I tried my level best to get it done. But I would like to have a better solution in all terms.

Given 2 huge numbers as seperate digits, store them in array and process them and calculate the sum of 2 numbers and store the result in an array and print the sum.

Input:

Number of digits:12

9 2 8 1 3 5 6 7 3 1 1 6
Number of digits:9
7 8 4 6 2 1 9 9 7

Output:

9 2 8 9 2 0 2 9 5 1 1 3

    #include<stdio.h>
    #include<stdlib.h>
    int main()
    {
        int num1[]={9,9,9,9,9,9,9,9,9};
        int num2[]={1,1,1};
        int num1_n=sizeof(num1)/sizeof(int);
        int num2_n=sizeof(num2)/sizeof(int);
        int z=num1_n+num2_n;
        
        int carry=0;
        int digi=0;
        int i=0;
        int j=0;
        int *num3;
        if(num1_n>num2_n)
        {
            num3=(int*)malloc(sizeof(int)*(num1_n+1));
            for(i=0;i<num1_n+1;i++)
            {
                num3[i]=0;
            }
            for(i=num1_n-1,j=num2_n-1;i>=0 || j>=0 ;i--)
            {
                num3[i+1]=num3[i+1]+num1[i]+num2[j];
                printf("i %d j %d num3[i] %d num1[i] %d num2[i] %d\n",i,j,num3[i],num1[i],num2[j]);
                if(num3[i+1]>9)
                {
                    carry=1;
                    num3[i+1]=num3[i+1]%10;
                }
                if(carry==1)
                {
                    num3[i]=1;
                    carry=0;
                }
                if(j>=0)
                {
                    j--;
                }
            }
        }
        for(i=0;i<num1_n+1;i++)
        {
            printf("%d ",num3[i]);
        }
        return 0;
    }