return 0;
}