高精度加法

题目链接：

https://www.acwing.com/activity/content/problem/content/825/

代码模版：

#include <iostream>
#include <vector>using namespace std;// C = A + B
vector<int> add(vector<int> &A, vector<int> &B)
{vector<int> C;int t = 0;  // 进位for (int i = 0; i < A.size() || i < B.size(); i++){if (i < A.size()) t += A[i];if (i < B.size()) t += B[i];C.push_back(t % 10);t /= 10;}if (t) C.push_back(1);return C;
}int main()
{string a, b;vector<int> A, B;cin >> a >> b;  // a = "123456"for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');  // A = [6, 5, 4, 3, 2, 1]for (int i = b.size() - 1; i >= 0; i--) B.push_back(b[i] - '0');auto C = add(A, B);for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);return 0;
}

高精度减法

题目链接：

https://www.acwing.com/activity/content/problem/content/826/

代码模版：

#include <iostream>
#include <vector>using namespace std;// 判断是否有 A >= B
bool cmp(vector<int> &A, vector<int> &B)
{if (A.size() != B.size()) return A.size() > B.size();for (int i = A.size() - 1; i >= 0; i--)if (A[i] != B[i])return A[i] > B[i];return true;
}// C = A - B
vector<int> sub(vector<int> &A, vector<int> &B)
{vector<int> 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();  // 去掉前导0return C;
}int main()
{string a, b;vector<int> A, B;cin >> a >> b;  // a = "123456"for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');  // A = [6, 5, 4, 3, 2, 1]for (int i = b.size() - 1; i >= 0; i--) B.push_back(b[i] - '0');if (cmp(A, B)){auto C = sub(A, B);for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);}else{auto C = sub(B, A);printf("-");for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);}return 0;
}

高精度乘法

题目链接：

https://www.acwing.com/problem/content/795/

代码模版：

#include <iostream>
#include <vector>using namespace std;// C = A * b
vector<int> mul(vector<int> &A, int b)
{vector<int> C;int t = 0;  // 进位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()) C.pop_back();return C;
}int main()
{string a;int b;vector<int> A;cin >> a >> b;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;
}

高精度除法

题目链接：

https://www.acwing.com/problem/content/796/

代码模版：

#include <iostream>
#include <vector>
#include <algorithm>using namespace std;// A / b，商是C，余数是r
vector<int> div(vector<int> &A, int b, int &r)  // r是引用
{vector<int> C;  // 商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()) C.pop_back();return C;
}int main()
{string a;int b;vector<int> A;cin >> a >> b;for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');int r;auto C = div(A, b, r);for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);cout << endl << r << endl;return 0;
}