diff --git "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257/\347\254\254\345\207\240\345\244\251/solution.md" "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257/\347\254\254\345\207\240\345\244\251/solution.md"
index 3e11bdb95318a6f07f85161b3eb309da037dd7ad..850f185c487021262b7bda39a786a86d2ac6df06 100644
--- "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257/\347\254\254\345\207\240\345\244\251/solution.md"
+++ "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257/\347\254\254\345\207\240\345\244\251/solution.md"
@@ -44,9 +44,9 @@ int main()
     int L_m_d[12] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
     int nonL_m_d[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
 
-    if (is_leap(y)) //判断闰年
+    if (is_leap(y)) 
     {
-        for (int i = 0; i < (m - 1); i++) //记录1-（m-1）月天数
+        for (int i = 0; i < (m - 1); i++) 
         {
             ans += L_m_d[i];
         }
@@ -76,9 +76,9 @@ int main()
     int L_m_d[12] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
     int nonL_m_d[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
 
-    if (is_leap(y)) //判断闰年
+    if (is_leap(y)) 
     {
-        for (int i = 0; i < m; i++) //记录1-（m-1）月天数
+        for (int i = 0; i < m; i++) 
         {
             ans += L_m_d[i];
         }
@@ -106,9 +106,9 @@ int main()
     int L_m_d[12] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
     int nonL_m_d[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
 
-    if (is_leap(y)) //判断闰年
+    if (is_leap(y)) 
     {
-        for (int i = 0; i < (m - 1); i++) //记录1-（m-1）月天数
+        for (int i = 0; i < (m - 1); i++) 
         {
             ans += L_m_d[i];
         }
@@ -136,9 +136,9 @@ int main()
     int L_m_d[12] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
     int nonL_m_d[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
 
-    if (is_leap(y)) //判断闰年
+    if (is_leap(y)) 
     {
-        for (int i = 0; i < m; i++) //记录1-（m-1）月天数
+        for (int i = 0; i < m; i++) 
         {
             ans += L_m_d[i];
         }
diff --git "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257/\351\242\240\345\200\222\347\232\204\344\273\267\347\211\214/solution.cpp" "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257/\351\242\240\345\200\222\347\232\204\344\273\267\347\211\214/solution.cpp"
index 7f775b8b9b07f61ab97c4efec4e754b391ff2c79..14d84f3491b2020b9013dcde0bf1c49427daf13b 100644
--- "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257/\351\242\240\345\200\222\347\232\204\344\273\267\347\211\214/solution.cpp"
+++ "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257/\351\242\240\345\200\222\347\232\204\344\273\267\347\211\214/solution.cpp"
@@ -1,84 +1,55 @@
 #include <iostream>
-#include <sstream>
-#include <vector>
 using namespace std;
-
-void i2s(int num, string &str)
-{                    //将数字转换为字符串
-    stringstream ss; //用法类似于cout
-    ss << num;       //>>是将那个num传入到stringstream中
-    ss >> str;       //用stringstream 类的函数str() 将对象所包含的内容赋给一个string对象。
-}
-
-void s2i(string &str, int &num)
-{
-    stringstream ss; //将字符串反转为数字
-    ss << str;
-    ss >> num;
-}
-char to(char x)
-{ //字符的翻转
-    if (x == '6')
-        return '9';
-    else if (x == '9')
-        return '6';
-    else
-        return x; //x:指0,1,2,5,8，这几个数字反转后还是本身
-}
-
-string reserve(const string &str)
-{               //reserve作用类似旋转180°
-    string ans; //结果
-    for (int i = 3; i >= 0; i--)
-    {                                      //扫描，从最后一位开始转换之后
-        ans.insert(ans.end(), to(str[i])); //把最后一个字符插入
-    }
-    return ans;
-}
-struct price
-{                //定义一个价格的结构体
-    int a, b, c; //a:原始价格 ，b：翻转颠倒价格，c：原始价格与颠倒价格的差
-};
-vector<price> v1; //存储-200多的
-vector<price> v2; //存储+800多的
 int main()
 {
-    cout << reserve << endl; //在这个地方可以测试原题中给出的数据
