高精度乘法模板
- 高精度乘低精度
// C = A * b, A >= 0, b >= 0
vector<int> mul(vector<int> &A, int b)
{
vector<int> C;
int t = 0;
for (int i = 0; i < A.size() || t; i ++ )
{
if (i < A.size()) t += A[i] * b;
C.push_back(t % 10);
t /= 10;
}
while (C.size() > 1 && C.back() == 0) C.pop_back();
return C;
}
https://www.acwing.com/blog/content/14887/
有高精度*高精度