Find (a^b)%m where ‘a’ is very large

• Difficulty Level : Medium
• Last Updated : 17 Aug, 2022

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
Recommended Practice

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, 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'#includeusing namespace std; // utility function to calculate a%munsigned 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) % munsigned 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

Java

 // Java program to find (a^b) mod m for a large 'a' public class GFG {         // utility function to calculate a%m    static int aModM(String s, int mod)    {        int number = 0;        for (int i = 0; i < s.length(); i++)        {                         // (s[i]-'0') gives the digit            // value and form the number            number = (number * 10 );            int x = Character.getNumericValue(s.charAt(i));            number = number + x;            number %= mod;        }                 return number;    }     // Returns find (a^b) % m    static int ApowBmodM(String a, int b, int m)    {                 // Find a%m        int ans = aModM(a, m);        int mul = ans;             // now multiply ans by b-1 times        // and take mod with m        for (int i = 1; i < b; i++)            ans = (ans * mul) % m;             return ans;    }     // Driver code    public static void main(String args[])    {        String a = "987584345091051645734583954832576";        int b = 3, m = 11;        System.out.println(ApowBmodM(a, b, m));    }} // This code is contributed by Sam007

Python3

 # 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) % mdef ApowBmodM(a, b, m):     # 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 casea = "987584345091051645734583954832576"b, m = 3, 11print (ApowBmodM(a, b, m))

C#

 // C# program to find (a^b) mod m// for a large 'a'using System; class GFG {     // utility function to calculate a%mstatic int aModM(string s, int mod){    int number = 0;    for (int i = 0; i < s.Length; i++)    {                 // (s[i]-'0') gives the digit        // value and form the number        number = (number * 10 );        int x = (int)(s[i] - '0');        number = number + x;        number %= mod;    }    return number;} // Returns find (a^b) % mstatic int ApowBmodM(string a, int b,                              int m){         // Find a%m    int ans = aModM(a, m);    int mul = ans;     // now multiply ans by b-1 times    // and take mod with m    for (int i = 1; i < b; i++)        ans = (ans * mul) % m;     return ans;} // Driver Codepublic static void Main(){    string a = "987584345091051645734583954832576";    int b=3, m=11;    Console.Write(ApowBmodM(a, b, m));     }} // This code is contributed by Sam007



Javascript



Output

10

Time Complexity: O(len(a)+b)

Auxiliary Space: O(1)

Efficient Approach: The above multiplications can be reduced to log b by using fast modular exponentiation where we calculate result by the binary representation of exponent (b). If the set bit is 1 we multiply current value of base to result and square the value of base for each recursive call.

Recursive Code:

C++14

 // C++ program to find (a^b) mod m for a large 'a', with an// efficient approach.#include using namespace std;typedef long long ll; // Reduce the number B to a small number// using Fermat Littlell MOD(string num, int mod){    ll res = 0;     for (int i = 0; i < num.length(); i++)        res = (res * 10 + num[i] - '0') % mod;     return res;} ll ModExponent(ll a, ll b, ll m){    ll result;    if (a == 0)        return 0;    else if (b == 0)        return 1;    else if (b & 1) {        result = a % m;        result = result * ModExponent(a, b - 1, m);    }    else {        result = ModExponent(a, b / 2, m);        result = ((result * result) % m + m) % m;    }    return (result % m + m) % m;} int main(){    // String input as b is very large    string a = "987584345091051645734583954832576";    // String input as b is very large    ll b = 3;    ll m = 11;    ll remainderA = MOD(a, m);     cout << ModExponent(remainderA, b, m);     return 0;}

