Skip to content
Related Articles

Related Articles

C/C++ Program to Find remainder of array multiplication divided by n
  • Last Updated : 09 Aug, 2019

Given multiple numbers and a number n, the task is to print the remainder after multiply all the number divide by n.

Examples:

Input : arr[] = {100, 10, 5, 25, 35, 14}, 
            n = 11
Output : 9
100 x 10 x 5 x 25 x 35 x 14 = 61250000 % 11 = 9

Naive approach: First multiple all the number then take % by n then find the remainder, But in this approach, if the number is maximum of 2^64 then it give the wrong answer.

Approach that avoids overflow : First take a remainder or individual number like arr[i] % n. Then multiply the remainder with current result. After multiplication, again take remainder to avoid overflow. This works because of distributive properties of modular arithmetic. ( a * b) % c = ( ( a % c ) * ( b % c ) ) % c




// C++ program to find
// remainder when all
// array elements are
// multiplied.
#include <iostream>
using namespace std;
  
// Find remainder of arr[0] * arr[1] *
// .. * arr[n-1]
int findremainder(int arr[], int len, int n)
{
    int mul = 1;
  
    // find the individual remainder
    // and multiple with mul.
    for (int i = 0; i < len; i++)
        mul = (mul * (arr[i] % n)) % n;
  
    return mul % n;
}
  
// Driver code
int main()
{
    int arr[] = { 100, 10, 5, 25, 35, 14 };
    int len = sizeof(arr) / sizeof(arr[0]);
    int n = 11;
  
    // print the remainder of after
    // multiple all the numbers
    cout << findremainder(arr, len, n);
}
Output:
9

Please refer complete article on Find remainder of array multiplication divided by n for more details!

Want to learn from the best curated videos and practice problems, check out the C Foundation Course for Basic to Advanced C.

My Personal Notes arrow_drop_up
Recommended Articles
Page :