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  • + 0 comments

    Working with dynamic arrays in C really highlights the importance of managing memory efficiently, much like how careful planning is crucial in plumbing marketing services to achieve consistent results.

  • + 2 comments
    #include <stdio.h>
    #include <string.h>
    #include <math.h>
    #include <stdlib.h>
    
    int main() {
    
        /* Enter your code here. Read input from STDIN. Print output to STDOUT */ 
        int n;
        scanf("%d",&n);
        
        int *arr = (int*)malloc(n * sizeof(int));
        int sum = 0;
        for (int i=0; i<n ; i++)
        {
            scanf("%d",&arr[i]);
            sum+=arr[i];
        }   
        printf("%d",sum);
        return 0;
    }
    
  • + 0 comments
    int main() {
        int n;
        scanf("%d", &n);
        int array[n], sum=0;
        for(int i=0; i<n; i++) {
            scanf("%d", &array[i]);
            sum += array[i];
        }  
        printf("%d", sum);
        return 0;
    }
    
  • + 0 comments

    int main() {

    /* Enter your code here. Read input from STDIN. Print output to STDOUT */
    int arr[1000] = {0},n, sum = 0;
    scanf("%d",&n);
    for(int i = 1;i <= n;i++){
        scanf("%d",&arr[i]);
    }
    for(int i = 1;i <= n;i++){
        sum = sum + arr[i];
    }
    printf("%d",sum);
    return 0;
    

    }

  • + 0 comments

    For C

    #include<stdio.h>
    #include<stdlib.h>
    
    int main()
    {
        int n;
        scanf("%d", &n);
        
        int *arr;
        arr = (int *) malloc(n * sizeof(int));
        for(int i=0; i<n; i++)
        {
            scanf("%d", &arr[i]);
        }
        
        int sum = 0;
        for(int j=0; j<n; j++)
        {
            sum = sum + arr[j];
        }
        
        free(arr);
        
        printf("%d\n", sum);
        
        return 0;
    }