Another way of thinking about exponentiation

In the introduction of the algorithm, it is mentioned that in seeking a certain number of powers, we follow the thought of reducing the law.

Here is another approach, by the exponential y binary decomposition of the exponentiation, such as X^y, when y=5, the x^5=x^101,x*=x equivalent to solve the x^10...0....0:, the algorithm time complexity is O (log (y)), the code is as follows.

#include <iostream>int len (char *arr) {int count=0;for (int i=0;arr[i]!= ' + '; i++) {count++;} return count;} void Main (int argc,char* argv[]) {int x=0,y=0;int result=1;printf ("pls input") numbers:\n ("%d", scanf); &x ( "%d", &y), Char Ch[256];itoa (y,ch,2), for (int i=len (CH) -1;i>=0;i--) {if (ch[i]== ' 1 ') {printf ("%d\n", x); result*= x;}　　　X*=x;}   print ("result:%d", result); System ("Pause");}

　　

Another way of thinking about exponentiation