Java

 // Java program to find (a^b) mod m for a large 'a', with an// efficient approach.public class GFG{     // Reduce the number B to a small number  // using Fermat Little  static long MOD(String num, long mod)  {    long res = 0;    for (int i = 0; i < num.length(); i++) {      res = (res * 10 + num.charAt(i) - '0') % mod;    }    return res;  }   // Calculate the ModExponent of the given large number  // 'a'  static long ModExponent(long a, long b, long m)  {    long result;    if (a == 0) {      return 0;    }    else if (b == 0) {      return 1;    }    else if (b % 2 != 0) {      result = a % m;      result = result * ModExponent(a, b - 1, m);    }    else {      result = ModExponent(a, b / 2, m);      result = ((result * result) % m + m) % m;    }    return (result % m + m) % m;  }  public static void main(String[] args)  {     // String input as a is very large    String a = "987584345091051645734583954832576";    long b = 3;    long m = 11;    long remainderA = MOD(a, m);    System.out.println(ModExponent(remainderA, b, m));  }} // The code is contributed by Gautam goel (gautamgoel962)

Python3

 # Python3 program to find (a^b) mod m# for a large 'a' # Utility function to calculate a%mdef MOD(s, mod):     res = 0    for i in range(len(s)):        res = (res * 10 + int(s[i])) % mod    return res # Returns find (a^b) % mdef ModExponent(a, b, m):     if (a == 0):        return 0     elif (b == 0):        return 1     elif (b % 2 != 0):        result = a % m        result = result * ModExponent(a, b - 1, m)     else:        result = ModExponent(a, b / 2, m)        result = ((result * result) % m + m) % m     return (result % m + m) % m # Driver Codea = "987584345091051645734583954832576"b = 3m = 11remainderA = MOD(a, m)print(ModExponent(remainderA, b, m)) # This code is contributed by phasing17

C#

 // C# program to find (a^b) mod m for a large 'a', with an// efficient approach. using System;using System.Collections.Generic; public class GFG {     // Reduce the number B to a small number    // using Fermat Little    static long MOD(string num, long mod)    {        long res = 0;        for (int i = 0; i < num.Length; i++) {            res = (res * 10 + num[i] - '0') % mod;        }        return res;    }     // Calculate the ModExponent of the given large number    // 'a'    static long ModExponent(long a, long b, long m)    {        long result;        if (a == 0) {            return 0;        }        else if (b == 0) {            return 1;        }        else if (b % 2 != 0) {            result = a % m;            result = result * ModExponent(a, b - 1, m);        }        else {            result = ModExponent(a, b / 2, m);            result = ((result * result) % m + m) % m;        }        return (result % m + m) % m;    }     // Driver Code    public static void Main(string[] args)    {         // String input as a is very large        string a = "987584345091051645734583954832576";        long b = 3;        long m = 11;         // Function Call        long remainderA = MOD(a, m);        Console.WriteLine(ModExponent(remainderA, b, m));    }} // The code is contributed by phasing17

Javascript



Output

10

Time Complexity: O(len(a)+ log b)

Auxiliary Space: O(log b)

Space Efficient Iterative Code:

C++14

 // C++ program to find (a^b) mod m for a large 'a'#includeusing namespace std;typedef long long ll; // utility function to calculate a%m and b%mll aModM(string s, ll mod){    ll number = 0;    for (ll 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) % mll ApowBmodM(ll x, ll y,ll m){    ll res=1;     while(y)    {        if(y&1)        res=(res*x)%m;        y=y>>1;        x=((x*x)%m+m)%m;    }    return (res%m+m)%m;} // Driver program to run the caseint main(){    string a = "987584345091051645734583954832576";    ll b=3;    ll m=11;            // Find a%m    ll x=aModM(a,m);    cout << ApowBmodM(x,b,m);    return 0;}

Java

 // Java program to find (a^b) mod m for a large 'a'import java.util.*; class GFG {     // utility function to calculate a%m and b%m    static long aModM(String s, long mod)    {        long number = 0;        for (int i = 0; i < s.length(); i++) {             // (s[i]-'0') gives the digit value and form            // the number            number = (number * 10 + (s.charAt(i) - '0'));            number %= mod;        }        return number;    }     // Returns find (a^b) % m    static long ApowBmodM(long x, long y, long m)    {        long res = 1;         while (y > 0) {            if ((y & 1) != 0)                res = (res * x) % m;            y = y >> 1;            x = ((x * x) % m + m) % m;        }        return (res % m + m) % m;    }     // Driver program to run the case    public static void main(String[] args)    {        String a = "987584345091051645734583954832576";        long b = 3;        long m = 11;         // Find a%m        long x = aModM(a, m);        System.out.println(ApowBmodM(x, b, m));    }} // This code is contributed by phasing17

