高精度乘法

      #include <iostream>
      #include <vector>
      using namespace std;
      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;
      }
      int main()
      {
          string a;
         int b;
          cin >> a >> b;
          vector<int> A;
         for (int i = a.size() - 1; i >= 0; i -- ) A.push_back(a[i] - '0');
         auto C = mul(A, b);
         for (int i = C.size() - 1; i >= 0; i -- ) printf("%d", C[i]);
         return 0;
      }
  
 

同样，模拟即可，注意本题是高精*低精。

文章来源: blog.csdn.net，作者：irrationality，版权归原作者所有，如需转载，请联系作者。

原文链接：blog.csdn.net/weixin_54227557/article/details/120595091