-                             //枚举所有可以颠倒的四位数
-                             //	将其颠倒过来，与原来的数值做差，将-200多与800多记录下来，分别记录在两个集合中
-                             //遍历两个集合，-200多和800多两两求和，结果为558的即为正确答案
-    for (int i = 1000; i < 10000; i++)
+    int num1[7] = {0, 1, 2, 5, 6, 8, 9}; //价牌包含的数字
+    int num2[7] = {0, 1, 2, 5, 9, 8, 6}; //价牌翻转对应的数字
+    int profit1[1111][2];                //利润1
+    int profit2[1111][2];                //利润2
+    int before_reverse;                  //颠倒前
+    int after_reverse;                   //颠倒后
+    int i = 0;
+    int j = 0;
+    for (int a = 1; a < 7; a++) //从1开始，第一位数字不可以是0
     {
-        string str;
-        i2s(i, str);
-        if (str.find('3') != string::npos || str.find('4') != string::npos || str.find('7') != string::npos || str.rfind('0') == 3)
-            //str里面是i的字符串形式，里面的3,4,7，还有最后一位也就是数组第4位不能为0
-            continue;
-        string r = reserve(str); //翻转的字符
-        int r_int;
-        s2i(r, r_int);        //r_int就是翻转后的价格，i是原始价格
-        int plus = r_int - i; //plus就是价格差
-        if (plus > -300 && plus < -200)
-        { //价格范围
-            price p = {i, r_int, plus};
-            v1.push_back(p);
-        }
-        else if (plus > 800 && plus < 900)
-        { //价格范围
-            price p = {i, r_int, plus};
-            v2.push_back(p);
-        }
-        //此时v1存储-200多的价格，v2存储+800多
-        for (int i = 0; i < v1.size(); i++)
+        for (int b = 0; b < 7; b++)
         {
-            for (int j = 0; j < v2.size(); j++)
+            for (int c = 0; c < 7; c++)
             {
-                if (v1[i].c + v2[j].c == 558)
+                for (int d = 0; d < 7; d++)
                 {
-                    printf("%d %d %d %d %d %d\n", v1[i].a, v1[i].b, v1[i].c, v2[j].a, v2[j].b, v2[j].c);
+                    before_reverse = num1[a] * 1000 + num1[b] * 100 + num1[c] * 10 + num1[d];
+                    after_reverse = num2[d] * 1000 + num2[c] * 100 + num2[b] * 10 + num2[a];
+                    //因为是颠倒之后所以第一位可以是0
+                    if (after_reverse - before_reverse > -300 && after_reverse - before_reverse < -200)
+                    //在200-300范围内，我们以正数为准
+                    {
+                        profit1[i][0] = before_reverse;
+                        profit1[i][1] = after_reverse - before_reverse;
+                        i++;
+                    }
+                    else if (after_reverse - before_reverse > 800 && after_reverse - before_reverse < 900)
+                    //在800-900范围内
+                    {
+                        profit1[j][0] = before_reverse;
+                        profit2[j][1] = after_reverse - before_reverse;
+                        j++;
+                    }
                 }
             }
         }
     }
+    int answer = 0;
+    for (int a = 0; a < i; a++)
+    {
+        for (int b = 0; b < j; b++)
+        {
+            if (profit1[a][1] + profit2[b][1] == 558)
+            {
+                answer = profit1[a][0];
+            }
+        }
+    }
+    cout << answer << endl;
     return 0;
-}
+}
\ No newline at end of file
diff --git "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257/\351\242\240\345\200\222\347\232\204\344\273\267\347\211\214/solution.md" "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257/\351\242\240\345\200\222\347\232\204\344\273\267\347\211\214/solution.md"
index 95c4f32719e819d400c9c6b62069a159718887fe..f6531270ee294d07fd7163eb8977b3258b215f2b 100644
--- "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257/\351\242\240\345\200\222\347\232\204\344\273\267\347\211\214/solution.md"
+++ "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257/\351\242\240\345\200\222\347\232\204\344\273\267\347\211\214/solution.md"
@@ -14,7 +14,8 @@
 ## aop
 ### before
 ```cpp
-
