C：在大数组中查找和计算重复项

Question

我参加了C编程课程，在我们的家庭作业中，我们的教授要求我们编写一个程序，生成一个千位数组并存储随机数（范围从5到15）。然后，它需要显示每个数字出现的次数（以及其他计算）。我已经设法做到这一切，但效率非常低。正如您在我的代码中所看到的，我已经使用了一堆 if 语句，我可能已经使用了 for 循环。能帮我改进一下这段代码吗？我无法使用指针或重新排列数组。谢谢！

main()

{     的printf（＆＃34; -------------------------------------------- ---------------------------- \ n＆＃34）;     printf（＆＃34;这个程序生成一个带有随机值的一千个数字数组\ n＆＃34;             ＆＃34;范围从5到15.它还计算所有这些数字的总和，\ n＆＃34;             ＆＃34;他们的平均值，并计算每个数字。\ n＆＃34;）;     的printf（＆＃34; -------------------------------------------- ---------------------------- \ n \ n＆＃34）;

int i,count5=0,count6=0, count7=0, count8=0, count9=0,
count10=0, count11=0, count12=0, count13=0, count14=0, count15=0, sum = 0;
float avg;

int a[1000] = {0};
srand(time(NULL));

for(i=0; i<=999; i++)
{
    a[i] = (rand()%11)+5;

    if(a[i] == 5)
    {
        count5 = count5++;
    }
    if(a[i] == 6)
    {
        count6 = count6++;
    }
    if(a[i] == 7)
    {
        count7 = count7++;
    }
    if(a[i] == 8)
    {
        count8 = count8++;
    }
    if(a[i] == 9)
    {
        count9 = count9++;
    }
    if(a[i] == 10)
    {
        count10 = count10++;
    }
    if(a[i] == 11)
    {
        count11 = count11++;
    }
    if(a[i] == 12)
    {
        count12 = count12++;
    }
    if(a[i] == 13)
    {
        count13 = count13++;
    }
    if(a[i] == 14)
    {
        count14 = count14++;
    }
    if(a[i] == 15)
    {
        count15 = count15++;
    }

    sum = sum + a[i];

}

printf("Sum of entire array = %d\n",sum);

avg = sum/1000.0;

printf("Average of values in array = %.2f\n\n",avg);

printf("Fives:        %d\n",count5);
printf("Sixes:        %d\n",count6);
printf("Sevens:       %d\n",count7);
printf("Eights:       %d\n",count8);
printf("Nines:        %d\n",count9);
printf("Tens:         %d\n",count10);
printf("Elevens:      %d\n",count11);
printf("Twelves:      %d\n",count12);
printf("Thirteens:    %d\n",count13);
printf("Fourteens:    %d\n",count14);
printf("Fifteens:     %d\n\n",count15);

}