Python3

 # Python3 program to find (a^b) mod m for a large 'a' # utility function to calculate a%m and b%mdef aModM(s, mod):     number = 0;    for i in range(len(s)):        # int(s[i]) gives the digit value and form        # the number        number = (number * 10 + int(s[i]));        number %= mod;         return number;  # Returns find (a^b) % mdef ApowBmodM(x, y, m):         res = 1;     while (y > 0):        if (y & 1):            res = (res * x) % m;        y = y >> 1;        x = ((x * x) % m + m) % m;         return (res % m + m) % m;  # Driver program to run the casea = "987584345091051645734583954832576";b = 3;m = 11; # Find a%mx = aModM(a, m);print(ApowBmodM(x, b, m)); # This code is contributed by phasing17

C#

 // C# program to find (a^b) mod m for a large 'a'using System; class GFG{   // utility function to calculate a%m and b%m  static long aModM(string s, long mod)  {    long number = 0;    for (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  static long ApowBmodM(long x, long y, long m)  {    long res = 1;     while (y > 0) {      if ((y & 1) != 0)        res = (res * x) % m;      y = y >> 1;      x = ((x * x) % m + m) % m;    }    return (res % m + m) % m;  }   // Driver program to run the case  public static void Main(string[] args)  {    string a = "987584345091051645734583954832576";    long b = 3;    long m = 11;     // Find a%m    long x = aModM(a, m);    Console.WriteLine(ApowBmodM(x, b, m));  }} // This code is contributed by phasing17

Javascript

 // JavaScript program to find (a^b) mod m for a large 'a' // utility function to calculate a%m and b%mfunction aModM(s, mod){    let number = 0;    for (var i = 0; i < s.length; i++) {        // parseInt(s[i]) gives the digit value and form        // the number        number = (number * 10 + parseInt(s[i]));        number %= mod;    }    return number;} // Returns find (a^b) % mfunction ApowBmodM(x, y, m){    let res = 1;     while (y) {        if (y & 1)            res = (res * x) % m;        y = y >> 1;        x = ((x * x) % m + m) % m;    }    return (res % m + m) % m;} // Driver program to run the caselet a = "987584345091051645734583954832576";let b = 3;let m = 11; // Find a%mlet x = aModM(a, m);console.log(ApowBmodM(x, b, m));  // This code is contributed by phasing17

Output

10

Time Complexity: O(len(a)+ log b)

Auxiliary Space: O(1)

Case: When both ‘a’ and ‘b’ are very large.

We can also implement the same approach if both ‘a’ and ‘b’ was very large. In that case, we would have first took mod of it with m using our aModM function. Then pass it to the above ApowBmodM recursive or iterative function to get the required result.

Recursive Code:

C++14

 #include using namespace std;typedef long long ll; // Reduce the number B to a small number    // using Fermat Littlell MOD(string num,int mod){    ll res=0;         for(int i=0;i

Java

 /*package whatever //do not write package name here */import java.io.*;class GFG {   // Reduce the number B to a small number  // using Fermat Little.  static long MOD(String num,int mod)  {    long res = 0;     for(int i = 0; i < num.length(); i++)    {      res = (res * 10 + num.charAt(i) - '0') % mod;    }     return res;  }   static long ModExponent(long a,long b,long m){    long result = 0;     if(a == 0)      return 0;    else if(b == 0)      return 1;    else if((b&1) == 1){      result = a % m;      result = result*ModExponent(a, b - 1, m);    }    else{      result = ModExponent(a, b/2, m);      result = ((result % m)*(result % m)) % m;    }    return (result % m + m) % m;  }   public static void main (String[] args) {     // String input as b is very large    String a = "987584345091051645734583954832576";     // String input as b is very large    String b = "467687655456765756453454365476765";    int m = 1000000007;    long remainderA = MOD(a,m);    long remainderB = MOD(b,m);     System.out.println(ModExponent(remainderA, remainderB, m));   }} // This code is contributed by aadityapburujwale

