使用变量、数组和函数进行准确的求和平均值计算

问：

我是编程新手，但我的代码包括一个数组，一直给出 0 作为数组的总和。我的代码是：

/*y = (1 / (p * sqrt(2π))) * e^((-1/2) * ((a - x) / p)^2)

在这个等式中：

我试图从一侧（总和和平均值）解决它，但事情对我来说并不顺利。

/* 
a represents the grade or score achieved by a student.
y represents the probability density of obtaining that grade.
x represents the mean (average) grade.
p represents the standard deviation, which measures the spread or variability of the grades.*/
#include <stdio.h>
#include <math.h>
float curve(int a, float p, float x);
int main()
{
    int i = 0, grade[100], sum = 0, n = 0, success = 0, fail = 0, low = 100, high = 0;
    float avg, sd;
    printf("ente the grade:\n");
    scanf("%d", &grade[i]);
    while (grade[i] != -1)
    { // grade[i] >=60? success++:fail++;
        // if (grade[i]>high)
        // high=grade[i];
        // if (grade[i]<low)
        // low=grade[i];
        n++;
        i++;
        sum = sum + (grade[i]);
        printf("ente the grade:\n");
        scanf("%d", &grade[i]);
    }
    for (i = 0; i < n; i++)
    {
        printf("%d\n", grade[i] + 10);
    }
    avg = sum / n;
    // sd=pow(garde)
    printf("the sum of the class is %d\n", sum);
    printf("the average of the class is %f\n", avg);
    printf("the number of the student who passed the exam is %d\n", success);
    printf("the number of the student who failed the exam is %d\n", fail);
    printf("the highest mark in the class is %d\n", high);
    printf("the lowest mark in the class is %d\n", low);
    if (avg < 50)
    {
        printf("raise suggested\n");
        printf("the new average of the class is %f\n", (avg + 10));
    }
    else if (avg > 50 && avg < 60)
    {
        printf("curve suggested\n");
    }
    return 0;
}
float curve(int a, float p, float x)
{
    float y;
    y = (1 / (p * sqrt(2 * (3.14159)))) * pow((2.718), ((-1 / 2) * ((a - x) / p) * ((a - x) / p)));
    return y;
}