update exercises

data/1.算法初阶/1.蓝桥杯/分类计数/solution.md

@@ -26,7 +26,8 @@
 ## aop
 ### before
 ```cpp
-
+#include <iostream>
+using namespace std;
 ```
 ### after
 ```cpp
@@ -35,21 +36,122 @@
 
 ## 答案
 ```cpp
+int main(int argc, char **argv)
+{
+    string str;
+    cin >> str;
+    int A = 0, a = 0, number = 0;
+    int len = str.length();
+    for (int i = 0; i < len; i++)
+    {
+        if (str[i] <= '9' && str[i] >= '0')
+        {
+            number++;
+        }
+        if (str[i] <= 'Z' && str[i] >= 'A')
+        {
+            A++;
+        }
+        if (str[i] <= 'z' && str[i] >= 'a')
+        {
+            a++;
+        }
+    }
+    cout << A << endl;
+    cout << a << endl;
+    cout << number << endl;
+    return 0;
+}
 
 ```
 ## 选项
 
 ### A
 ```cpp
-
+int main(int argc, char **argv)
+{
+    string str;
+    cin >> str;
+    int A = 0, a = 0, number = 0;
+    int len = str.length();
+    for (int i = 0; i < len; i++)
+    {
+        if (str[i] < '9' && str[i] > '0')
+        {
+            number++;
+        }
+        if (str[i] < 'Z' && str[i] > 'A')
+        {
+            A++;
+        }
+        if (str[i] < 'z' && str[i] > 'a')
+        {
+            a++;
+        }
+    }
+    cout << A << endl;
+    cout << a << endl;
+    cout << number << endl;
+    return 0;
+}
 ```
 
 ### B
 ```cpp
-
+int main(int argc, char **argv)
+{
+    string str;
+    cin >> str;
+    int A = 0, a = 0, number = 0;
+    int len = str.length();
+    for (int i = 0; i < len; i++)
+    {
+        if (str[i] < 71 && str[i] > 60)
+        {
+            number++;
+        }
+        if (str[i] < 132 && str[i] > 101)
+        {
+            A++;
+        }
+        if (str[i] < 172 && str[i] > 141)
+        {
+            a++;
+        }
+    }
+    cout << A << endl;
+    cout << a << endl;
+    cout << number << endl;
+    return 0;
+}
 ```
 
 ### C
 ```cpp
-
+int main(int argc, char **argv)
+{
+    string str;
+    cin >> str;
+    int A = 0, a = 0, number = 0;
+    int len = str.length();
+    for (int i = 0; i < len; i++)
+    {
+        if (str[i] < 39 && str[i] > 30)
+        {
+            number++;
+        }
+        if (str[i] < 132 && str[i] > 101)
+        {
+            A++;
+        }
+        if (str[i] <= 'z' && str[i] >= 'a')
+        {
+            a++;
+        }
+    }
+    cout << A << endl;
+    cout << a << endl;
+    cout << number << endl;
+    return 0;
+}
 ```

data/1.算法初阶/1.蓝桥杯/分配口罩/solution.md

@@ -14,30 +14,80 @@ masks = [9090400, 8499400, 5926800, 8547000, 4958200, 4422600, 5751200, 4175600,
 ## aop
 ### before
 ```cpp
+#include <iostream>
+#include <algorithm>
+#include <cmath>
+using namespace std;
+long int masks[15] = {9090400, 8499400, 5926800, 8547000, 4958200,
+                      4422600, 5751200, 4175600, 6309600,
+                      5865200, 6604400, 4635000, 10663400, 8087200, 4554000};
 
+long ans = 1000000000;
 ```
 ### after
 ```cpp
+int main()
+{
+    dfs(0, 0, 0);
+    cout << ans;
+}
 
 ```
 
 ## 答案
 ```cpp
+void dfs(int n, long h1, long h2)
+{
+    if (n == 15)
+    {
+        ans = min(ans, abs(h1 - h2));
+        return;
+    }
+    dfs(n + 1, h1 + masks[n], h2);
+    dfs(n + 1, h1, h2 + masks[n]);
+}
 
 ```
 ## 选项
 
 ### A
 ```cpp
-
+void dfs(int n, long h1, long h2)
+{
+    if (n == 15)
+    {
+        ans = min(ans, abs(h1 - h2));
+        return;
+    }
+    dfs(n + 1, h1 + masks[n + 1], h2);
+    dfs(n + 1, h1, h2 + masks[n + 1]);
+}
 ```
 
 ### B
 ```cpp
-
+void dfs(int n, long h1, long h2)
+{
+    if (n == 15)
+    {
+        ans = min(ans, abs(h1 - h2));
+        return;
+    }
+    dfs(n + 1, h1 + masks[n + 1], h2);
+    dfs(n + 1, h1, h2 + masks[n]);
+}
 ```
 
 ### C
 ```cpp
-
+void dfs(int n, long h1, long h2)
+{
+    if (n == 15)
+    {
+        ans = min(ans, abs(h1 - h2));
+        return;
+    }
+    dfs(n, h1 + masks[n], h2);
+    dfs(n, h1, h2 + masks[n]);
+}
 ```

data/1.算法初阶/1.蓝桥杯/斐波那契/solution.md

@@ -30,7 +30,7 @@ Fibonacci数列的递推公式为：Fn=Fn-1+Fn-2，其中F1=F2=1。
 ## aop
 ### before
 ```cpp
-
+#include <stdio.h>
 ```
 ### after
 ```cpp
@@ -39,21 +39,89 @@ Fibonacci数列的递推公式为：Fn=Fn-1+Fn-2，其中F1=F2=1。
 
