# Exponential Squaring (Fast Modulo Multiplication)

Given two numbers base and exp, we need to compute baseexp under Modulo 10^9+7

Examples:

Input : base = 2, exp = 2
Output : 4

Input  : base = 5, exp = 100000
Output : 754573817


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

In competitions, for calculating large powers of a number we are given a modulus value(a large prime number) because as the values of is being calculated it can get very large so instead we have to calculate (%modulus value.)
We can use the modulus in our naive way by using modulus on all the intermediate steps and take modulus at the end,
but in competitions it will definitely show TLE. So, what we can do.
The answer is we can try exponentiation by squaring which is a fast method for calculating exponentiation of a number.
Here we will be discussing two most common/important methods:

1. Basic Method(Binary Exponentiation)
2. -ary method.

Binary Exponentiation

As described in this article we will be using following formula to recursively calculate (%modulus value):

## C++

 // C++ program to compute exponential   // value under modulo using binary   // exponentiation.  #include  using namespace std;     #define N 1000000007 // prime modulo value     long int exponentiation(long int base,                          long int exp)  {      if (exp == 0)          return 1;         if (exp == 1)          return base % N;         long int t = exponentiation(base, exp / 2);      t = (t * t) % N;         // if exponent is even value      if (exp % 2 == 0)          return t;         // if exponent is odd value      else         return ((base % N) * t) % N;  }     // Driver Code  int main()  {      long int base = 5;      long int exp = 100000;         long int modulo = exponentiation(base, exp);      cout << modulo << endl;      return 0;  }     // This Code is contributed by mits

## Java

 // Java program to compute exponential value under modulo  // using binary exponentiation.  import java.util.*;  import java.lang.*;  import java.io.*;     class exp_sq {      static long N = 1000000007L; // prime modulo value      public static void main(String[] args)      {          long base = 5;          long exp = 100000;             long modulo = exponentiation(base, exp);          System.out.println(modulo);      }         static long exponentiation(long base, long exp)      {          if (exp == 0)              return 1;             if (exp == 1)              return base % N;             long t = exponentiation(base, exp / 2);          t = (t * t) % N;             // if exponent is even value          if (exp % 2 == 0)              return t;             // if exponent is odd value          else             return ((base % N) * t) % N;      }  }

## Python3

 # Python3 program to compute   # exponential value under  # modulo using binary   # exponentiation.     # prime modulo value  N = 1000000007;         # Function code   def exponentiation(bas, exp):      if (exp == 0):          return 1;      if (exp == 1):          return bas % N;             t = exponentiation(bas, int(exp / 2));      t = (t * t) % N;             # if exponent is      # even value      if (exp % 2 == 0):          return t;                 # if exponent is      # odd value      else:          return ((bas % N) * t) % N;     # Driver code  bas = 5;  exp = 100000;     modulo = exponentiation(bas, exp);  print(modulo);     # This code is contributed  # by mits

## C#

 // C# program to compute exponential  // value under modulo using binary   // exponentiation.  using System;     class GFG {             // prime modulo value      static long N = 1000000007L;             // Driver code      public static void Main()      {          long bas = 5;          long exp = 100000;             long modulo = exponentiation(bas, exp);          Console.Write(modulo);      }         static long exponentiation(long bas, long exp)      {          if (exp == 0)              return 1;             if (exp == 1)              return bas % N;             long t = exponentiation(bas, exp / 2);          t = (t * t) % N;             // if exponent is even value          if (exp % 2 == 0)              return t;             // if exponent is odd value          else             return ((bas % N) * t) % N;      }  }     // This code is contributed by nitin mittal.

## PHP

 

Output :

754573817


-ary method:

In this algorithm we will be expanding the exponent in base (k>=1), which is somehow similar to above method except we are not using recursion this method uses comparatively less memory and time.

## C++

 // C++ program to compute exponential value using (2^k)  // -ary method.  #include  using namespace std;     #define N 1000000007L; // prime modulo value     long exponentiation(long base, long exp)  {      long t = 1L;      while (exp > 0)       {             // for cases where exponent          // is not an even value          if (exp % 2 != 0)              t = (t * base) % N;             base = (base * base) % N;          exp /= 2;      }      return t % N;  }     // Driver code  int main()  {      long base = 5;      long exp = 100000;         long modulo = exponentiation(base, exp);      cout << (modulo);      return 0;  }     // This code is contributed by Rajput-Ji

## Java

 // Java program to compute exponential value using (2^k)  // -ary method.  import java.util.*;  import java.lang.*;  import java.io.*;     class exp_sq {      static long N = 1000000007L; // prime modulo value      public static void main(String[] args)      {          long base = 5;          long exp = 100000;             long modulo = exponentiation(base, exp);          System.out.println(modulo);      }         static long exponentiation(long base, long exp)      {          long t = 1L;          while (exp > 0) {                 // for cases where exponent              // is not an even value              if (exp % 2 != 0)                  t = (t * base) % N;                 base = (base * base) % N;              exp /= 2;          }          return t % N;      }  }

## Python3

 # Python3 program to compute   # exponential value   # using (2^k) -ary method.     # prime modulo value  N = 1000000007;      def exponentiation(bas, exp):      t = 1;      while(exp > 0):              # for cases where exponent          # is not an even value          if (exp % 2 != 0):              t = (t * bas) % N;             bas = (bas * bas) % N;          exp = int(exp / 2);      return t % N;     # Driver Code   bas = 5;  exp = 100000;     modulo = exponentiation(bas,exp);  print(modulo);     # This code is contributed  # by mits

## C#

 // C# program to compute   // exponential value   // using (2^k) -ary method.  using System;     class GFG  {  // prime modulo value  static long N = 1000000007L;      static long exponentiation(long bas,                              long exp)  {      long t = 1L;      while (exp > 0)       {             // for cases where exponent          // is not an even value          if (exp % 2 != 0)              t = (t * bas) % N;             bas = (bas * bas) % N;          exp /= 2;      }      return t % N;  }     // Driver Code      public static void Main ()  {      long bas = 5;      long exp = 100000;         long modulo = exponentiation(bas,                                    exp);      Console.WriteLine(modulo);  }  }     //This code is contributed by ajit

## PHP

  0)       {             // for cases where exponent          // is not an even value          if ($exp % 2 != 0)   $t = ($t * $bas) % $N;     $bas = ($bas * $bas) % $N;   $exp = (int)$exp / 2;   }   return $t % $N;  }    // Driver Code  $bas = 5;  $exp = 100000;    $modulo = exponentiation($bas,   $exp);  echo (\$modulo);     // This code is contributed  // by ajit  ?>

Output :

754573817


Applications:
Besides fast calculation of this method have several other advantages, like it is used in cryptography, in calculating Matrix Exponentiation et cetera.

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.