update exercises

.gitignore

@@ -11,4 +11,5 @@ leetcode_template.md
 auto_gen_leetcode_tem.py
 leetcode_helper.py
 lanqiao_helper.py
-lanqiao_template.md
\ No newline at end of file
+lanqiao_template.md
+./src/__pycache__/*.pyc
\ No newline at end of file

data/1.算法初阶/1.蓝桥杯/分巧克力/solution.cpp

@@ -2,8 +2,6 @@
 #include <string.h>
 int co[100100][2];
 
-//n块巧克力，分成x*x型的。
-//返回巧克力的块数
 int coun(int n, int x)
 {
     int sum = 0;
@@ -14,16 +12,15 @@ int coun(int n, int x)
     return sum;
 }
 
-//二分求解
 int binary(int n, int k)
 {
     int l = 0, r = 100000;
     while (l < r)
     {
         int mid = (r - l) / 2 + l;
-        if (coun(n, mid) < k)
+        if (coun(n, mid) > k)
             r = mid - 1;
-        else if (coun(n, mid) > k)
+        else if (coun(n, mid) < k)
             l = mid + 1;
         else
             return mid;

data/1.算法初阶/1.蓝桥杯/分巧克力/solution.md

@@ -29,30 +29,107 @@
 ## aop
 ### before
 ```cpp
+#include <stdio.h>
+#include <string.h>
+int co[100100][2];
+
+int coun(int n, int x)
+{
+    int sum = 0;
+    for (int i = 0; i < n; i++)
+    {
+        sum += (co[i][0] / x) * (co[i][1] / x);
+    }
+    return sum;
+}
 
 ```
 ### after
 ```cpp
-
+int main()
+{
+    int n;
+    int k;
+    scanf("%d%d", &n, &k);
+    for (int i = 0; i < n; i++)
+        scanf("%d%d", &co[i][0], &co[i][1]);
+    printf("%d\n", binary(n, k));
+}
 ```
 
 ## 答案
 ```cpp
-
+int binary(int n, int k)
+{
+    int l = 0, r = 100000;
+    while (l < r)
+    {
+        int mid = (r - l) / 2 + l;
+        if (coun(n, mid) < k)
+            r = mid - 1;
+        else if (coun(n, mid) > k)
+            l = mid + 1;
+        else
+            return mid;
+    }
+    return r;
+}
 ```
 ## 选项
 
 ### A
 ```cpp
-
+int binary(int n, int k)
+{
+    int l = 0, r = 100000;
+    while (l < r)
+    {
+        int mid = (r - l) / 2 + l;
+        if (coun(n, mid) < k)
+            r = mid;
+        else if (coun(n, mid) > k)
+            l = mid;
+        else
+            return mid;
+    }
+    return r;
+}
 ```
 
 ### B
 ```cpp
-
+int binary(int n, int k)
+{
+    int l = 0, r = 100000;
+    while (l < r)
+    {
+        int mid = (r - l) / 2 + l;
+        if (coun(n, mid) > k)
+            r = mid;
+        else if (coun(n, mid) < k)
+            l = mid;
+        else
+            return mid;
+    }
+    return r;
+}
 ```
 
 ### C
 ```cpp
-
+int binary(int n, int k)
+{
+    int l = 0, r = 100000;
+    while (l < r)
+    {
+        int mid = (r - l) / 2 + l;
+        if (coun(n, mid) > k)
+            r = mid - 1;
+        else if (coun(n, mid) < k)
+            l = mid + 1;
+        else
+            return mid;
+    }
+    return r;
+}
 ```

data/1.算法初阶/1.蓝桥杯/分数/solution.cpp

 #include <bits/stdc++.h>
 using namespace std;
-long pow_2(int b) //快速幂运算
+long pow_2(int b)
 {
     long x = 2;
     long res = 1;
     while (b > 0)
     {
         if (b & 1)
-            res *= x;
-        b >>= 1; //右移一位
+            res = x;
+        b >>= 1;
         x = x * x;
     }
     return res;
 }
-int gcd(long a, long b) //求出最大公约数
+int gcd(long a, long b)
 {
     if (b == 0)
         return a;
@@ -21,6 +21,6 @@ int gcd(long a, long b) //求出最大公约数
 }
 int main()
 {
-    cout << gcd(pow_2(20) - 1, pow_2(19)) << endl;     //最大公约数
-    cout << pow_2(20) - 1 << "/" << pow_2(19) << endl; //输出分子分母
+    cout << gcd(pow_2(20) - 1, pow_2(19)) << endl;
+    cout << pow_2(20) - 1 << "/" << pow_2(19) << endl;
 }

