高精度加法 —— 模板题 AcWing 791. 高精度加法// C A B, A 0, B 0 vectorint add(vectorint A, vectorint B) { if (A.size() B.size()) return add(B, A); vectorint C; int t 0; for (int i 0; i A.size(); i ) { t A[i]; if (i B.size()) t B[i]; C.push_back(t % 10); t / 10; } if (t) C.push_back(t); return C; }高精度减法 —— 模板题 AcWing 792. 高精度减法// C A - B, 满足A B, A 0, B 0 vectorint sub(vectorint A, vectorint B) { vectorint C; for (int i 0, t 0; i A.size(); i ) { t A[i] - t; if (i B.size()) t - B[i]; C.push_back((t 10) % 10); if (t 0) t 1; else t 0; } while (C.size() 1 C.back() 0) C.pop_back(); return C; }高精度乘低精度 —— 模板题 AcWing 793. 高精度乘法// C A * b, A 0, b 0 vectorint mul(vectorint A, int b) { vectorint 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; }高精度除以低精度 —— 模板题 AcWing 794. 高精度除法// A / b C ... r, A 0, b 0 vectorint div(vectorint A, int b, int r) { vectorint 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; }
高精度加法 —— 模板题 AcWing 791. 高精度加法// C A B, A 0, B 0 vectorint add(vectorint A, vectorint B) { if (A.size() B.size()) return add(B, A); vectorint C; int t 0; for (int i 0; i A.size(); i ) { t A[i]; if (i B.size()) t B[i]; C.push_back(t % 10); t / 10; } if (t) C.push_back(t); return C; }高精度减法 —— 模板题 AcWing 792. 高精度减法// C A - B, 满足A B, A 0, B 0 vectorint sub(vectorint A, vectorint B) { vectorint C; for (int i 0, t 0; i A.size(); i ) { t A[i] - t; if (i B.size()) t - B[i]; C.push_back((t 10) % 10); if (t 0) t 1; else t 0; } while (C.size() 1 C.back() 0) C.pop_back(); return C; }高精度乘低精度 —— 模板题 AcWing 793. 高精度乘法// C A * b, A 0, b 0 vectorint mul(vectorint A, int b) { vectorint 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; }高精度除以低精度 —— 模板题 AcWing 794. 高精度除法// A / b C ... r, A 0, b 0 vectorint div(vectorint A, int b, int r) { vectorint 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; }
