Super Pow in C++

Suppose we have to calculate a^b mod 1337 where a is one positive integer and b is an extremely large positive integer given in the form of an array. So if a = 2 and b = [1,0] then the output will be 1024

To solve this, we will follow these steps −

• Define powerMod() method this takes base and power

• m := 1337, ret := 1

• while power is not 0

  • if power is odd, then ret := ret * base mod m

  • base := base^2 mod m

  • power := power / 2

• return ret

• Define superPower(), this takes a and b

• if size of b = 0, then return 1

• last := last element of b

• delete last element from b

• return powerMod(superpower(a, b), 10) * powerMod(a, last)) mod 1337

Example(C++)

Let us see the following implementation to get a better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
class Solution {
   public:
   int powerMod(lli base, lli power){
      lli mod = 1337;
      lli ret = 1;
      while(power){
         if(power & 1) ret = (ret * base) % mod;
         base = (base * base) % mod;
         power >>= 1;
      }
      return ret;
   }
   int superPow(int a, vector<int>& b) {
      if(b.size() == 0) return 1;
      int last = b.back();
      b.pop_back();
      return (powerMod(superPow(a, b), 10) * powerMod(a, last)) % 1337;
   }
};
main(){
   Solution ob;
   vector<int> v = {1,0};
   cout << (ob.superPow(2, v));
}

Input

2
[1,0]

Output

1024

Arnab Chakraborty

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Updated on: 02-May-2020

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