 ## 答案
 ```cpp
+int main()
+{
+    int n, b;
+    scanf("%d", &n);
+    int a[n];
+    a[0] = a[1] = 1;
 
+    for (int i = 2; i < n; i++)
+    {
+        a[i] = (a[i - 1] + a[i - 2]) % 10007;
+        b = a[i];
+    }
+    if (n > 2)
+        printf("%d", b);
+    else
+        printf("1");
+    return 0;
+}
 ```
 ## 选项
 
 ### A
 ```cpp
+int main()
+{
+    int n, b;
+    scanf("%d", &n);
+    int a[n];
+    a[0] = a[1] = 1;
 
+    for (int i = 2; i < n; i++)
+    {
+        a[i] = (a[i - 1] + a[i - 2]) % 10007;
+        b = a[i] + 1;
+    }
+    if (n > 2)
+        printf("%d", b);
+    else
+        printf("1");
+    return 0;
+}
 ```
 
 ### B
 ```cpp
+int main()
+{
+    int n, b;
+    scanf("%d", &n);
+    int a[n];
+    a[0] = a[1] = 1;
 
+    for (int i = 2; i < n; i++)
+    {
+        a[i] = (a[i - 1] + a[i - 2]) % 10007;
+        b = a[i - 1];
+    }
+    if (n > 2)
+        printf("%d", b);
+    else
+        printf("1");
+    return 0;
+}
 ```
 
 ### C
 ```cpp
+int main()
+{
+    int n, b;
+    scanf("%d", &n);
+    int a[n];
+    a[0] = a[1] = 1;
 
+    for (int i = 2; i < n; i++)
+    {
+        a[i] = (a[i - 1] + a[i - 2]) % 10007;
+        b = a[i - 2];
+    }
+    if (n > 2)
+        printf("%d", b);
+    else
+        printf("1");
+    return 0;
+}
 ```

data/1.算法初阶/1.蓝桥杯/斐波那契数列最大公约数/config.json

-{
-  "node_id": "569d5e11c4fc5de7844053d9a733c5e8",
-  "keywords": [],
-  "children": [],
-  "export": []
-}
\ No newline at end of file

data/1.算法初阶/1.蓝桥杯/斐波那契数列最大公约数/desc.md

-#### 问题描述
-斐波那契数列满足 F1 = F2 = 1，从 F3 开始有 Fn = Fn−1 + Fn−2。请你计算 GCD(F2020, F520)，其中 GCD(A, B) 表示 A 和 B 的最大公约数。
\ No newline at end of file

data/1.算法初阶/1.蓝桥杯/斐波那契数列最大公约数/solution.java

-
-import java.math.BigInteger;
-
-public class BigNumber {
-    public static void main(String[] args) {
-        BigInteger a = new BigInteger("1");
-        BigInteger b = new BigInteger("1");
-        BigInteger[] fib = new BigInteger[2030];
-        fib[1] = a;
-        fib[2] = b;
-        for (int x = 3; x < fib.length; x++) {
-            fib[x] = fib[x - 1].add(fib[x - 2]);
-        }
-        a = fib[2020];
-        b = fib[520];
-        BigInteger temp;
-        while (b != BigInteger.ZERO) {
-            temp = a.mod(b);
-            a = b;
-            b = temp;
-        }
-        System.out.println(a);
-    }
-}

data/1.算法初阶/1.蓝桥杯/斐波那契数列最大公约数/solution.md

-# 斐波那契数列最大公约数
-#### 问题描述
-斐波那契数列满足 F1 = F2 = 1，从 F3 开始有 Fn = Fn−1 + Fn−2。请你计算 GCD(F2020, F520)，其中 GCD(A, B) 表示 A 和 B 的最大公约数。
-
-## aop
-### before
-```cpp
-
-```
-### after
-```cpp
-
-```
-
-## 答案
-```cpp
-
-```
-## 选项
-
-### A
-```cpp
-
-```
-
-### B
-```cpp
-
-```
-
-### C
-```cpp
-
-```