C语言中的平均值，最大值和最小值

So I'm coding in C, and I need to come up with code that will take n numbers from the user, and find their minimum, maximum, average, and sum of squares for for their values. So far I have the average and sum of squares portion, but the minimum and maximum is biting me.

我用C编写代码，我需要从用户那里得到n个数字，然后找到它们的最小，最大值，平均值，和它们的值的平方和。到目前为止，我有平方部分的平均值和总和，但是最小和最大值让我很不爽。

Keep in mind I'm at a very rudimentary level, and I have not reached arrays yet. All I know are logical operators, functions, loops, and the use of the stdlib.h, math.h, and stdio.h libraries.

请记住，我还处于非常初级的阶段，而且还没有到达数组。我只知道逻辑运算符、函数、循环和stdlib的使用。h,数学。h,头。h库。

This is what I have so far. The average function gave me a lot of problems when I tried to put float and double during compiling, so multiply it by a 1.0 fixed that. I have everything, just the minimum and maximum. I keep getting the last entry as my maximum, and a 0 for my minimum.

这是我目前所拥有的。当我试图在编译过程中放入浮点和双精度浮点数时，平均函数给我带来了很多问题，所以乘以1。0修正了这个问题。我有所有东西，只有最小值和最大值。最后一项作为最大值，最小值为0。

#include<stdio.h>
int main()
{
    float average;
    int i, n, count=0, sum=0, squaresum=0, num, min, max;


    printf("Please enter the number of numbers you wish to evaluate\n");
    scanf_s("%d",&n);

    printf("Please enter %d numbers\n",n);

    while(count<n)
    {
        min=0;
        max=0;

            if(num>max)
               max=num;
            if(num<min)
               min=num;

            scanf_s("%d",&num);

        sum = sum+num;
        squaresum = squaresum + (num*num);

        count++;
    }
        average = 1.0*sum/n;

    printf("Your average is %.2f\n",average);
    printf("The sum of your squares is %d\n",squaresum);    



    printf("Your maximum number is %d\n",max);
    printf("Your minimum number is %d\n",min);

return(0);
}

7 个解决方案

#1

Your algorithm is not quite right. Below is the correct implementation:

你的算法不太对。以下是正确的实施:

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

int main(void)
{
    float average;
    int n, num, count = 0, sum = 0, squaresum = 0;
    int min = INT_MAX, max = INT_MIN;
    bool gotAnswer = false;

    /* Don't Let User Enter Wrong Input */
    while(!gotAnswer)
    {
        printf("Please enter the number of numbers you wish to evaluate: ");
        if(scanf_s("%d", &n) != 1)
        {
            /* User Entered Wrong Input; Clean Up stdin Stream*/
            while(getchar() != '\n')
            {
                 continue;
            }
        }
        else
        {
            /* User Input Was Good */
            gotAnswer = true;
        }
    }

    /* Clear stdin Stream Just In Case */
    while(getchar() != '\n')
        continue;

    while(count < n)
    {
        /* Don't Let User Enter Wrong Input */
        gotAnswer = false;
        printf("Enter number %d: ", count + 1);
        if(scanf_s("%d", &num) != 1)
        {
            /* User Entered Wrong Input; Clean Up stdin Stream */
            while(getchar() != '\n')
                continue;

            /* Let User Try Again */
            continue;
        }
        else
        {
            /* User Input Was Correct */
            gotAnswer = true;

            /* Clear stdin Stream Just In Case */
            while(getchar() != '\n')
                continue;
        }

        if(num > max)
            max = num;
        if(num < min)
            min = num;

        sum += num;
        squaresum += num * num;
        count++;
    }

    average = 1.0 * sum / n;

    printf("Your average is %.2f\n", average);
    printf("The sum of your squares is %d\n", squaresum);    
    printf("Your maximum number is %d\n", max);
    printf("Your minimum number is %d\n", min);

    system("pause");
    return 0;
}

I've added error checking and recovery. Please ask if you have any questions about the logic.

我添加了错误检查和恢复。如果对逻辑有任何疑问，请提出来。

#2

The way your code is currently written, min has to start out at a high value (not 0), or the code won't work. The best value to choose is the maximum possible value for an int.

