Problem Analysis for Sid and his Dream Home

PROBLEM: Sid and his Dream Home

This problem can be solved as follows:
Instead of using the traditional brute force technique to solve this problem.
This problem can be solved easily using the method of long multiplication and using the concept of strings.
Both numbers are taken as strings, converted to numbers and multiplication is carried out and then output is again a string.

Problem setter's solution in C:

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#include<string.h>
#define MAX 10000

char * multiply(char [],char[]);
int main()
{
    char a[MAX];
    char b[MAX];
    char *c;
    int la,lb;
    int i,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",a);
        scanf("%s",b);
        c = multiply(a,b);
        printf("%s\n",c);
    }
    return 0;
}

char * multiply(char a[],char b[]){
    static char mul[MAX];
    char c[MAX];
    char temp[MAX];
    int la,lb;
    int i,j,k=0,x=0,y;
    long int r=0;
    long sum = 0;
    la=strlen(a)-1;
    lb=strlen(b)-1;

        for(i=0;i<=la;i++){
                a[i] = a[i] - 48;
        }

        for(i=0;i<=lb;i++){
                b[i] = b[i] - 48;
        }

    for(i=lb;i>=0;i--){
         r=0;
         for(j=la;j>=0;j--){
             temp[k++] = (b[i]*a[j] + r)%10;
             r = (b[i]*a[j]+r)/10;
         }
         temp[k++] = r;
         x++;
         for(y=0;y<x;y++){
             temp[k++] = 0;
         }
    }
    k=0;
    r=0;
    for(i=0;i<la+lb+2;i++){
         sum =0;
         y=0;
         for(j=1;j<=lb+1;j++){
             if(i <= la+j){
                 sum = sum + temp[y+i];
             }
             y += j + la + 1;
         }
    if(((((sum+r) %10)==0)&&i<(la+lb+1))||((((sum+r)%10)!=0)))
         c[k++] = (sum+r) %10;
         r = (sum+r)/10;
    }
    c[k] = r;
    j=0;
    for(i=k-1;i>=0;i--){
         mul[j++]=c[i] + 48;
    }
    mul[j]='\0';
    return mul;
}