Fast Power algorithm

Fast Power Algorithm idea: iterative/binary

We know a formula: a*b%c= (a%c*b%c)%c

If ab%c is required:

First, iteration

When B is odd: ab%c= ((A2) b/2*a)%c, Kee K=a2%c, that is (kb/2%c*a)%c

When B is even: ab%c= (A2) b/2%c, remember K=a2%c, that is to beg kb/2%c

Then the problem turns to Kb/2%c,

We assume that

Make the k1=a2%c, the problem turns to seek k1b/2%c

Make the k2=k12%c=a4%c, the problem turns to seek k2b/4%c

Make the k3=k22%c=a8%c, the problem turns to seek k3b/8%c

......

This process can be iterated, a start k=a, each b=b/2,k=k*k%c, whenever B is odd, the time of the K (also not squared) in the answer, to the last, b=1 (finally B must first = 1, then =0), the answer will be in, and then b=0 on the end of the algorithm.

#include <stdio.h>int main () {    int a,b,c,k,ans;    scanf ("%d%d%d", &a,&b,&c);    Ans=1;    K=a;    while (b) {        if (b%2)//b is an odd            ans=ans*k%c;        k=k*k%c;        B=B/2;    }    printf ("%d", ans);    return 0;}

Ii. binary

Suppose b= (18) 10 = (10010) 2

That Ab%c=a (2*a) (10000) 2%c=a2*a16%c

Each cycle is equivalent to the bits from right to left of B, and if it's 1, multiply K,

What is K? is a value of this bits of B, for example, to the second position from right to left of 10010, K=a (Ten) 2=a2, when the fifth position, K=a (10000) 2=a16

So each k=k*k%c,k change is: k=a (1) 2%c,k=a (+) 2%c,k=a (2), ...

#include <stdio.h>int main () {    int a,b,c,k,ans;    scanf ("%d%d%d", &a,&b,&c);    Ans=1;    K=a;    while (b) {        if (b&1)//b The last digit of the binary is 1            ans=ans*k%c;        k=k*k%c;        b=b>>1;//b binary minus the last    }    printf ("%d", ans);    return 0;}

　　

Fast Power algorithm