Add Sub Multiply very large number stored as string

(Flipkart Machine coding round)
Very large number like 2994320121 and 125346232 is given in character string.
You need to add, subtract and multiply two numbers and store number in character array only.

Logic
1) First way to create a "long long"number from String and add/multiply/subtract.
    And convert this result in to character array.

2) keep numbers in Character array only and try to add/sub/multiply index wise.
See code for Add+Multiply
You can create for Subtract.

Thanks to Dhaval for suggesting this approach and Article

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

char * addition(char a[],char b[]){

    static char add[MAX];
    char c[MAX];
    int temp;
    int la,lb;
    int i,j,k=0,x=0,y;
    long int car=0;
    long sum = 0;
    la=strlen(a)-1;
    lb=strlen(b)-1;
    //printf("\n la=%d lb=%d ",la,lb);
    for(i=0;i<=la;i++){ a[i] = a[i] - 48; }
    for(i=0;i<=lb;i++){ b[i] = b[i] - 48; }

   if(la>lb){
       y=0;
       k=la;

   }
   else{
       k=lb;
       y=1;
   }
    //printf("\nAddition called ");
    for(i=la,j=lb;i>=0,j>=0;i--,j--){
        //printf("Addition called ");
        c[k--]=(car+a[i]+b[j])%10;
        car=(car+a[i]+b[j])/10;
    }//for
    printf("\nWithout carry %s",c);

    if(y=0){
        j=0;
        for(i=0;i<la+1;i++){
        add[j++]=c[i] + 48;
        }
        add[j]='\0';
    }
    else{
        j=0;
        for(i=0;i<lb+1;i++){
        add[j++]=c[i] + 48;
        }
        add[j]='\0';
    }
    return add;

}


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;
         }
         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;
}

int main(){
    char a[MAX]={'9','5','5','5','5' };
    char b[MAX]={'3','3','3','3','3','3','3','3','3','3'};
    char *c;
    int la,lb;
    //int i = 0;
    int i = 1;
    //int i = 2;

    printf("Answer of two numbers : ");
    switch(i){
        case 0:  c = multiply(a,b); break;
        case 1:  c = addition(a,b); break;
        //case 2:  c = sub(a,b); break;
        default: break;
    }
    //c = multiply(a,b);
    printf("%s",c);
    return 0;
}

You can view code at http://ideone.com/Kd2Rv1