您的代码当前编写的方式，min必须以高值(而不是0)开始，否则代码将无法工作。选择的最佳值是int的最大可能值。

You should also consider whether or not you want to reset these variable each time through the loop.

您还应该考虑是否希望每次通过循环重置这些变量。

#3

Enter the first num outside the loop and assign that to max min

在循环外部输入第一个num并将其分配给最大最小值

scanf("%d",&num);
max = min = num;  

Change your while loop to infinite loop

改变你的while循环到无限循环。

while(1) {...} 

and now check for the condition that whether your counter count is equal to n is or not to break out from the infinite loop

现在检查计数是否等于n的条件是否从无限循环中跳出

if(count == n)
    break;  

Full code after modification:

修改后完整代码:

#include<stdio.h>
int main()
{
    float average;
    int i, n, count=0, sum=0, squaresum=0, num, min, max;

    printf("Please enter the number of numbers you wish to evaluate\n");
    scanf_s("%d",&n);

   printf("Please enter %d numbers\n",n);

   scanf_s("%d",&num);
   max = min = num;

   while(1)
   {
        if(num>max)
           max=num;
        if(num<min)
           min=num;

     sum = sum+num;
     squaresum = squaresum + (num*num);

     count++;
     if(count == n)
        break;
     scanf_s("%d",&num);

   }
    average = 1.0*sum/n;

printf("Your average is %.2f\n",average);
printf("The sum of your squares is %d\n",squaresum);



printf("Your maximum number is %d\n",max);
printf("Your minimum number is %d\n",min);

return(0);
}

#4

Assume your first number in the list as the minimum and maximum. Compare every next character with the current minimum and the current maximum and update accordingly.

假设列表中的第一个数字为最小和最大值。将每个下一个字符与当前最小字符和当前最大字符进行比较，并相应地更新。

#5

your while loop should look like

你的while循环应该是这样的

   min=3;
   max=0;
while(count<n)
    { 

     scanf("%d",&num);
    if(num>max)
    max=num;
    if(num<min)
    min=num;

    sum = sum+num;
    squaresum = squaresum + (num*num);

    count++;
}

And I agree with Robert Harvey♦.. You must set min

和我同意罗伯特·哈维♦. .必须设置最小

#6

Add a boolean, moved giving the values min, max 0 are the start of loop

添加一个布尔值，移动给值最小，最大值0是循环的开始

#include<stdio.h>
int main()
{
    float average;
    int i, n, count=0, sum=0, squaresum=0, num, min, max;

    bool first = true;
    printf("Please enter the number of numbers you wish to evaluate\n");
    scanf_s("%d",&n);

    printf("Please enter %d numbers\n",n);
        min=0;
        max=0;

    while(count<n)
    {
            scanf_s("%d",&num);

        if (first) {
           first = false;
           min = max = num;
        }
            if(num>max)
               max=num;
            if(num<min)
               min=num;


        sum = sum+num;
        squaresum = squaresum + (num*num);

        count++;
    }
        average = 1.0*sum/n;

    printf("Your average is %.2f\n",average);
    printf("The sum of your squares is %d\n",squaresum);    



    printf("Your maximum number is %d\n",max);
    printf("Your minimum number is %d\n",min);

return(0);
}

Should also consider to check the return value of scanf

还应该考虑检查scanf的返回值吗

#7

There're some issues in your code:

你的代码中有一些问题:

1. Where num is read? You should do it before min and max
2. num在哪里读书?你应该在最小和最大之前做
3. When while loop executes first time you should just assign num to max and min.
4. 当while循环第一次执行时，应该将num分配给max和min。

Something like that:

这样的:

  int min = 0;
  int max = 0;

  // If your compiler supports C99 standard you can put
  // bool first_time = true;
  int first_time = 1;

  while (count < n) {
    printf("Please, enter the next number\n");
    scanf_s("%d", &num); 

    // If your compiler supports C99 you can put it easier:
    // if (first_time) { 
    if (first_time == 1) {
      first_time = 0;
      max = num;
      min = num;
    }
    else {
      if(num > max)
        max = num;

      if(num < min)
        min = num;
    }
  ...