+#include <iostream>
+using namespace std;
 ```
 ### after
 ```cpp
@@ -23,21 +24,229 @@
 
 ## 答案
 ```cpp
-
+int main()
+{
+    int num1[7] = {0, 1, 2, 5, 6, 8, 9}; //价牌包含的数字
+    int num2[7] = {0, 1, 2, 5, 9, 8, 6}; //价牌翻转对应的数字
+    int profit1[1111][2];                //利润1
+    int profit2[1111][2];                //利润2
+    int before_reverse;                  //颠倒前
+    int after_reverse;                   //颠倒后
+    int i = 0;
+    int j = 0;
+    for (int a = 1; a < 7; a++) //从1开始，第一位数字不可以是0
+    {
+        for (int b = 0; b < 7; b++)
+        {
+            for (int c = 0; c < 7; c++)
+            {
+                for (int d = 0; d < 7; d++)
+                {
+                    before_reverse = num1[a] * 1000 + num1[b] * 100 + num1[c] * 10 + num1[d];
+                    after_reverse = num2[d] * 1000 + num2[c] * 100 + num2[b] * 10 + num2[a];
+                    //因为是颠倒之后所以第一位可以是0
+                    if (after_reverse - before_reverse > -300 && after_reverse - before_reverse < -200)
+                    //在200-300范围内，我们以正数为准
+                    {
+                        profit1[i][0] = before_reverse;
+                        profit1[i][1] = after_reverse - before_reverse;
+                        i++;
+                    }
+                    else if (after_reverse - before_reverse > 800 && after_reverse - before_reverse < 900)
+                    //在800-900范围内
+                    {
+                        profit1[j][0] = before_reverse;
+                        profit2[j][1] = after_reverse - before_reverse;
+                        j++;
+                    }
+                }
+            }
+        }
+    }
+    int answer = 0;
+    for (int a = 0; a < i; a++)
+    {
+        for (int b = 0; b < j; b++)
+        {
+            if (profit1[a][1] + profit2[b][1] == 558)
+            {
+                answer = profit1[a][0];
+            }
+        }
+    }
+    cout << answer << endl;
+    return 0;
+}
 ```
 ## 选项
 
 ### A
 ```cpp
-
+int main()
+{
+    int num1[7] = {0, 1, 2, 5, 6, 8, 9}; //价牌包含的数字
+    int num2[7] = {0, 1, 2, 5, 9, 8, 6}; //价牌翻转对应的数字
+    int profit1[1111][2];                //利润1
+    int profit2[1111][2];                //利润2
+    int before_reverse;                  //颠倒前
+    int after_reverse;                   //颠倒后
+    int i = 0;
+    int j = 0;
+    for (int a = 1; a < 7; a++) //从1开始，第一位数字不可以是0
+    {
+        for (int b = 0; b < 7; b++)
+        {
+            for (int c = 0; c < 7; c++)
+            {
+                for (int d = 0; d < 7; d++)
+                {
+                    before_reverse = num1[a] * 1000 + num1[b] * 100 + num1[c] * 10 + num1[d];
+                    after_reverse = num2[d] * 1000 + num2[c] * 100 + num2[b] * 10 + num2[a];
+                    //因为是颠倒之后所以第一位可以是0
+                    if (after_reverse - before_reverse > -300 && after_reverse - before_reverse < -200)
+                    //在200-300范围内，我们以正数为准
+                    {
+                        profit1[i][0] = before_reverse;
+                        profit1[i][1] = after_reverse - before_reverse;
+                        i++;
+                    }
+                    else if (after_reverse - before_reverse > 800 && after_reverse - before_reverse < 900)
+                    //在800-900范围内
+                    {
+                        profit1[j][0] = before_reverse;
+                        profit2[j][1] = after_reverse - before_reverse;
+                        j++;
+                    }
+                }
+            }
+        }
+    }
+    int answer = 0;
+    for (int a = 0; a < i; a++)
+    {
+        for (int b = 0; b < j; b++)
+        {
+            if (profit1[a][1] + profit2[b][1] == 558)
+            {
+                answer = profit1[b][1];
+            }
+        }
+    }
