#93 High precision in addition to low precision
template
// A>=0, b>0, A/b = C...r
vector<int> div(vector<int>& A, int& b, int& r) {
vector<int> C;
r = 0;
for (int i = A.size() - 1; i >= 0; i--) {
r = r * 10 + A[i];
C.push_back(r / b);
r %= b;
}
reverse(C.begin(), C.end());
while (C.size() > 1 && C.back() == 0) C.pop_back();
return C;
}
train of thought
1. Multiply the remainder r of the previous round by 10+A[i]
md, don't know how to describe
