Find (a^b)%m where ‘a’ is very large

2.8

Given three numbers a, b and m where 1<=b,m<=10^6 and ‘a’ may be very large and contains upto 10^6 digits. The task is to find (a^b)%m.

Examples:

Input  : a = 3, b = 2, m = 4
Output : 1
Explanation : (3^2)%4 = 9%4 = 1

Input : a = 987584345091051645734583954832576, b = 3, m = 11
Output: 10

This problem is basically based on modular arithmetic. We can write (a^b) % m as (a%m) * (a%m) * (a%m) * … (a%m), b times. Below is a algorithm to solve this problem :

  • Since ‘a’ is very large so read ‘a’ as string.
  • Now we try to reduce ‘a’. We take modulo of ‘a’ by m once (See this article) i.e; ans = a % m , in this way now ans=a%m lies between integer range 1 to 10^6 i.e; 1 <= a%m <= 10^6.
  • Now multiply ans by b-1 times and simultaneously take mod of intermediate multiplication result with m because intermediate multiplication of ans may exceed range of integer and it will produce wrong answer.

C++

// C++ program to find (a^b) mod m for a large 'a'
#include<bits/stdc++.h>
using namespace std;

// utility function to calculate a%m
unsigned int aModM(string s, unsigned int mod)
{
    unsigned int number = 0;
    for (unsigned int i = 0; i < s.length(); i++)
    {
        // (s[i]-'0') gives the digit value and form
        // the number
        number = (number*10 + (s[i] - '0'));
        number %= mod;
    }
    return number;
}

// Returns find (a^b) % m
unsigned int ApowBmodM(string &a, unsigned int b,
                                  unsigned int m)
{
    // Find a%m
    unsigned int ans = aModM(a, m);
    unsigned int mul = ans;

    // now multiply ans by b-1 times and take
    // mod with m
    for (unsigned int i=1; i<b; i++)
        ans = (ans*mul) % m;

    return ans;
}

// Driver program to run the case
int main()
{
    string a = "987584345091051645734583954832576";
    unsigned int b=3, m=11;
    cout << ApowBmodM(a, b, m);
    return 0;
}

Python

# Python program to find (a^b) mod m for a large 'a'
def aModM(s, mod):
    number = 0

    # convert string s[i] to integer which gives
    # the digit value and form the number
    for i in range(len(s)):
        number = (number*10 + int(s[i]))
        number = number % m

    return number

# Returns find (a^b) % m
def ApowBmodM(a, b, mod):

    # Find a%m    
    ans = aModM(a, m)
    mul = ans

    # now multiply ans by b-1 times and take
    # mod with m
    for i in range(1,b):
        ans = (ans*mul) % m
        
    return ans


# Driver program to run the case
a = "987584345091051645734583954832576"
b, m = 3, 11
print ApowBmodM(a, b, m)


Output:

10

This article is contributed by Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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