Super Pow
<p>Your task is to calculate <code>a<sup>b</sup></code> mod <code>1337</code> where <code>a</code> is a positive integer and <code>b</code> is an extremely large positive integer given in the form of an array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> a = 2, b = [3]
<strong>Output:</strong> 8
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> a = 2, b = [1,0]
<strong>Output:</strong> 1024
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> a = 1, b = [4,3,3,8,5,2]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= a &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>1 &lt;= b.length &lt;= 2000</code></li>
	<li><code>0 &lt;= b[i] &lt;= 9</code></li>
	<li><code>b</code> does not contain leading zeros.</li>
</ul>
Math; Divide and Conquer
class Solution {
public:
    int superPow(int a, vector<int>& b) {
        using ll = long long;
        const int mod = 1337;
        ll ans = 1;
        auto qpow = [&](ll a, int n) {
            ll ans = 1;
            for (; n; n >>= 1) {
                if (n & 1) {
                    ans = ans * a % mod;
                }
                a = a * a % mod;
            }
            return (int) ans;
        };
        for (int i = b.size() - 1; ~i; --i) {
            ans = ans * qpow(a, b[i]) % mod;
            a = qpow(a, 10);
        }
        return ans;
    }
};