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- /* this code can add two 50 digit number that mean 100000000000000000000009991 + 1088888888888214124 */
- //Just increase the array length size//
- #include <iostream>
- #include <string>
- using namespace std;
- int getnum(char c_num){
- int res = c_num - '0';
- return res;
- }
- void print_array(int *arr,int len){
- for(int i = 0;i<len;i++){
- cout << arr[i] << " ";
- }
- cout << endl;
- return;
- }
- int find_small(int a,int b){
- if(a < b) return a;
- return b;
- }
- int main(){
- int first_arr,sec_arr;
- cout << "First Array Len:";cin >> first_arr;
- cout << "Second Array Len:";cin >> sec_arr;
- cout << "Enter Your First Array:";char a_arr[10],b_arr[10];
- cin >> a_arr;
- cout << "Enter your Second Array:";
- cin >> b_arr;
- int a[10],b[10],sum[20];
- for(int i = 0;i<first_arr;i++){
- a[i] = getnum(a_arr[i]);
- }
- for(int i = 0;i< sec_arr;i++){
- b[i] = getnum(b_arr[i]);
- }
- cout << "int arr_a = ";print_array(a,first_arr);cout << "\n\n";
- cout << "int arr_b = ";print_array(b,sec_arr);cout << "\n\n";
- int g ,flag = 0,sm;
- if(first_arr == sec_arr){
- g = first_arr;
- flag = 1;
- }
- else if (first_arr > sec_arr ){
- g = first_arr;
- }
- else g = sec_arr;
- sm = find_small(first_arr,sec_arr);
- int j,k,r,c,s,q,temp;
- for(j = first_arr-1, k = sec_arr-1,r=0;g > 0;j--,k--,r++,g--,sm--){
- cout << "a[i] = " << a[j] <<"b[i] = " << b[k] << "sm = " << sm;
- if(sm > 0){
- s = a[j] + b[k];
- } else{
- s = a[j];
- }
- cout << " s = " << s;
- cout << " j = " << j << " k = " << k << " r = " << r << " greatest = " << g<<endl;
- if (s > 9){
- q = s%10;temp = s/10;
- a[j-1] = a[j-1] + temp;
- sum[r] = q;
- if (g == 1){sum[++r] = temp;}
- }
- else sum[r] = s;
- }
- print_array(sum,r);
- cout << "Rev === == \n\n";
- /*Rev THE ARRAY*/
- for(int i = r-1;i >= 0;i--){
- cout << sum[i];
- }
- //
- cout << endl;
- }
- /*output */
- 1.give the first array length big or equal to the second array length.
- 2.dont forget to increase the array size of the code.
- 3.contactthedead1998@gmail.com //=> if you want me to modify the code
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