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  1. #include <stdio.h>
  2. #include <math.h>
  3.  
  4. double pow2(double x, int n) {
  5.  
  6.        double ans = 1; 
  7.  
  8.        if(n == 0) return 1;
  9.  
  10.        if(n < 0) x = 1 / x;
  11.  
  12.        while(n) {
  13.  
  14.           if(n % 2) ans *= x;
  15.  
  16.           x *= x;
  17.  
  18.           n /= 2;
  19.        }
  20.  
  21.        return ans;
  22. }
  23.  
  24. double pow2_rec(double x, int n) {
  25.  
  26.        double t;
  27.  
  28.        if(n == 0) return 1;       
  29.  
  30.        t = pow2_rec(x, n / 2);
  31.  
  32.        if( (n & 1) == 0) return t * t;
  33.        //if( n % 2 == 0) return t * t;
  34.  
  35.              else {
  36.  
  37.                    if(n > 0) return x * t * t; 
  38.  
  39.                    else 
  40.  
  41.                             return  (t * t) / x;
  42.               }
  43. }	
  44.  
  45. int main(int argc, char const *argv[])
  46. {
  47. 	printf("%f\n", pow(3.14, -5));
  48.  
  49. 	printf("%f\n", pow2(3.14, -5));
  50.  
  51. 	printf("%f\n", pow2_rec(3.14, -5));
  52.  
  53. 	return 0;
  54. }
Success #stdin #stdout 0s 4376KB
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Standard input is empty
stdout
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0.003276
0.003276
0.003276
https://ideone.com/Mygt0h
Pow in logN Complexity time and O(1)
language:
C (gcc 8.3)
created:
3 years ago
visibility:
secret

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