add exercises

data/1.算法初阶/5.蓝桥杯-模拟/18.螺旋折线/config.json

+{
+  "node_id": "algorithm-95fa7ac83e7d4750b7aae3d4c2b0e797",
+  "keywords": [
+    "蓝桥杯",
+    "螺旋折线"
+  ],
+  "children": [],
+  "export": [
+    "solution.json"
+  ]
+}
\ No newline at end of file

data/1.算法初阶/5.蓝桥杯-模拟/18.螺旋折线/desc.md

+如图所示的螺旋折线经过平面上所有整点恰好一次。  
+对于整点(X, Y)，我们定义它到原点的距离dis(X, Y)是从原点到(X, Y)的螺旋折线段的长度。  
+
+例如```dis(0, 1)=3, dis(-2, -1)=9```
+
+给出整点坐标(X, Y)，你能计算出dis(X, Y)吗？  
+
+#### 输入格式
+```
+X和Y
+
+对于40%的数据，-1000 <= X, Y <= 1000
+对于70%的数据，-100000 <= X， Y <= 100000
+对于100%的数据, -1000000000 <= X, Y <= 1000000000
+```
+#### 输出格式
+```
+输出dis(X, Y)
+```
+#### 样例输入
+```
+0 1
+```
+#### 样例输出
+```
+3
+```
\ No newline at end of file

data/1.算法初阶/5.蓝桥杯-模拟/18.螺旋折线/solution.cpp

+#include <iostream>
+#include <algorithm>
+using namespace std;
+
+int main()
+{
+    int x, y;
+    cin >> x >> y;
+    long long sum = 0;
+
+    if ((y < x) && (-y <= x))
+    {
+        sum += x > 1 ? (long long)(4 * (2 + (abs(x) - 1) * (abs(x) - 2))) : 0; //求前n个正方形周长的等差数列和
+        sum += (long long)x > 0 ? (4 * x + (x - y)) : (y - x);
+    }
+    else
+    {
+        sum += y > 1 ? (long long)(4 * (2 + (abs(y) - 1) * (abs(y) - 2))) : 0; //求前n个正方形周长的等差数列和
+        sum += (long long)y > 0 ? (2 * y + y + x) : (6 * y - y - x);
+    }
+
+    cout << sum << endl;
+    return 0;
+}

data/1.算法初阶/5.蓝桥杯-模拟/18.螺旋折线/solution.json

+{
+  "type": "code_options",
+  "author": "CSDN.net",
+  "source": "solution.md",
+  "exercise_id": "f4968aaaa7a34063a534a664765f429d"
+}
\ No newline at end of file

data/1.算法初阶/5.蓝桥杯-模拟/18.螺旋折线/solution.md

+# 螺旋折线
+
+如图所示的螺旋折线经过平面上所有整点恰好一次。  
+对于整点(X, Y)，我们定义它到原点的距离dis(X, Y)是从原点到(X, Y)的螺旋折线段的长度。  
+
+例如`dis(0, 1)=3, dis(-2, -1)=9`
+
+给出整点坐标(X, Y)，你能计算出dis(X, Y)吗？  
+
+
+**输入格式**
+
+```
+X和Y
+
+对于40%的数据，-1000 <= X, Y <= 1000
+对于70%的数据，-100000 <= X， Y <= 100000
+对于100%的数据, -1000000000 <= X, Y <= 1000000000
+```
+
+**输出格式**
+
+```
+输出dis(X, Y)
+```
+
+**样例输入**
+
+```
+0 1
+```
+
+**样例输出**
+
+```
+3
+```
+
+以下程序实现了该功能，请你补全空白处内容：
+
+```cpp
+#include <bits/stdc++.h>
+using namespace std;
+
+int main()
+{
+    int x, y;
+    cin >> x >> y;
+    long long sum = 0;
+
+    if ((y < x) && (-y <= x))
+    {
+        sum += x > 1 ? (long long)(4 * (2 + (abs(x) - 1) * (abs(x) - 2))) : 0;
+        sum += (long long)x > 0 ? (4 * x + (x - y)) : (y - x);
+    }
+    else
+    {
+        sum += y > 1 ? (long long)(4 * (2 + (abs(y) - 1) * (abs(y) - 2))) : 0;
+        __________________
+    }
+
+    cout << sum << endl;
+    return 0;
+}
+```
+
+## aop
+
+### before
+
+```cpp
+
+```
+
+### after
+
+```cpp
+
+```
+
+## 答案
+
+```cpp
+sum += (long long)y > 0 ? (2 * y + y + x) : (6 * y - y - x);
+```
+
+## 选项
+
+### A
+
+```cpp
+sum += (long long)y > 0 ? (2 * y + x) : (6 * y - y - x);
+```
+
+### B
+
+```cpp
+sum += (long long)y > 0 ? (2 * y + y + x) : (6 * y - x);
+```
+
+### C
+
+```cpp
+sum += (long long)y > 0 ? (2 * y + x) : (6 * y - x);
+```

