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C/C++ Program to Find remainder of array multiplication divided by n

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Write a C/C++ program for a given multiple numbers and a number n, the task is to print the remainder after multiplying all the numbers divided by n.

Examples:

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

Input : arr[] = {100, 10},
n = 5
Output : 0
Explanation: 100 x 10 = 1000 % 5 = 0

C/C++ Program to Find remainder of array multiplication divided by n using 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.

Below is the implementation of the above approach:

CPP




// c++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
int findremainder(int arr[], int len, int n)
{
    long long int product = 1;
    for (int i = 0; i < len; i++) {
        product = product * arr[i];
    }
    return product % n;
}
 
// Drivers Code
int main()
{
    int arr[] = { 100, 10, 5, 25, 35, 14 };
    int len = sizeof(arr) / sizeof(arr[0]);
    int n = 11;
    cout << findremainder(arr, len, n);
    return 0;
}


Output

9

Time Complexity: O(n)
Auxiliary Space: O(1)

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

Below is the implementation of the above approach:

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

Time Complexity: O(len), where len is the size of the given array
Auxiliary Space: O(1)

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



Last Updated : 18 Oct, 2023
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