Python3

 # Python3 program to implement the approach # Reduce the number B to a small number# using Fermat Littledef MOD(num, mod):    res = 0;     for i in range(len(num)):        res = (res * 10 + int(num[i])) % mod;     return res; def ModExponent(a, b, m):    if (a == 0):        return 0;    elif (b == 0):        return 1;    elif (b & 1):        result = a % m;        result = result * ModExponent(a, b - 1, m);    else:        b = b // 2        result = ModExponent(a, b, m);        result = ((result % m) * (result % m)) % m;         return (result % m + m) % m;  # String input as b is very largea = "987584345091051645734583954832576"; # String input as b is very largeb = "467687655456765756453454365476765"; m = 1000000007;remainderA = (MOD(a, m));remainderB = (MOD(b, m)); print(ModExponent(remainderA, remainderB, m)); # This code is contributed by phasing17

C#

 // C# program to implement the approachusing System;using System.Collections.Generic; class GFG {   // Reduce the number B to a small number  // using Fermat Little.  static long MOD(string num, int mod)  {    long res = 0;     for (int i = 0; i < num.Length; i++) {      res = (res * 10 + num[i] - '0') % mod;    }     return res;  }   static long ModExponent(long a, long b, long m)  {    long result = 0;     if (a == 0)      return 0;    else if (b == 0)      return 1;    else if ((b & 1) == 1) {      result = a % m;      result = result * ModExponent(a, b - 1, m);    }    else {      result = ModExponent(a, b / 2, m);      result = ((result % m) * (result % m)) % m;    }    return (result % m + m) % m;  }   public static void Main(string[] args)  {     // String input as b is very large    string a = "987584345091051645734583954832576";     // String input as b is very large    string b = "467687655456765756453454365476765";    int m = 1000000007;    long remainderA = MOD(a, m);    long remainderB = MOD(b, m);     Console.WriteLine(      ModExponent(remainderA, remainderB, m));  }} // This code is contributed by phasing17

Javascript

 // JavaScript program to implement the approach // Reduce the number B to a small number// using Fermat Littlefunction MOD(num, mod){    let res = 0;     for (var i = 0; i < num.length; i++)        res = (res * 10 + parseInt(num[i])) % mod;     return res;} function ModExponent(a, b, m){    let result;    if (a == 0n)        return 0n;    else if (b == 0n)        return 1n;    else if (b & 1n) {        result = a % m;        result = result * ModExponent(a, b - 1n, m);    }    else {        b = b / 2n - (b % 2n);        result = ModExponent(a, b, m);        result = ((result % m) * (result % m)) % m;    }    return (result % m + m) % m;} // String input as b is very largelet a = "987584345091051645734583954832576"; // String input as b is very largelet b = "467687655456765756453454365476765"; let m = 1000000007;let remainderA = BigInt(MOD(a, m));let remainderB = BigInt(MOD(b, m)); console.log(ModExponent(remainderA, remainderB, BigInt(m))); // This code is contributed by phasing17

Output

546081867

Time Complexity: O(len(a)+len(b)+log b)

Auxiliary Space: O(log b)

Space Efficient Iterative Code:

C++14

 // C++ program to find (a^b) mod m for a large 'a'#include using namespace std;typedef long long ll; // utility function to calculate a%m and b%mll aModM(string s, ll mod){    ll number = 0;    for (ll 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) % mll ApowBmodM(string& a, string& b, ll m){    ll res = 1;     // Find a%m    ll x = aModM(a, m);     // Find b%m    ll y = aModM(b, m);     while (y) {        if (y & 1)            res = (res * x) % m;        y = y >> 1;        x = ((x % m) * (x % m)) % m;    }    return (res % m + m) % m;} // Driver program to run the caseint main(){    string a = "987584345091051645734583954832576";    string b = "467687655456765756453454365476765";    ll m = 1000000007;    cout << ApowBmodM(a, b, m);    return 0;}

