acwing演算法第二課
阿新 • • 發佈:2022-04-13
一、高精度加法--A+B 106
大整數從個位(小位)開始存在數組裡
// C = A + B, A >= 0, B >= 0 vector<int> add(vector<int> &A, vector<int> &B) { if (A.size() < B.size()) return add(B, A); vector<int> 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; } 作者:yxc 連結:https://www.acwing.com/blog/content/277/ 來源:AcWing 著作權歸作者所有。商業轉載請聯絡作者獲得授權,非商業轉載請註明出處。
二、高精度減法
和加法差不多,但要先判斷A是否大於B,要分情況討論。
1 vector<int> sub(vector<int> &A, vector<int> &B) 2 { 3 vector<int> C; 4 for (int i = 0, t = 0; i < A.size(); i ++ ) 5 { 6 t = A[i] - t; 7 if (i < B.size()) t -= B[i]; 8 C.push_back((t + 10) % 10); 9 if (t < 0) t = 1; 10 else t = 0; 11 } 12 13 while (C.size() > 1&& C.back() == 0) C.pop_back(); 14 return C; 15 } 16 17 作者:yxc 18 連結:https://www.acwing.com/blog/content/277/ 19 來源:AcWing 20 著作權歸作者所有。商業轉載請聯絡作者獲得授權,非商業轉載請註明出處。
三、高精度乘法
高精度乘低精度:
1 // C = A * b, A >= 0, b >= 0 2 vector<int> mul(vector<int> &A, int b) 3 { 4 vector<int> C; 5 6 int t = 0; 7 for (int i = 0; i < A.size() || t; i ++ ) 8 { 9 if (i < A.size()) t += A[i] * b; 10 C.push_back(t % 10); 11 t /= 10; 12 } 13 14 while (C.size() > 1 && C.back() == 0) C.pop_back(); 15 16 return C; 17 } 18 19 作者:yxc 20 連結:https://www.acwing.com/blog/content/277/ 21 來源:AcWing 22 著作權歸作者所有。商業轉載請聯絡作者獲得授權,非商業轉載請註明出處。
四、高精度除法
高精度除以低精度:
1 // A / b = C ... r, A >= 0, b > 0 2 vector<int> div(vector<int> &A, int b, int &r) 3 { 4 vector<int> C; 5 r = 0; 6 for (int i = A.size() - 1; i >= 0; i -- ) 7 { 8 r = r * 10 + A[i]; 9 C.push_back(r / b); 10 r %= b; 11 } 12 reverse(C.begin(), C.end()); 13 while (C.size() > 1 && C.back() == 0) C.pop_back(); 14 return C; 15 } 16 17 作者:yxc 18 連結:https://www.acwing.com/blog/content/277/ 19 來源:AcWing 20 著作權歸作者所有。商業轉載請聯絡作者獲得授權,非商業轉載請註明出處。
如何將高精度整數以字串寫入,再讀到陣列中:
五、字首和
一維:
S[i] = a[1] + a[2] + ... a[i] a[l] + ... + a[r] = S[r] - S[l - 1]
二維:
S[i, j] = 第i行j列格子左上部分所有元素的和 以(x1, y1)為左上角,(x2, y2)為右下角的子矩陣的和為: S[x2, y2] - S[x1 - 1, y2] - S[x2, y1 - 1] + S[x1 - 1, y1 - 1]
六、差分
一維:
給區間[l, r]中的每個數加上c:B[l] += c, B[r + 1] -= c
1 #include <iostream> 2 3 using namespace std; 4 5 const int N = 100010; 6 7 int n, m; 8 int a[N], b[N]; 9 10 void insert(int l, int r, int c) 11 { 12 b[l] += c; 13 b[r + 1] -= c; 14 } 15 16 int main() 17 { 18 scanf("%d%d", &n, &m); 19 for (int i = 1; i <= n; i ++ ) scanf("%d", &a[i]); 20 21 for (int i = 1; i <= n; i ++ ) insert(i, i, a[i]); 22 23 while (m -- ) 24 { 25 int l, r, c; 26 scanf("%d%d%d", &l, &r, &c); 27 insert(l, r, c); 28 } 29 30 for (int i = 1; i <= n; i ++ ) b[i] += b[i - 1]; 31 32 for (int i = 1; i <= n; i ++ ) printf("%d ", b[i]); 33 34 return 0; 35 } 36 37 作者:yxc 38 連結:https://www.acwing.com/activity/content/code/content/39799/ 39 來源:AcWing 40 著作權歸作者所有。商業轉載請聯絡作者獲得授權,非商業轉載請註明出處。
二維:
1 #include <iostream> 2 3 using namespace std; 4 5 const int N = 1010; 6 7 int n, m, q; 8 int a[N][N], b[N][N]; 9 10 void insert(int x1, int y1, int x2, int y2, int c) 11 { 12 b[x1][y1] += c; 13 b[x2 + 1][y1] -= c; 14 b[x1][y2 + 1] -= c; 15 b[x2 + 1][y2 + 1] += c; 16 } 17 18 int main() 19 { 20 scanf("%d%d%d", &n, &m, &q); 21 22 for (int i = 1; i <= n; i ++ ) 23 for (int j = 1; j <= m; j ++ ) 24 scanf("%d", &a[i][j]); 25 26 for (int i = 1; i <= n; i ++ ) 27 for (int j = 1; j <= m; j ++ ) 28 insert(i, j, i, j, a[i][j]); 29 30 while (q -- ) 31 { 32 int x1, y1, x2, y2, c; 33 cin >> x1 >> y1 >> x2 >> y2 >> c; 34 insert(x1, y1, x2, y2, c); 35 } 36 37 for (int i = 1; i <= n; i ++ ) 38 for (int j = 1; j <= m; j ++ ) 39 b[i][j] += b[i - 1][j] + b[i][j - 1] - b[i - 1][j - 1]; 40 41 for (int i = 1; i <= n; i ++ ) 42 { 43 for (int j = 1; j <= m; j ++ ) printf("%d ", b[i][j]); 44 puts(""); 45 } 46 47 return 0; 48 } 49 50 作者:yxc 51 連結:https://www.acwing.com/activity/content/code/content/39800/ 52 來源:AcWing 53 著作權歸作者所有。商業轉載請聯絡作者獲得授權,非商業轉載請註明出處。
淺淺記錄一下y總的課,順便搞個板子。