LeetCode:72. 超级次方

class Solution {

private:

int base=1337;

//求a的k次幂并与base求模

int mypow(int a,int k){

int res=1;

a%=base;

for(int i=0;i<k;i++){

res*=a;

res%=base;

}

return res;

}

public:

int superPow(int a, vector<int>& b) {

if(b.empty()) return 1;

int last=b.back();

b.pop_back();

int part1=mypow(a,last);

int part2=mypow(superPow(a,b),10);

return (part1*part2)%base;

}

};