Modulo power for large numbers represented as strings

Given two numbers sa and sb represented as strings, find ab % MOD where MOD is 1e9 + 7. The numbers a and b can contain upto 106 digits.

Examples:

Input : sa = 2, sb = 3
Output : 8

Input : sa = 10000000000000000000000000000000000000000000
sb = 10000000000000000000000000000000000000000000
Output : 494234546

As a and b very large (may contain upto 10^6 digits each). So what we can do, we apply Fermat’s little theorem and property of modulo to reduce a and b.
Reduce a:
As we know,

(ab) % MOD = ((a % MOD)b) % MOD

Reduce b:
How to reduce b, We have already discuss in Find (a^b)%m where ‘b’ is very large

Now finally we have both a and b are in range of 1<=a, b<=10^9+7. Hence we can now use our modular exponentiation to calculate required answer.

C++

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// CPP program to find (a^b) % MOD where a and
// b may be very large and represented as strings.
#include <bits/stdc++.h>
using namespace std;
  
#define ll long long int
const ll MOD = 1e9 + 7;
  
// Returns modulo exponentiation for two numbers
// represented as long long int. It is used by
// powerStrings(). Its complexity is log(n)
ll powerLL(ll x, ll n)
{
    ll result = 1;
    while (n) {
        if (n & 1)
            result = result * x % MOD;
        n = n / 2;
        x = x * x % MOD;
    }
    return result;
}
  
// Returns modulo exponentiation for two numbers
// represented as strings. It is used by
// powerStrings()
ll powerStrings(string sa, string sb)
{
    // We convert strings to number 
  
    ll a = 0, b = 0;
  
    // calculating  a % MOD
    for (int i = 0; i < sa.length(); i++)
        a = (a * 10 + (sa[i] - '0')) % MOD;
  
    // calculating  b % (MOD - 1)
    for (int i = 0; i < sb.length(); i++)
        b = (b * 10 + (sb[i] - '0')) % (MOD - 1);
  
    // Now a and b are long long int. We
    // calculate a^b using modulo exponentiation
    return powerLL(a, b);
}
  
int main()
{
    // As numbers are very large
    // that is it may contains upto
    // 10^6 digits. So, we use string.
    string sa = "2", sb = "3";
  
    cout << powerStrings(sa, sb) << endl;
    return 0;
}

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Java

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// Java program to find (a^b) % MOD 
// where a and b may be very large 
// and represented as strings. 
import java.util.*;
  
class GFG 
{
  
    static long MOD = (long) (1e9 + 7);
  
    // Returns modulo exponentiation for two numbers 
    // represented as long long int. It is used by 
    // powerStrings(). Its complexity is log(n) 
    static long powerLL(long x, long n) 
    {
        long result = 1;
        while (n > 0
        {
            if (n % 2 == 1
            {
                result = result * x % MOD;
            }
            n = n / 2;
            x = x * x % MOD;
        }
        return result;
    }
  
    // Returns modulo exponentiation for
    // two numbers  represented as strings.  
    // It is used by powerStrings() 
    static long powerStrings(String sa, String sb) 
    {
        // We convert strings to number 
        long a = 0, b = 0;
  
        // calculating a % MOD 
        for (int i = 0; i < sa.length(); i++) 
        {
            a = (a * 10 + (sa.charAt(i) - '0')) % 
                                               MOD;
        }
  
        // calculating b % (MOD - 1) 
        for (int i = 0; i < sb.length(); i++) 
        {
            b = (b * 10 + (sb.charAt(i) - '0')) %
                                        (MOD - 1);
        }
  
        // Now a and b are long long int. We 
        // calculate a^b using modulo exponentiation 
        return powerLL(a, b);
    }
  
    // Driver code
    public static void main(String[] args)
    {
          
        // As numbers are very large 
        // that is it may contains upto 
        // 10^6 digits. So, we use string. 
        String sa = "2", sb = "3";
        System.out.println(powerStrings(sa, sb));
    }
  
// This code is contributed by Rajput-JI

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Python3

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# Python3 program to find (a^b) % MOD 
# where a and b may be very large 
# and represented as strings.
MOD = 1000000007;
  
