高精度乘法 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 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)

排序 重载运算符 1 2 3 4 5 bool operator < (const struct &a, const struct &b){ return your own rule; } sort(struct a, struct a + n); map 1 2 map<typeA,typeB> m; 映射typeA到typeB 方法 例子 [A] m[A]; //A映射的值 find m.find