calculate 2^n for large N

Question

I want to calculate 2^N for large value of N i.e. upto 10^9 with mod operation. Used an iterative approach but its taking a lot of time for the input 10^8.

for(long i=1;i<=n;i++){ res=(res*2)%1000000007 }

Its taking time in calculating for n=10^8 and above.

`2^n` is equivalent to `1 << n`. That's `n`-long binary number, which gives us 1 megabyte of data filled with zeros. Well, there's gotta be another way. — Bartek Banachewicz, Nov 03 '13 at 16:53
Your algorithm is O(N). Of course it takes along time with large N. — Barmar, Nov 03 '13 at 16:54
@BartekBanachewicz But he's calculating the modulus at every step, so it never gets to be millions of bits long. — Barmar, Nov 03 '13 at 16:55

score 0 · Answer 1 · answered Nov 03 '13 at 16:58

You can increase the speed by squaring as well as multiplying by two - e.g. 2^4 can be reached as 2*2 => 4, 4*4 => 16.

Start with 2. If N is even, square what you have and halve N. If N is odd, multiply by two and decrement N. Apply the modulus as you go.

You may have to be careful the squaring doesn't overflow.

calculate 2^n for large N

1 Answers1