problem 38

/*
Take the number 192 and multiply it by each of 1, 2, and 3:

192 × 1 = 192
192 × 2 = 384
192 × 3 = 576
By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)

The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).

What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n > 1?
*/
#include <iostream>
#include<string>
#include <algorithm> 
#include<sstream>
#include <bits/stdc++.h> 

using namespace std;

void check_pandigital(int num){
  char perfect_pandigital[9] = {'1', '2', '3', '4', '5', '6', '7', '8', '9'};
  string final_number = to_string(num * 1);

  if (num < 10000){

    for(int j = 2; j < 3; j++){
     final_number += to_string(num * j);
    }

    int size = final_number.size();
    char nums[size];
    strcpy(nums , final_number.c_str()); 
    std::sort(nums, nums + size); 
    bool check = true;
    for(int k = 0; k < 9; k++){
      if(nums[k] != perfect_pandigital[k]){
       check = false;
      }
    }
    if (check){
      cout << num << final_number << endl;
    }
  }
  
}

int main() {
 for(int i = 9000; i  < 10000; i++){
 check_pandigital(i);
 }
 return 0;
}