# Find (a^b)%m where ‘a’ is very large

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: Please solve it on “PRACTICE” first, before moving on to the solution.

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 (See this article) 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;
}
```

## Python

```# 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, mod):

# 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)
```

Output:

```10
```

This article is contributed by Shashank Mishra ( Gullu ). 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 write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

# GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
2.8 Average Difficulty : 2.8/5.0
Based on 7 vote(s)

Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.