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, 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;
} |
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) % m def 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 case a = "987584345091051645734583954832576"
b, m = 3, 11
print (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%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 = (int)(s[i] - '0');
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 a = "987584345091051645734583954832576";
int b=3, m=11;
Console.Write(ApowBmodM(a, b, m));
} } // This code is contributed by Sam007 |
PHP
<?php // PHP program to find (a^b) // mod m for a large 'a' // utility function to // calculate a%m function aModM($s, $mod)
{ $number = 0;
for ($i = 0; $i < strlen($s); $i++)
{
// (s[i]-'0') gives the digit
// value and form the number
$number = ($number * 10 +
($s[$i] - '0'));
$number %= $mod;
}
return $number;
} // Returns find (a^b) % m function ApowBmodM($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 = 1; $i < $b; $i++)
$ans = ($ans * $mul) % $m;
return $ans;
} // Driver code
$a = "987584345091051645734583954832576";
$b = 3;
$m = 11;
echo ApowBmodM($a, $b, $m);
return 0;
// This code is contributed by nitin mittal. ?> |
Javascript
<script> // JavaScript program to find (a^b) mod m // for a large 'a' // Utility function to calculate a%m function aModM(s, mod)
{ let number = 0;
for(let i = 0; i < s.length; i++)
{
// (s[i]-'0') gives the digit
// value and form the number
number = (number * 10 );
let x = (s[i] - '0');
number = number + x;
number %= mod;
}
return number;
} // Returns find (a^b) % m function ApowBmodM(a, b, m)
{ // Find a%m
let ans = aModM(a, m);
let mul = ans;
// Now multiply ans by b-1 times
// and take mod with m
for(let i = 1; i < b; i++)
ans = (ans * mul) % m;
return ans;
} // Driver Code let a = "987584345091051645734583954832576";
let b = 3, m = 11; document.write(ApowBmodM(a, b, m)); // This code is contributed by souravghosh0416 </script> |
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 <bits/stdc++.h> using namespace std;
typedef long long ll;
// Reduce the number B to a small number // using Fermat Little ll 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%m def MOD(s, mod):
res = 0
for i in range(len(s)):
res = (res * 10 + int(s[i])) % mod
return res
# Returns find (a^b) % m def 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 Code a = "987584345091051645734583954832576"
b = 3
m = 11
remainderA = 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
<script> // JavaScript program to find (a^b) mod m // for a large 'a' // Utility function to calculate a%m function MOD(s, mod)
{ var res = 0;
for (var i = 0; i < s.length; i++) {
res = (res * 10 + (s[i] - '0')) % mod;
}
return res;
} // Returns find (a^b) % m function ModExponent(a, b, m)
{ var 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 let a = "987584345091051645734583954832576";
let b = 3, m = 11; var remainderA = MOD(a, m);
document.write(ModExponent(remainderA, b, m)); // This code is contributed by shinjanpatra. </script> |
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' #include<bits/stdc++.h> using namespace std;
typedef long long ll;
// utility function to calculate a%m and b%m ll 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) % m ll 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 case int 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%m def 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) % m def 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 case a = "987584345091051645734583954832576";
b = 3;
m = 11;
# Find a%m x = 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%m function 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) % m function 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 case let a = "987584345091051645734583954832576";
let b = 3; let m = 11; // Find a%m let 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 <bits/stdc++.h> using namespace std;
typedef long long ll;
// Reduce the number B to a small number // using Fermat Little
ll 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%m)*(result%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
string b = "467687655456765756453454365476765";
ll m = 1000000007;
ll remainderA = MOD(a,m);
ll remainderB = MOD(b,m);
cout << ModExponent(remainderA, remainderB, m);
return 0;
} |
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 Little def 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 large a = "987584345091051645734583954832576";
# String input as b is very large b = "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 approach using 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 Little function 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 large let a = "987584345091051645734583954832576";
// String input as b is very large let 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 <bits/stdc++.h> using namespace std;
typedef long long ll;
// utility function to calculate a%m and b%m ll 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) % m ll 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 case int 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) % m def 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 case a = "987584345091051645734583954832576"
b = "467687655456765756453454365476765"
m = 1000000007
print(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%m function 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) % m function 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 case let 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)
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