如何读取一个数字数组，将该数组拆分为2，然后找到这两个单独数组的平均值？

Question

My program so far asks to input a list of n numbers, then it finds the average of that entire list. How can I split that list in half, and then find the average of the two split up arrays?

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

int main() {
    int n, i;
    float num[100], sum = 0.0, average;
    printf("Enter the total amount of numbers: ");
    scanf("%d", &n);

    for (i = 0; i < n; ++i) {
        scanf("%f", &num[i]);
        sum += num[i];
    }
    average = sum / n;
    printf("The average is = %.3f", average);
    return 0;
} 

Here is what my code does so far:

Enter the total amount of numbers:10  
10   
8  
9   
15  
12    
2  
3  
8  
7  
11  

The average is: 8.500

What I want it to do is:

Enter the total amount of numbers: 10  
10  
8  
9  
15  
12  
2  
3  
8  
7  
11  

The average of the first half of the array is: 10.8  
The average of the second half of the array is: 6.2

Answer 1

Simple changing to have array of sum and array of average - each having two members - would do (this way, you only need a single loop):

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

int main(){
    int n, i;
    float num[100], sum[2] = {0,0}, average[2]; //have arrays here
    printf("Enter the total amount of numbers: ");
    scanf("%d",&n);

   for(i=0; i<n; ++i)
   {
      scanf("%f",&num[i]);
      sum[(i*2)/n] +=num[i]; //note the trick here
   }
   average[0]=sum[0]/n;
   average[1]=sum[1]/n;
   printf("The average of the first half of the array is = %.3f\n", average[0]);
   printf("The average of the second half of the array is = %.3f", average[1]);
   return 0;
} 

Answer 2

There are several ways to sum and average 1/2 your array, all essentially do the same thing in different ways. You can sum/average as you read each 1/2 of the array, or you can wait and iterate of each half at the end to produce the same numbers. You could also separate your array into two separate arrays (which is quite inefficient, since you already have the data in memory, you need only address the values you want to segregate with proper indexing).

Before looking at the ways, there is one improvement you need to make in the way you use scanf . All of the scanf family members return the total number of successful conversions based on the format-string you provide. You need to check the return and validate that you have the number of successful conversions specified in your format-string . Always.

One approach to the split is to read all numbers and then iterate over 1/2 the array to get each of the sum and average values desired.

#include <stdio.h>

int main (void) {
    int n, i;
    float num[100], sum[2] = {0.0}, average[2] = {0.0};

    printf("Enter the total amount of numbers: ");
    if (scanf("%d", &n) != 1) {  /* validate your input */
        fprintf (stderr, "error: invalid input.\n");
        return 1;
    }

    for (i = 0; i < n; i++) {
        printf (" enter sum[%2d] ", i);
        if (scanf("%f", &num[i]) != 1) { /* validate input */
            fprintf (stderr, "error: invalid input.\n");
            return 1;
        }
    }

    /* sum/average 1st-half */
    for (i = 0; i < n/2; ++i)
        sum[0] += num[i];
    average[0] = sum[0] * 2 / n;

    /* sum/average 2nd-half */
    for (i = n/2; i < n; ++i)
        sum[1] += num[i];
    average[1] = sum[1] * 2 / n;

    printf("\nThe 1st-half average is = %.3f\n", average[0]);
    printf("The 2nd-half average is = %.3f\n\n", average[1]);

    return 0;
}

( note: you do not need to use arrays for sum or average if you just move your printf statements immediately after each calculation).

The second approach uses two loops. The first from i = 0; i < n/2 i = 0; i < n/2 , the second from i = n/2; i < n i = n/2; i < n . Again, no need for sum or average arrays if you move the printf to immediately following the sum and average calculations:

#include <stdio.h>

int main (void) {
    int n, i;
    float num[100], sum[2] = {0.0}, average[2] = {0.0};

    printf("Enter the total amount of numbers: ");
    if (scanf("%d", &n) != 1) {  /* validate your input */
        fprintf (stderr, "error: invalid input.\n");
        return 1;
    }

    /* read/sum/average 1st-half */
    for (i = 0; i < n/2; ++i) {
        printf (" enter sum[%2d] ", i);
        if (scanf("%f", &num[i]) != 1) {  /* validate input */
            fprintf (stderr, "error: invalid input.\n");
            return 1;
        }
        sum[0] += num[i];
    }
    average[0] = sum[0] * 2/ n;

    /* read/sum/average 2nd-half */
    for (i = n/2; i < n; ++i) {
        printf (" enter sum[%2d] ", i);
        if (scanf("%f", &num[i]) != 1) {  /* validate input */
            fprintf (stderr, "error: invalid input.\n");
            return 1;
        }
        sum[1] += num[i];
    }
    average[1] = sum[1] * 2 / n;

    printf("\nThe 1st-half average is = %.3f\n", average[0]);
    printf("The 2nd-half average is = %.3f\n\n", average[1]);

    return 0;
} 

Output

Either way the resulting output is the same:

$ ./bin/array_split
Enter the total amount of numbers: 8
 enter sum[ 0] 1
 enter sum[ 1] 2
 enter sum[ 2] 3
 enter sum[ 3] 4
 enter sum[ 4] 5
 enter sum[ 5] 6
 enter sum[ 6] 7
 enter sum[ 7] 8

The 1st-half average is = 2.500
The 2nd-half average is = 6.500

Or, for your example numbers:

$ ./bin/array_split
Enter the total amount of numbers: 10
 enter sum[ 0] 10
 enter sum[ 1] 8
 enter sum[ 2] 9
 enter sum[ 3] 15
 enter sum[ 4] 12
 enter sum[ 5] 2
 enter sum[ 6] 3
 enter sum[ 7] 8
 enter sum[ 8] 7
 enter sum[ 9] 11

The 1st-half average is = 10.800
The 2nd-half average is = 6.200