Super Pow
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:
a = 2
b = [3]
Result: 8
Example2:
a = 2
b = [1,0]
Result: 1024 解析：
每一位一位的的求，当前位为之前所有位的结果的10次方再乘以a的当前位次方。
代码：
class Solution {
public:
int spow(int a,int b)
{
int base=1337;
a%=1337;
int res=1;
for (int i=0; i<b; i++)
{
res*=a;
res%=base;
}
return res;
}
int superPow(int a, vector<int>& b) {
int ans=1;
int base=1337;
for (int i=0; i<b.size(); i++)
{
ans=(spow(ans,10)*spow(a,b[i]))