data/1.算法初阶/1.蓝桥杯/分数/solution.md

@@ -6,30 +6,89 @@
 ## aop
 ### before
 ```cpp
-
+#include <bits/stdc++.h>
+using namespace std;
 ```
 ### after
 ```cpp
-
+int gcd(long a, long b)
+{
+    if (b == 0)
+        return a;
+    return gcd(b, a % b);
+}
+int main()
+{
+    cout << gcd(pow_2(20) - 1, pow_2(19)) << endl;
+    cout << pow_2(20) - 1 << "/" << pow_2(19) << endl;
+}
 ```
 
 ## 答案
 ```cpp
-
+long pow_2(int b)
+{
+    long x = 2;
+    long res = 1;
+    while (b > 0)
+    {
+        if (b & 1)
+            res *= x;
+        b >>= 1;
+        x = x * x;
+    }
+    return res;
+}
 ```
 ## 选项
 
 ### A
 ```cpp
-
+long pow_2(int b)
+{
+    long x = 2;
+    long res = 1;
+    while (b > 0)
+    {
+        if (b & 1)
+            res *= x;
+        b <<= 1;
+        x = x * x;
+    }
+    return res;
+}
 ```
 
 ### B
 ```cpp
-
+long pow_2(int b)
+{
+    long x = 2;
+    long res = 1;
+    while (b > 0)
+    {
+        if (b && 1)
+            res *= x;
+        b <<= 1;
+        x = x * x;
+    }
+    return res;
+}
 ```
 
 ### C
 ```cpp
-
+long pow_2(int b)
+{
+    long x = 2;
+    long res = 1;
+    while (b > 0)
+    {
+        if (b & 1)
+            res = x;
+        b >>= 1;
+        x = x * x;
+    }
+    return res;
+}
 ```

data/1.算法初阶/1.蓝桥杯/分糖果/solution.cpp

@@ -2,40 +2,40 @@
 using namespace std;
 int main()
 {
-    int n; //小朋友数量
+    int n;
     cin >> n;
-    int *Candy = new int[n]; //存储每个小朋友的糖果数
+    int *Candy = new int[n];
     for (int i = 0; i < n; i++)
     {
         cin >> Candy[i];
     }
-    int total = 0; //补发的糖果总数
+    int total = 0;
     while (1)
     {
-        bool flag = true; //判断每人糖果数量是否相同
+        bool flag = true;
         for (int i = 1; i < n; i++)
         {
             if (Candy[0] != Candy[i])
-            { //不相同
+            {
                 flag = false;
                 break;
             }
         }
-        if (flag == true) //数量相同，退出循环
+        if (flag == true)
             break;
-        int temp = Candy[n - 1]; //记录最后一个小朋友手中的糖果
+        int temp = Candy[n - 1];
         for (int i = n - 1; i >= 0; i--)
         {
             if (i == 0)
-            {                                       //如果是第一个小朋友
-                Candy[0] = Candy[0] / 2 + temp / 2; //最后一个小朋友获得最初第一个人的一半
+            {
+                Candy[0] = Candy[0] / 2 + temp / 2;
             }
             else
-                Candy[i] = Candy[i] / 2 + Candy[i - 1] / 2; //　每个小朋友都把自己的糖果分一半给左手边的孩子
-            if (Candy[i] % 2 != 0)
-            { //如果此时手中为奇数，则补糖果
+                Candy[i] = Candy[i] / 2 + Candy[i - 1] / 2;
+            if (Candy[i] % 2 == 0)
+            {
                 Candy[i] += 1;
-                total += 1; //补发数量加1
+                total += 1;
             }
         }
     }