# Returns modulo exponentiation 
# for two numbers represented as
# long long int. It is used by
# powerStrings(). Its complexity 
# is log(n)
def powerLL(x, n):
  
    result = 1;
    while (n): 
        if (n & 1):
            result = result * x % MOD;
        n = int(n / 2);
        x = x * x % MOD;
    return result;
  
# Returns modulo exponentiation 
# for two numbers represented as 
# strings. It is used by powerStrings()
def powerStrings(sa, sb):
      
    # We convert strings to number 
    a = 0;
    b = 0;
  
    # calculating a % MOD
    for i in range(len(sa)):
        a = (a * 10 + (ord(sa[i]) - 
                       ord('0'))) % MOD;
  
    # calculating b % (MOD - 1)
    for i in range(len(sb)):
        b = (b * 10 + (ord(sb[i]) - 
                       ord('0'))) % (MOD - 1);
  
    # Now a and b are long long int. 
    # We calculate a^b using modulo 
    # exponentiation
    return powerLL(a, b);
  
# Driver code
  
# As numbers are very large
# that is it may contains upto
# 10^6 digits. So, we use string.
sa = "2";
sb = "3";
  
print(powerStrings(sa, sb));
      
# This code is contributed by mits

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C#

// C# program to find (a^b) % MOD where a and b
// may be very large and represented as strings.
using System;

class GFG
{
static long MOD = (long) (1e9 + 7);

// Returns modulo exponentiation for two numbers
// represented as long long int. It is used by
// powerStrings(). Its complexity is log(n)
static long powerLL(long x, long n)
{
long result = 1;
while (n > 0)
{
if (n % 2 == 1)
{
result = result * x % MOD;
}
n = n / 2;
x = x * x % MOD;
}
return result;
}

// Returns modulo exponentiation for
// two numbers represented as strings.
// It is used by powerStrings()
static long powerStrings(String sa, String sb)
{
// We convert strings to number
long a = 0, b = 0;

// calculating a % MOD
for (int i = 0; i < sa.Length; i++) { a = (a * 10 + (sa[i] - '0')) % MOD; } // calculating b % (MOD - 1) for (int i = 0; i < sb.Length; i++) { b = (b * 10 + (sb[i] - '0')) % (MOD - 1); } // Now a and b are long long int. We // calculate a^b using modulo exponentiation return powerLL(a, b); } // Driver code public static void Main(String[] args) { // As numbers are very large // that is it may contains upto // 10^6 digits. So, we use string. String sa = "2", sb = "3"; Console.WriteLine(powerStrings(sa, sb)); } } // This code is contributed by 29AjayKumar [tabby title="PHP"]

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<?php
// PHP program to find (a^b) % MOD 
// where a and b may be very large 
// and represented as strings.
$MOD = 1000000007;
  
// Returns modulo exponentiation 
// for two numbers represented as
// long long int. It is used by
// powerStrings(). Its complexity 
// is log(n)
function powerLL($x, $n)
{
    global $MOD;
    $result = 1;
    while ($n
    {
        if ($n & 1)
            $result = $result * $x % $MOD;
        $n = (int)$n / 2;
        $x = $x * $x % $MOD;
    }
    return $result;
}
  
// Returns modulo exponentiation 
// for two numbers represented as 
// strings. It is used by powerStrings()
function powerStrings($sa, $sb)
{
    global $MOD;
      
    // We convert strings to number 
    $a = 0;
    $b = 0;
  
    // calculating a % MOD
    for ($i = 0; $i < strlen($sa); $i++)
        $a = ($a * 10 + ($sa[$i] - 
                     '0')) % $MOD;
  
    // calculating b % (MOD - 1)
    for ($i = 0; $i < strlen($sb); $i++)
        $b = ($b * 10 + ($sb[$i] - '0')) % 
                        ($MOD - 1);
  
    // Now a and b are long long int. 
    // We calculate a^b using modulo 
    // exponentiation
    return powerLL($a, $b);
}
  
// Driver code
  
// As numbers are very large
// that is it may contains upto
// 10^6 digits. So, we use string.
$sa = "2";
$sb = "3";
  
echo powerStrings($sa, $sb);
      
// This code is contributed by mits
?>

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Output:

8


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