Java

 /*package whatever //do not write package name here */import java.io.*;class GFG {   // utility function to calculate a%m and b%m  static long aModM(String s, long mod){    long number = 0;    for (int i = 0; i < s.length(); i++)    {       // (s.charAt(i)-'0') gives the digit value and form      // the number      number = (number * 10 + (s.charAt(i) - '0'));      number %= mod;    }     return number;  }   // Returns find (a^b) % m  static long ApowBmodM(String a, String b, long m)  {    long res = 1;     // Find a%m    long x = aModM(a, m);     // Find b%m    long y = aModM(b, m);     while (y>0) {      if ((y & 1)==1)        res = (res * x) % m;      y = y >> 1;      x = ((x % m) * (x % m)) % m;    }    return (res % m + m) % m;  }   public static void main (String[] args) {     String a = "987584345091051645734583954832576";    String b = "467687655456765756453454365476765";    long m = 1000000007;    System.out.println(ApowBmodM(a, b, m));  }} // This code is contributed by aadityapburujwale

Python3

 # Python3 program to find (a^b) mod m for a large 'a' # utility function to calculate a%m and b%m  def aModM(s, mod):    number = 0    for i in range(len(s)):         # (s[i]-'0') gives the digit value and form        # the number        number = (number * 10 + (int(s[i])))        number %= mod     return number  # Returns find (a^b) % mdef ApowBmodM(a, b, m):    res = 1     # Find a%m    x = aModM(a, m)     # Find b%m    y = aModM(b, m)     while (y > 0):        if (y & 1):            res = (res * x) % m        y = y >> 1        x = ((x % m) * (x % m)) % m     return (res % m + m) % m  # Driver program to run the casea = "987584345091051645734583954832576"b = "467687655456765756453454365476765"m = 1000000007print(ApowBmodM(a, b, m)) # This code is contributed by phasing17

Javascript

 // JavaScript program to find (a^b) mod m for a large 'a' // utility function to calculate a%m and b%mfunction aModM(s, mod){    let number = 0n;    for (let i = 0; i < s.length; i++) {        // (s[i]-'0') gives the digit value and form        // the number        number = (number * 10n + BigInt(parseInt(s[i])));        number %= mod;    }    return number;} // Returns find (a^b) % mfunction ApowBmodM(a, b, m){    let res = 1n;     // Find a%m    let x = BigInt(aModM(a, m));     // Find b%m    let y = BigInt(aModM(b, m));     while (y > 0n) {        if (y & 1n)            res = (res * x) % m;        y = y >> 1n;        x = ((x % m) * (x % m)) % m;    }    return (res % m + m) % m;} // Driver program to run the caselet a = "987584345091051645734583954832576";let b = "467687655456765756453454365476765";let m = 1000000007n;console.log(ApowBmodM(a, b, m));  // This code is contributed by phasing17

C#

 // C# program to find (a^b) mod m for a large 'a' using System;using System.Collections.Generic; class GFG {    // utility function to calculate a%m and b%m    static long aModM(string s, long mod)    {        long number = 0;        for (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    static long ApowBmodM(string a, string b, long m)    {        long res = 1;         // Find a%m        long x = aModM(a, m);         // Find b%m        long y = aModM(b, m);         while (y != 0) {            if ((y & 1) != 0)                res = (res * x) % m;            y = y >> 1;            x = ((x % m) * (x % m)) % m;        }        return (res % m + m) % m;    }     // Driver program to run the case    public static void Main(string[] args)    {        string a = "987584345091051645734583954832576";        string b = "467687655456765756453454365476765";        long m = 1000000007;        Console.WriteLine(ApowBmodM(a, b, m));    }} // This code is contributed by phasing17

Output

546081867

Time Complexity: O(len(a)+len(b)+ log b)

Auxiliary Space: O(1)

This article is contributed by Shashank Mishra ( Gullu ) and improved by prophet1999. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.