Modular Exponentiation#273

Modular Exponentiation (Editorial)

Exponentiation is a mathematical operation, written as $x^{n}$ x^n, involving two numbers, the base $x$ and the exponent or power $n$ . When $n$ is a positive integer, exponentiation corresponds to repeated multiplication of the base: that is, $x^{n}$ x^n is the product of multiplying $n$ bases: $x^{n} = x \cdot x \cdot ... \cdot x$ x^n = x \cdot x \cdot ... \cdot x.

How can we compute $x^{n}$ x^n for given $x$ and $n$ ? One option is to use one loop with a time complexity of $O (n)$ . The linear time algorithm will pass the time limit because $n \leq 1 0^{7}$ n \le 10^7.

Alternatively, you can use fast exponentiation:

x^{n} m o d m = ⎩ ⎨ ⎧ (x^{2})^{n /2}, i f n i s e v e n, x \cdot x^{n - 1}, i f n i s o dd, 1, n = 0

x^n~mod~m = \begin{cases} (x^2)^{n/2}, if~n~is~even,\\ x \cdot x^{n-1}, if~n~is~odd, \\ 1, n = 0 \end{cases}

Algorithm implementation

Let's use the $64$ -bit integer type long long. Read the input data.

scanf("%lld %lld %lld",&x,&n,&m);

Compute the value of $x^{n} m o d m$ x^n~mod~m using a loop.

res = 1;
for (i = 1; i <= n; i++)
  res = (res * x) % m;

Print the answer.

printf("%lld\n",res);

Algorithm implementation — Fast Exponentiation

The PowMod function finds the value of $x^{n} m o d m$ x^n~mod~m.

long long PowMod(long long x, long long n, long long m)
{
  if (n == 0) return 1;
  if (n % 2 == 0) return PowMod((x * x) % m, n / 2, m);
  return (x * PowMod(x, n - 1, m)) % m;
}

The main part of the program. Read the input data.

scanf("%lld %lld %lld",&x,&n,&m);

Compute and print the answer.

res = PowMod(x,n,m);
printf("%lld\n",res);

Java implementation — linear complexity

import java.util.*;

public class Main
{
  public static void main(String[] args)
  {
    Scanner con = new Scanner(System.in);
    long x = con.nextLong(),
         n = con.nextLong(),
         m = con.nextLong();
    long res = 1;
    for (int i = 1; i <= n; i++) res = (res * x) % m;
    System.out.println(res);
    con.close();
  }
}

Java

Java implementation — O(log2n)

import java.util.*;

public class Main
{
  static long PowMod(long x, long n, long m)
  {
    if (n == 0) return 1;
    if (n % 2 == 0) return PowMod((x * x) % m, n / 2, m);
    return (x * PowMod(x, n - 1, m)) % m;
  }
	
  public static void main(String[] args)
  {
    Scanner con = new Scanner(System.in);
    long x = con.nextLong();
    long n = con.nextLong();
    long m = con.nextLong();
    long res = PowMod(x,n,m);
    System.out.println(res);
    con.close();
  }
}

Java

Java implementation — powMod

import java.util.*;
import java.math.*;

public class Main
{
  public static void main(String[] args)
  {
    Scanner con = new Scanner(System.in);	  
    BigInteger x = con.nextBigInteger();
    BigInteger n = con.nextBigInteger();
    BigInteger m = con.nextBigInteger();
    System.out.println(x.modPow(n,m));    
    con.close();
  }
}

Java

Python implementation — pow

Read the input data.

x, n, m = map(int, input().split())

Compute and print the answer. Use the built-in function pow to evaluate the expression $x^{n} m o d m$ x^n~mod~m.

print(pow(x, n, m))

Python implementation — loop

Read the input data.

x, n, m = map(int, input().split())

Compute the value of $x^{n} m o d m$ x^n~mod~m using a loop.

res = 1
for i in range(1, n + 1):
  res = (res * x) % m

Print the answer.

print(res)