372. Super Pow

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Your task is to calculate ab mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array.

Example1:

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a = 2
b = [3]
Result: 8

Example2:

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a = 2
b = [1,0]
Result: 1024

解法1： 二分分割

参考了这篇解答1,主要的思想还是做二分分割。用到的性质是

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(ab)%x = (a%x)*(b%x) % x

然后判断下是否是偶数或者奇数。这里需要记忆一个高精度除法的计算方法。

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class Solution {
    public static int DIV = 1337;
    public void divide(int[] b, int divisor) {
        int temp = 0;
        for (int i = 0; i < b.length; i++) {
            b[i] += temp * 10;
            temp = b[i] % divisor;
            b[i] /= divisor;
        }
    }
    public boolean isZero(int[] b) {
        for (int i = 0; i < b.length; i++) {
            if (b[i] != 0) {
                return false;
            }
        }
        return true;
    }
    public int superPow(int a, int[] b) {
        if (isZero(b)) {
            return 1;
        }
        a %= Solution.DIV;  // take a % first
        boolean isEven = false;
        if (b[b.length - 1] % 2 == 0) {
            isEven = true;
        }
        divide(b, 2);
        int sub = superPow(a, b);
        sub %= Solution.DIV;
        sub *= sub;
        sub %= Solution.DIV;
        if (!isEven) {
            sub *= a;
            sub %= Solution.DIV;
        }
        return sub;
    }
}