Divisible by 37 for large numbers in C++ Program

C++Server Side ProgrammingProgramming

In this tutorial, we are going to write a program that checks whether the given large number is divisible by 37 or not.

We are going to use a little bit of math here. Let's see the steps to solve the problem.

• Initialize the number.

• If the length of the given number is not divisible by 3, then add zeroes at the beginning of the number to make length is divisible by 3.

• Divide the number into 3 digits groups and add them.

• If the resultant sum is divisible by 37, then the given number is divisible by 37.

• If the resultant sum is 4 digits number, then repeat the steps from 2.

• Print whether the given number is divisible by 37 or not.

Example

Let's see the code.

Live Demo

#include <bits/stdc++.h>
using namespace std;
bool isNumberDivisibleBy37(string number, int n) {
if (number == "0") {
return 0;
}
if (n % 3 == 1){
number = "00"+ number;
n += 2;
}
else if (n % 3 == 2){
number = "0"+ number;
n += 1;
}
int groups_sum = 0;
while (n != 0){
string group = number.substr(n - 3, n);
int group_value = (group[0] - '0') * 100 + (group[1] - '0') * 10 + (group[2] - '0') * 1;
groups_sum += group_value;
n = n - 3;
}
if (groups_sum >= 1000) {
string new_number = to_string(groups_sum);
return isNumberDivisibleBy37(new_number, new_number.length());
}
else {
return groups_sum % 37 == 0;
}
}
int main() {
string number = "4048675309";
if (isNumberDivisibleBy37(number, 10)) {
cout << "Yes" << endl;
}
else {
cout << "No" << endl;
}
return 0;
}

Output

If you run the above code, then you will get the following result.

Yes

Conclusion

If you have any queries in the tutorial, mention them in the comment section.

Published on 28-Jan-2021 06:45:26