7.高精度加法

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 vector<int> add(vector<int> &A, vector<int> &B) { //加上引用，不然很慢 
 4     vector<int> C; //答案 
 5     int t = 0; //进位 
 6     for (int i = 0; i < A.size() || i < B.size(); i++) {
 7         //让t来表示三个数的和 
 8         if (i < A.size()) {
 9             t += A[i];
10         }
11         if (i < B.size()) {
12             t += B[i];
13         }
14         C.push_back(t % 10);
15         t /= 10;
16     }
17     if (t) {
18         C.push_back(1);
19     }
20     return C;
21 }
22 int main() {
23     string a, b; //太长了，所以用字符串来读入 
24     cin >> a >> b;
25     vector<int> A, B; //然后存进vector 
26     for (int i = a.length() - 1; i >= 0; i--) {
27         A.push_back(a[i] - '0'); //字符变数字 
28     }
29     for (int i = b.length() - 1; i >= 0; i--) {
30         B.push_back(b[i] - '0');
31     }
32     vector<int> C = add(A, B);
33     for (int i = C.size() - 1; i >= 0; i--) {
34         cout << C[i];
35     }
36     return 0;
37 }

2020年9月29日更新： 高精度压位压8位的做法

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int base = 100000000;
 4 vector<int> add(vector<int> &A, vector<int> &B) {
 5     vector<int> C;
 6     int t = 0;
 7     for (int i = 0; i < A.size() || i < B.size(); i++) {
 8         if (i < A.size()) {
 9             t += A[i];
10         }
11         if (i < B.size()) {
12             t += B[i];
13         }
14         C.push_back(t % base);
15         t /= base;
16     }
17     if (t) {
18         C.push_back(t);
19     }
20     return C;
21 }
22 int main() {
23     string a, b;
24     cin >> a >> b;
25     vector<int> A, B, C;
26     for (int i = a.size() - 1, s = 0, j = 0, t = 1; i >= 0; i--) {
27         s += (a[i] - '0') * t;
28         j++;
29         t *= 10;
30         if (j == 8 || i == 0) {
31             A.push_back(s);
32             s = 0;
33             j = 0;
34             t = 1;
35         }
36     }
37     for (int i = b.size() - 1, s = 0, j = 0, t = 1; i >= 0; i--) {
38         s += (b[i] - '0') * t;
39         j++;
40         t *= 10;
41         if (j == 8 || i == 0) {
42             B.push_back(s);
43             s = 0; 
44             j = 0;
45             t = 1;
46         }
47     }
48     C = add(A, B);
49     cout << C.back();
50     for (int i = C.size() - 2; i >= 0; i--) {
51         printf("%08d", C[i]);
52     }
53     return 0;
54 }