data/1.算法初阶/7.蓝桥杯-动态规划/3.装饰珠/solution.md

@@ -46,6 +46,7 @@
 **样例说明**
 
 按照如下方式镶嵌珠子得到最大价值 20，括号内表示镶嵌的装饰珠的种类编号：
+
 ```
 1: (1)
 2: (1) (2)
@@ -54,6 +55,7 @@
 5: (1)
 6: (2)
 ```
+
 4 颗技能 1 装饰珠，4 颗技能 2 装饰珠 $W_1(4) + W_2(4) = 5 + 15 = 20。W_1(4)+W_2(4)=5+15=20$。
 
 以下<span style="color:red">错误</span>的一项是？

data_backup/3.蓝桥杯/75.螺旋折线/solution.md

@@ -3,7 +3,7 @@
 如图所示的螺旋折线经过平面上所有整点恰好一次。  
 对于整点(X, Y)，我们定义它到原点的距离dis(X, Y)是从原点到(X, Y)的螺旋折线段的长度。  
 
-例如```dis(0, 1)=3, dis(-2, -1)=9```
+例如`dis(0, 1)=3, dis(-2, -1)=9`
 
 给出整点坐标(X, Y)，你能计算出dis(X, Y)吗？  
 
@@ -36,6 +36,34 @@ X和Y
 3
 ```
 
+以下程序实现了该功能，请你补全空白处内容：
+
+```cpp
+#include <bits/stdc++.h>
+using namespace std;
+
+int main()
+{
+    int x, y;
+    cin >> x >> y;
+    long long sum = 0;
+
+    if ((y < x) && (-y <= x))
+    {
+        sum += x > 1 ? (long long)(4 * (2 + (abs(x) - 1) * (abs(x) - 2))) : 0;
+        sum += (long long)x > 0 ? (4 * x + (x - y)) : (y - x);
+    }
+    else
+    {
+        sum += y > 1 ? (long long)(4 * (2 + (abs(y) - 1) * (abs(y) - 2))) : 0;
+        __________________
+    }
+
+    cout << sum << endl;
+    return 0;
+}
+```
+
 ## aop
 
 ### before
@@ -53,7 +81,7 @@ X和Y
 ## 答案
 
 ```cpp
-
+sum += (long long)y > 0 ? (2 * y + y + x) : (6 * y - y - x);
 ```
 
 ## 选项
@@ -61,17 +89,17 @@ X和Y
 ### A
 
 ```cpp
-
+sum += (long long)y > 0 ? (2 * y + x) : (6 * y - y - x);
 ```
 
 ### B
 
 ```cpp
-
+sum += (long long)y > 0 ? (2 * y + y + x) : (6 * y - x);
 ```
 
 ### C
 
 ```cpp
-
+sum += (long long)y > 0 ? (2 * y + x) : (6 * y - x);
 ```