Middle-Term-Factorization

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
using namespace std;
void find_the_Number(int,int,int,int);
int gcd(int,int);
char problem[20], beg[20],mid[10],end[10];char extr;
int main(){
    int temp;
    cin >> problem; // 6x^2
    cin >> mid; // +5x
    cin >> end;//-6
    char *ptr;
    int a,b,c;
    a = strtol(problem, &ptr, 10); b = strtol(mid, &ptr, 10);c = strtol(end, &ptr, 10);
    for(int i=0;i<strlen(mid);i++){ // extracting the mid char
        if(mid[i] >= 97 && mid[i] <= 122){
            extr = mid[i];
            break;
        }
    }
    if(a == 0) a = 1;
    if(b == 0 ) b = 1;
    cout << " A = " <<a << " b = " << b << "C = "<<c<<endl;
    long int res = a*c;
    strcpy(beg,problem);
    strcat(problem,mid);//Making It The Total String 6x^2 + 5x - 6
    strcat(problem,end);//Same
    find_the_Number(res,b,a,c);
}
void find_the_Number(int number,int diff,int begNum,int endNum){
    bool negativeBit = false;
    int co =0; int flag=0;
    int arr[20];
    if (number < 0){
        negativeBit = true; number = -number;
    }
    for(int i=1; i <= number; ++i){
        if (number%i == 0) arr[co++] = i;
    }// (ax^2 + bx + c) //
    int a , b;
    for(int k = 0;k<co;k++){ // product should be equal to res,and the diffrence
        for(int s= 0;s<co;s++){ /*should be equal to b*/
            if(negativeBit){
                if(arr[k] - arr[s] == diff && arr[k] * arr[s] == number){
                a = arr[k]; b = -arr[s];
                flag = 1;
                break;}
                if(flag == 1)break;
            }
            else{
                if(arr[k] + arr[s] == diff && arr[k] * arr[s] == number){
                a = arr[k]; b = arr[s];
                flag = 1;
                break;}
                if(flag == 1)break;
            }
        }
    }
    system ("cls");
    cout << endl<<endl;
    cout << problem;cout << endl;
    cout << endl;
    printf("%s %+d%c %+d%c %s\n",beg,a,extr,b,extr,end);
    cout << endl;
    int com1 = gcd(begNum,a);
    int com2 = gcd(endNum,b);
    printf("%+d%c(%+d%c%+d)%+d(%+d%c%+d)\n",com1,extr,begNum/com1,extr,a/com1,com2,b/com2,extr,endNum/com2);
    cout << endl;
    printf("(%+d%c%+d)(%+d%c%+d)\n",com1,extr,com2,begNum/com1,extr,a/com1);
    cout << endl;
}
int gcd(int a ,int b){
    int rem = 10;
    while(rem != 0){
        rem = b%a;
        b = a;
        a = rem;
    }
    return b;
}