data/1.算法初阶/1.蓝桥杯/分糖果/solution.md

@@ -26,7 +26,8 @@
 ## aop
 ### before
 ```cpp
-
+#include <iostream>
+using namespace std;
 ```
 ### after
 ```cpp
@@ -35,21 +36,182 @@
 
 ## 答案
 ```cpp
-
+int main()
+{
+    int n;
+    cin >> n;
+    int *Candy = new int[n];
+    for (int i = 0; i < n; i++)
+    {
+        cin >> Candy[i];
+    }
+    int total = 0;
+    while (1)
+    {
+        bool flag = true;
+        for (int i = 1; i < n; i++)
+        {
+            if (Candy[0] != Candy[i])
+            {
+                flag = false;
+                break;
+            }
+        }
+        if (flag == true)
+            break;
+        int temp = Candy[n - 1];
+        for (int i = n - 1; i >= 0; i--)
+        {
+            if (i == 0)
+            {
+                Candy[0] = Candy[0] / 2 + temp / 2;
+            }
+            else
+                Candy[i] = Candy[i] / 2 + Candy[i - 1] / 2;
+            if (Candy[i] % 2 != 0)
+            {
+                Candy[i] += 1;
+                total += 1;
+            }
+        }
+    }
+    cout << total << endl;
+}
 ```
 ## 选项
 
 ### A
 ```cpp
-
+int main()
+{
+    int n;
+    cin >> n;
+    int *Candy = new int[n];
+    for (int i = 0; i < n; i++)
+    {
+        cin >> Candy[i];
+    }
+    int total = 0;
+    while (1)
+    {
+        bool flag = true;
+        for (int i = 1; i < n; i++)
+        {
+            if (Candy[0] != Candy[i])
+            {
+                flag = false;
+                break;
+            }
+        }
+        if (flag == true)
+            break;
+        int temp = Candy[n - 1];
+        for (int i = n; i >= 0; i--)
+        {
+            if (i == 0)
+            {
+                Candy[0] = Candy[0] / 2 + temp / 2;
+            }
+            else
+                Candy[i] = Candy[i] / 2 + Candy[i - 1] / 2;
+            if (Candy[i] % 2 != 0)
+            {
+                Candy[i] += 1;
+                total += 1;
+            }
+        }
+    }
+    cout << total << endl;
+}
 ```
 
 ### B
 ```cpp
-
+int main()
+{
+    int n;
+    cin >> n;
+    int *Candy = new int[n];
+    for (int i = 0; i < n; i++)
+    {
+        cin >> Candy[i];
+    }
+    int total = 0;
+    while (1)
+    {
+        bool flag = true;
+        for (int i = 1; i < n; i++)
+        {
+            if (Candy[0] != Candy[i])
+            {
+                flag = false;
+                break;
+            }
+        }
+        if (flag == true)
+            break;
+        int temp = Candy[n - 1];
+        for (int i = n; i >= 0; i--)
+        {
+            if (i == 0)
+            {
+                Candy[0] = Candy[0] / 2 + temp / 2;
+            }
+            else
+                Candy[i] = Candy[i] / 2 + Candy[i - 1] / 2;
+            if (Candy[i] % 2 == 0)
+            {
+                Candy[i] += 1;
+                total += 1;
+            }
+        }
+    }
+    cout << total << endl;
+}
 ```
 
 ### C
 ```cpp
+int main()
+{
+    int n;
+    cin >> n;
+    int *Candy = new int[n];
+    for (int i = 0; i < n; i++)
+    {
+        cin >> Candy[i];
+    }
+    int total = 0;
+    while (1)
+    {
+        bool flag = true;
+        for (int i = 1; i < n; i++)
+        {
+            if (Candy[0] != Candy[i])
+            {
+                flag = false;
+                break;
+            }
+        }
+        if (flag == true)
+            break;
+        int temp = Candy[n - 1];
+        for (int i = n - 1; i >= 0; i--)
+        {
+            if (i == 0)
+            {
+                Candy[0] = Candy[0] / 2 + temp / 2;
+            }
+            else
+                Candy[i] = Candy[i] / 2 + Candy[i - 1] / 2;
+            if (Candy[i] % 2 == 0)
+            {
+                Candy[i] += 1;
+                total += 1;
+            }
+        }
+    }
+    cout << total << endl;
+}
 
 ```