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69 lines (62 loc) · 1.47 KB
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// Source : https://oj.leetcode.com/problems/powx-n/
// Author : Hao Chen
// Date : 2014-06-25
/**********************************************************************************
*
* Implement pow(x, n).
*
*
**********************************************************************************/
#include <stdio.h>
#include <stdlib.h>
/*
* Basically, most people think this is very easy as below:
*
* double result = 1.0;
* for (int i=0; i<n; i++){
* result *=x;
* }
*
* However,
*
* 1) We need think about the `n` is negtive number.
*
* 2) We need more wisely deal with the following cases:
*
* pow(1, MAX_INT);
* pow(-1,BIG_INT);
* pow(2, BIG_INT);
*
* To deal with such kind case, we can use x = x*x to reduce the `n` more quickly
*
* so, if `n` is an even number, we can `x = x*x`, and `n = n>>1;`
* if `n` is an odd number, we can just `result *= x;`
*
*/
double pow(double x, int n) {
bool sign = false;
unsigned int exp = n;
if(n<0){
exp = -n;
sign = true;
}
double result = 1.0;
while (exp) {
if (exp & 1){
result *= x;
}
exp >>= 1;
x *= x;
}
return sign ? 1/result : result;
}
int main(int argc, char** argv){
double x=2.0;
int n = 3;
if (argc==3){
x = atof(argv[1]);
n = atoi(argv[2]);
}
printf("%f\n", pow(x, n));
return 0;
}