高精度算法总结

高精度加法

题目链接：

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

代码模版：

 1 #include <iostream>
 2 #include <vector>
 3 
 4 using namespace std;
 5 
 6 // C = A + B
 7 vector<int> add(vector<int> &A, vector<int> &B)
 8 {
 9     vector<int> C;
10     
11     int t = 0;  // 进位
12     for (int i = 0; i < A.size() || i < B.size(); i++)
13     {
14         if (i < A.size()) t += A[i];
15         if (i < B.size()) t += B[i];
16         C.push_back(t % 10);
17         t /= 10;
18     }
19     
20     if (t) C.push_back(1);
21     return C;
22 }
23 
24 int main()
25 {
26     string a, b;
27     vector<int> A, B;
28     
29     cin >> a >> b;  // a = "123456"
30     for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');  // A = [6, 5, 4, 3, 2, 1]
31     for (int i = b.size() - 1; i >= 0; i--) B.push_back(b[i] - '0');
32     
33     auto C = add(A, B);
34     
35     for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);
36     return 0;
37 }

高精度减法

题目链接：

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

代码模版：

 1 #include <iostream>
 2 #include <vector>
 3 
 4 using namespace std;
 5 
 6 // 判断是否有 A >= B
 7 bool cmp(vector<int> &A, vector<int> &B)
 8 {
 9     if (A.size() != B.size()) return A.size() > B.size();
10     for (int i = A.size() - 1; i >= 0; i--)
11         if (A[i] != B[i])
12             return A[i] > B[i];
13     return true;
14 }
15 
16 // C = A - B
17 vector<int> sub(vector<int> &A, vector<int> &B)
18 {
19     vector<int> C;
20     for (int i = 0, t = 0; i < A.size(); i++)
21     {
22         t = A[i] - t;
23         if (i < B.size()) t -= B[i];
24         C.push_back((t + 10) % 10);
25         if (t < 0) t = 1;
26         else t = 0;
27     }
28     
29     while (C.size() > 1 && C.back() == 0) C.pop_back();  // 去掉前导0
30     
31     return C;
32 }
33 
34 int main()
35 {
36     string a, b;
37     vector<int> A, B;
38     
39     cin >> a >> b;  // a = "123456"
40     for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');  // A = [6, 5, 4, 3, 2, 1]
41     for (int i = b.size() - 1; i >= 0; i--) B.push_back(b[i] - '0');
42     
43     if (cmp(A, B))
44     {
45         auto C = sub(A, B);
46         
47         for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);
48     }
49     else
50     {
51         auto C = sub(B, A);
52         
53         printf("-");
54         for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);
55     }
56     return 0;
57 }

高精度乘法

题目链接：

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

代码模版：

 1 #include <iostream>
 2 #include <vector>
 3 
 4 using namespace std;
 5 
 6 // C = A * b
 7 vector<int> mul(vector<int> &A, int b)
 8 {
 9     vector<int> C;
10     
11     int t = 0;  // 进位
12     for (int i = 0; i < A.size() || t; i++)
13     {
14         if (i < A.size()) t += A[i] * b;
15         C.push_back(t % 10);
16         t /= 10;
17     }
18     
19     while (C.size() > 1 && !C.back()) C.pop_back();
20     
21     return C;
22 }
23 
24 int main()
25 {
26     string a;
27     int b;
28     vector<int> A;
29     
30     cin >> a >> b;
31     for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');
32     
33     auto C = mul(A, b);
34     
35     for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);
36     return 0;
37 }

高精度除法

题目链接：

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

代码模版：

 1 #include <iostream>
 2 #include <vector>
 3 #include <algorithm>
 4 
 5 using namespace std;
 6 
 7 // A / b，商是C，余数是r
 8 vector<int> div(vector<int> &A, int b, int &r)  // r是引用
 9 {
10     vector<int> C;  // 商
11     r = 0;
12     for (int i = A.size() - 1; i >= 0; i--)
13     {
14         r = r * 10 + A[i];
15         C.push_back(r / b);
16         r %= b;
17     }
18     
19     reverse(C.begin(), C.end());
20     while (C.size() > 1 && !C.back()) C.pop_back();
21     
22     return C;
23 }
24 
25 int main()
26 {
27     string a;
28     int b;
29     vector<int> A;
30     
31     cin >> a >> b;
32     for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');
33     
34     int r;
35     auto C = div(A, b, r);
36     
37     for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);
38     cout << endl << r << endl;
39     
40     return 0;
41 }