高精度模版

Posted 2020-11-27 hhyx

 1 // 高精度加法
 2 // C = A + B, A >= 0, B >= 0
 3 vector<int> add(vector<int> &A, vector<int> &B)
 4 {
 5     if (A.size() < B.size()) return add(B, A);
 6 
 7     vector<int> C;
 8 
 9     int t = 0;
10     for (int i = 0; i < A.size(); i ++ )
11     {
12         t += A[i];
13         if (i < B.size()) t += B[i];
14         C.push_back(t % 10);
15         t /= 10;
16     }
17 
18     if (t) C.push_back(t);
19     return C;
20 }
21 
22 
23 // 高精度减法
24 // C = A - B, 满足A >= B, A >= 0, B >= 0
25 bool cmp(vector<int> &A, vector<int> &B) {
26     if (A.size() != B.size())
27         return A.size() > B.size();
28     for (int i = A.size() - 1; i >= 0; i--)
29         if (A[i] != B[i])
30             return A[i] > B[i];
31     return 1;
32 }
33 vector<int> sub(vector<int> &A, vector<int> &B) { //倒序传入
34     if (cmp(A, B))
35         return sub(A, B);
36     vector<int> C;
37     int t = 0; //表示借位
38     for (int i = 0; i < A.size(); i++) {
39         t = A[i] - t;
40         if (i < B.size())
41             t = t - B[i];
42         C.push_back((t + 10) % 10);
43         if (t < 0)
44             t = 1;
45         else
46             t = 0;
47     }
48     while (C.size() > 1 && C.back() == 0)
49         C.pop_back();
50     return C;
51 }
52 
53 
54 
55 // 高精度乘低精度
56 // C = A * b, A >= 0, b > 0
57 vector<int> mul(vector<int> &A, int b)
58 {
59     vector<int> C;
60     int t = 0;
61     for (int i = 0; i < A.size() || t; i ++ )
62     {
63         if (i < A.size()) t += A[i] * b;
64         C.push_back(t % 10);
65         t /= 10;
66     }
67 
68     return C;
69 }
70 // for(int i = 0;i<A.size();i++){
71 //     t+=A[i]*B;
72 //     C.push_back(t%10);
73 //     t/=10;
74 // }
75 // while(t){
76 //     C.push_back(t%10);
77 //     t/=10;
78 // }
79 
80 
81 // 高精度除以低精度
82 // A / b = C ... r, A >= 0, b > 0
83 vector<int> div(vector<int> &A, int b, int &r)
84 {
85     vector<int> C;
86     r = 0;
87     for (int i = A.size() - 1; i >= 0; i -- )
88     {
89         r = r * 10 + A[i];
90         C.push_back(r / b);
91         r %= b;
92     }
93     reverse(C.begin(), C.end());
94     while (C.size() > 1 && C.back() == 0) C.pop_back();
95     return C;
96 }
97 //高精度除法