Problem solving in C language

moduel-18_5_1.c

/*
You are given an integer n or –n .If you are given n , print n to –n , if you are given –n, print –n to n.
See the sample output for more clarification.
Sample Input :                                      Sample Output :
5                                                   5 4 3 2 1 0 -1 -2 -3 -4 -5
-4                                                  -4 -3 -2 -1 0 1 2 3 4  
*/

#include <stdio.h>

int main() {
    // Write C code here
    int i, n;
    scanf("%d", &n);
    
    if(n<0) {
        for(i=n;i<=-(n);i++) {
            if(i<=0) {
                printf("%d ", i);
            }
            if(i>0) {
                printf("%d ", i);
            }
            
        }
    }
    
    if(n>0) {
        for(i=n;i>=-n;i--) {
            if(i>=0) {
                printf("%d ", i);
            }
            if(i<0) {
                printf("%d ", i);
            }
        }
    }

    return 0;
}



/*
Print the following pattern
For example n = 4

Sample Output
   
   ****
  ****
 ****
****


*/
#include <stdio.h>

int main() {
    // Write C code here
    int i, j, k, n;
    scanf("%d", &n);


    for(i=1; i<=n; i++) {

        for(j=n-i; j>0; j--) {
            printf(" ");
        }
        for(k=0; k<n; k++) {
            printf("*");
        }

        printf("\n");
    }

    return 0;
}


/*
You are given a positive integer n and after that a positive integer array of size n. Now print the sum of last digit
of the given n integers.
Sample Input :                      Sample Output :
5                                   Sum = 12
22 44 143 101 12
*/
#include <stdio.h>

int main() {
    // Write C code here
    int i, j, k, n;
    scanf("%d", &n);
    int arr[n];
    int sum = 0;

    
    for(i=0; i<n; i++) {
        scanf("%d", &arr[i]);
    }
    
    for(i=0;i<n;i++) {
        sum = sum + arr[i] % 10;
    }
    
    printf("Sum = %d", sum);
    
    return 0;
}