高精度乘低精度

  1. //string版本
  2. // C = A * b, A >= 0, b > 0
  3. vector<int> mul(vector<int> &A, int b)
  4. {
  5.     vector<int> C;
  7.     int t = 0;
  8.     for (int i = 0; i < A.size() || t; i ++ )
  9.     {
  10.         if (i < A.size()) t += A[i] * b;
  11.         C.push_back(t % 10);
  12.         t /= 10;
  13.     }
  15.     while (C.size() > 1 && C.back() == 0) C.pop_back();
  17.     return C;
  18. }
  1. //一维数组版本
  2. void mul(int A[], int x)
  3. {
  4.     int t = 0;
  5.     for (int i = 0; i < len; i++)
  6.     {
  7.         A[i] = A[i] * x + t;
  9.         if (A[i] >= 10) t = 1;
  10.         else t = 0;
  12.         A[i] %= 10;
  13.     }
  14.     if (t) A[len++] = 1;
  15. }

例题：1170：计算2的N次方

高精度乘高精度

  1. vector<int> mul(vector<int> A, vector<int> B)
  2. {
  3.     vector<int> C(A.size() + B.size());
  5.     for (int i = 0; i < A.size(); i ++ )
  6.         for (int j = 0; j < B.size(); j ++ )
  7.             C[i + j] += A[i] * B[j];
  9.     for (int i = 0, t = 0; i < C.size() || t; i ++ )
  10.     {
  11.         t += C[i];
  12.         if (i >= C.size()) C.push_back(t % 10);
  13.         else C[i] = t % 10;
  14.         t /= 10;
  15.     }
  17.     while (C.size() > 1 && !C.back()) C.pop_back();
  19.     return C;
  20. }
  1. //一维数组版本
  2.     for (int i = a.size()- 1; i >= 0; i--) A[lena++] = a[i] - '0';
  3.     for (int i = b.size()- 1; i >= 0; i--) B[lenb++] = b[i] - '0';
  5.     for (int i = 0; i < lena; i++)
  6.         for (int j = 0; j < lenb; j++)
  7.             C[i + j] += A[i] * B[j];
  9.     for (int i = 0; i < lena + lenb; i++)
  10.     {
  11.         C[i + 1] += C[i] / 10;
  12.         C[i] %= 10;
  13.     }
  15.     int lenc = lena + lenb;
  16.     while (lenc > 0 && C[lenc] == 0) lenc--;
  17.     for (int i = lenc; i >= 0; i--) cout << C[i];
  18.     puts("");

例题：1174：大整数乘法