+    cout << answer << endl;
+    return 0;
+}
 ```
 
 ### B
 ```cpp
-
+int main()
+{
+    int num1[7] = {0, 1, 2, 5, 6, 8, 9}; //价牌包含的数字
+    int num2[7] = {0, 1, 2, 5, 9, 8, 6}; //价牌翻转对应的数字
+    int profit1[1111][2];                //利润1
+    int profit2[1111][2];                //利润2
+    int before_reverse;                  //颠倒前
+    int after_reverse;                   //颠倒后
+    int i = 0;
+    int j = 0;
+    for (int a = 1; a < 7; a++) //从1开始，第一位数字不可以是0
+    {
+        for (int b = 0; b < 7; b++)
+        {
+            for (int c = 0; c < 7; c++)
+            {
+                for (int d = 0; d < 7; d++)
+                {
+                    before_reverse = num1[a] * 1000 + num1[b] * 100 + num1[c] * 10 + num1[d];
+                    after_reverse = num2[d] * 1000 + num2[c] * 100 + num2[b] * 10 + num2[a];
+                    //因为是颠倒之后所以第一位可以是0
+                    if (after_reverse - before_reverse > -300 && after_reverse - before_reverse < -200)
+                    //在200-300范围内，我们以正数为准
+                    {
+                        profit1[i][0] = before_reverse;
+                        profit1[i][1] = after_reverse - before_reverse;
+                        i++;
+                    }
+                    else if (after_reverse - before_reverse > 800 && after_reverse - before_reverse < 900)
+                    //在800-900范围内
+                    {
+                        profit1[j][0] = before_reverse;
+                        profit2[j][1] = after_reverse - before_reverse;
+                        j++;
+                    }
+                }
+            }
+        }
+    }
+    int answer = 0;
+    for (int a = 0; a < i; a++)
+    {
+        for (int b = 0; b < j; b++)
+        {
+            if (profit1[a][1] + profit2[b][1] == 558)
+            {
+                answer = profit1[b][0];
+            }
+        }
+    }
+    cout << answer << endl;
+    return 0;
+}
 ```
 
 ### C
 ```cpp
-
+int main()
+{
+    int num1[7] = {0, 1, 2, 5, 6, 8, 9}; //价牌包含的数字
+    int num2[7] = {0, 1, 2, 5, 9, 8, 6}; //价牌翻转对应的数字
+    int profit1[1111][2];                //利润1
+    int profit2[1111][2];                //利润2
+    int before_reverse;                  //颠倒前
+    int after_reverse;                   //颠倒后
+    int i = 0;
+    int j = 0;
+    for (int a = 1; a < 7; a++) //从1开始，第一位数字不可以是0
+    {
+        for (int b = 0; b < 7; b++)
+        {
+            for (int c = 0; c < 7; c++)
+            {
+                for (int d = 0; d < 7; d++)
+                {
+                    before_reverse = num1[a] * 1000 + num1[b] * 100 + num1[c] * 10 + num1[d];
+                    after_reverse = num2[d] * 1000 + num2[c] * 100 + num2[b] * 10 + num2[a];
+                    //因为是颠倒之后所以第一位可以是0
+                    if (after_reverse - before_reverse > -300 && after_reverse - before_reverse < -200)
+                    //在200-300范围内，我们以正数为准
+                    {
+                        profit1[i][0] = before_reverse;
+                        profit1[i][1] = after_reverse - before_reverse;
+                        i++;
+                    }
+                    else if (after_reverse - before_reverse > 800 && after_reverse - before_reverse < 900)
+                    //在800-900范围内
+                    {
+                        profit1[j][0] = before_reverse;
+                        profit2[j][1] = after_reverse - before_reverse;
+                        j++;
+                    }
+                }
+            }
+        }
+    }
+    int answer = 0;
+    for (int a = 0; a < i; a++)
+    {
+        for (int b = 0; b < j; b++)
+        {
+            if (profit1[a][1] + profit2[b][1] == 558)
+            {
+                answer = profit1[0][b];
+            }
+        }
+    }
+    cout << answer << endl;
+    return 0;
+}
 ```