题解 | 牛牛的数学作业

#include <stdio.h>
#include <stdlib.h>

int minum(int arr[], int n);
double fanch(int arr[], int n);

int main(int argc, char* argv[])
{
    int i,j,T;
    scanf("%d", &T);
    int n, arr[1000] = {0};
    for(j = 0;j < T;j++){
        scanf("%d", &n);
        for (i = 0; i < n; i++) {
            scanf("%d", &arr[i]);
        }
        printf("%d %.3lf\n", minum(arr, n), fanch(arr, n));
    }
    return 0;
}

int minum(int arr[], int n)
{
    int i;
    int min = arr[0], max = arr[0];
    for (i = 0; i < n; i++) {
        if (arr[i] > max) max = arr[i];
        if (arr[i] < min) min = arr[i];
    }
    return max - min;
}

double fanch(int arr[], int n)
{
    double dum = 0.0, sum = 0.0;
    for (int i = 0; i < n; i++) {
        sum += arr[i];
    }
    dum = sum / n;
    double num = 0;
    for (int i = 0; i < n; i++) {
        num += ((arr[i] - dum) * (arr[i] - dum));
    }
    return num / n;
}

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