Lintcode: Fast Power

Calculate the a^n % b where a, b and n are all 32bit integers.



Example

For 2^31 % 3 = 2



For 100^1000 % 1000 = 0



Challenge

O(logn)

这道题跟Pow这道题很像

数学问题，要求O(log n)的时间复杂度，也就是每次去掉一半的计算量，先要找到对应的数学公式：
(a * b) % p = (a % p * b % p) % p

所以(a^n)%b = (a^n/2 * a^n/2 * a) % b = ((a^n/2 * a^n/2)%b * a%b) % b
注意int和long的转化，防止溢出。

比较严谨的做法：

 1 class Solution {

 2     /*

 3      * @param a, b, n: 32bit integers

 4      * @return: An integer

 5      */

 6     public int fastPower(int a, int b, int n) {

 7         // write your code here

 8         long ret = getPower(a, b, n);

 9         return (int)ret;

10     }

11     public long getPower(int a, int b, int n){

12         if(a == 0) return 0;

13         if(n == 0) return 1 % b;

14         if(n == 1) return a % b;

15          

16         long ret = getPower(a, b, n/2);

17         ret *= ret;

18         ret %= b;

19         if(n % 2 == 1){

20             ret = ret * (a % b);

21         }

22         return ret % b;

23     }

24 };

我当时的做法：

 1 class Solution {

 2     public int fastPower(int a, int b, int n) {

 3         if (n == 0) return 1%b;

 4         long half = fastPower(a, b, n/2);

 5         if (n % 2 == 0) {

 6             long temp = half * half;

 7             return (int)(temp % b);

 8         }

 9         else {

10             long temp = (half * half)%b * (a%b);

11             return (int)(temp % b);

12         }

13     }

14 };

 1 class Solution {

 2     /*

 3      * @param a, b, n: 32bit integers

 4      * @return: An integer

 5      */

 6     public int fastPower(int a, int b, int n) {

 7         long ret = helper(a, b, n);

 8         return (int)ret;

 9     }

10     

11     public long helper(int a, int b, int n) {

12         if (n == 0) return 1%b;

13         long half = helper(a, b, n/2);

14         if (n % 2 == 0) {

15             return half * half % b;

16         }

17         else {

18             return ((half * half % b) * a%b)%b;

19         }

20     }

21 };