Exponents in C (n^m)

Question

http://z-trening.com/tasks.php?show_task=5000000069&lang=uk

#include <stdio.h>

int main(void)
{
    long long o, s, m, i; 
    long long rez = 1;

    scanf("%lld %lld %lld", &o, &s, &m);
    o = o % m;
    for (i = 0; i < s; i++){
        rez = (rez * o) % m;
    }
    printf("%lld", rez);
    return 0;
}

It works for 10 out of 20 tasks. Is there any faster way to raise o^s ?

Why do you care about how fast your code works given that it is not correct? How does it help you to get the wrong answer faster? — David Heffernan, Jan 30 '14 at 09:24
_"It works for 10 out of 20 tasks"_ That should be the first thing on your todo list. if it works for 20/20 tasks, then worry about the speed... — Elias Van Ootegem, Jan 30 '14 at 09:26
@David, Elias "Works" should probably be understood as "finishes within the alotted time". The code is correct, it just doesn't scale well for large values of `s`. — Klas Lindbäck, Jan 30 '14 at 09:55
It works for 10 out of 20 because for the last 10 tasks it gets TLE. It in c language — user31415, Jan 30 '14 at 10:00

Tamás Zahola · Accepted Answer · 2014-01-30T10:10:14.133

Yes, there is a faster method: modular exponentiation. http://en.wikipedia.org/wiki/Modular_exponentiation

Repeated multiplication (your algorithm) runs in exponential time, while modular exponentiation runs in polynomial time. It works like this:

Let's say, you want to compute A^B mod M. First, write B in binary:

B = bn,bn-1,...,b1,b0

This means, that

B = bn * 2^n + bn-1 * 2^(n-1) + ... + b1 * 2 + b0

Substituting this in the expression A^B:

A^B = A^(2^n)^bn * A^(2^(n-1))^bn-1 * ... * A^2^b1 * A^b0

A^(2^n) can be computed recursively:

A^(2^n) = (A^(2^(n-1)))^2

Now, the trick is, to use this identity to compute A^(2^i) for each i using repeated squaring modulo M. The usual identities of multiplication and exponentiation hold for modular arithmetics too, so this is perfectly legal. The full algorithm:

input: A,B,M
output: A^B mod M

result = 1
while(B != 0){
    if(B is odd){
        result = result * A mod M
    }
    B = B / 2
    A = A * A mod M
}
return result

Please, read the help-section: link only answers are discouraged (if not regarded as invalid). Sure wiki isn't likely to be gone tomorrow, but at least paste (and credit!) the gist of what applies here into your answer — Elias Van Ootegem, Jan 30 '14 at 09:27

score 1 · Answer 2 · answered Jan 30 '14 at 10:08

An easy way to reduce the number of calculations uses the equalities:

a^(b+c) = a^b*a^c     (1)

and

(x*y)%z = ((x%z)*(y%z))%z   (2)

These two equalities can be used quickly compute (o^1)%m, (o^2)%m, (o^4)%m, (o^8)%m, ...:

o2n = (on * on)%m

The problem can now be solved with a loop that iterates once for every bit in s, which means that the complexity has been reduced from O(s) to O(log(s).

long long s, o;
int m;

// Input s,o,m (code omitted)

int res = 1, on = o%m;  // Assuming int is at least 32 bits
for (i = 0; i < 35; i++) {  // 2^35 is larger than the largest s allowed.
  if (s & 1 << i) {
    res = res * on;
  }
  on = (on*on)%m;
}
printf("%d\n", res);

Exponents in C (n^m)

2 Answers2