diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/30.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/30.exercises/solution.md"
index 1e47166754216293c10fb74ee450a5988a04cfff..555211e1130c2c47ccb63f3c01777a409e724ec9 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/30.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/30.exercises/solution.md"
@@ -2,7 +2,7 @@
 
 1，2，4，2，3，6，12，6，3，....求第n项值
 
-## template
+以下程序实现了这一功能，请你填补空白处内容：
 
 ```cpp
 #include <stdio.h>
@@ -50,16 +50,70 @@ int main()
 }
 ```
 
-## 答案
+## template
 
 ```cpp
-for (i = 2; i <= (*y); i++)
+#include <stdio.h>
+int fun(int n, int *x, int *y)
+{
+    int sum = 0, i;
+    int size = 0;
+    int dd = 1;
+    for (i = 1; i <= (*x); i++)
+    {
+        sum += (2 * i - 1);
+    }
+    if (sum == n)
+    {
+        *y = 2 * (*x) - 1;
+        return (*x);
+    }
+    else if (sum > n)
     {
-        if (i <= (*x))
-            dd *= 2;
-        else
-            dd /= 2;
+        (*y) = n - (sum - (2 * (*x) - 1));
+        size = 2 * (*x) - 1;
+        dd = (*x);
+        for (i = 2; i <= (*y); i++)
+            for (i = 2; i <= (*y); i++)
+            {
+                if (i <= (*x))
+                    dd *= 2;
+                else
+                    dd /= 2;
+            }
+        return dd;
     }
+    else
+    {
+        (*x)++;
+        return fun(n, x, y);
+    }
+}
+int main()
+{
+    int n;
+    int row = 1, col = 0;
+    int val;
+    row = 1;
+    col = 0;
+    printf("请输入n：");
+    scanf("%d", &n);
+    val = fun(n, &row, &col);
+    printf("第%d项是：%d\n", n, val);
+    return 0;
+}
+```
+
+## 答案
+
+```cpp
+for (i = 2; i <= (*y); i++)
+{
+    if (i <= (*x))
+        dd *= 2;
+    else
+        dd /= 2;
+}
 ```
 
 ## 选项
@@ -68,34 +122,34 @@ for (i = 2; i <= (*y); i++)
 
 ```cpp
 for (i = 2; i <= (*y); i++)
-    {
-        if (i <= (*x))
-            dd /= 2;
-        else
-            dd *= 2;
-    }
+{
+    if (i <= (*x))
+        dd /= 2;
+    else
+        dd *= 2;
+}
 ```
 
 ### B
 
 ```cpp
 for (i = 2; i <= (*y); i++)
-    {
-        if (i <= (*x))
-            dd += 2;
-        else
-            dd -= 2;
-    }
+{
+    if (i <= (*x))
+        dd += 2;
+    else
+        dd -= 2;
+}
 ```
 
 ### C
 
 ```cpp
 for (i = 2; i <= (*y); i++)
-    {
-        if (i <= (*x))
-            dd -= 2;
-        else
-            dd += 2;
-    }
+{
+    if (i <= (*x))
+        dd -= 2;
+    else
+        dd += 2;
+}
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/31.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/31.exercises/solution.md"
index b0ff0d9ab30906a350fa5064d9b50dc6220703f2..fa225fa65f155d2240160a92148dd9ef6d58a5d1 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/31.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/31.exercises/solution.md"
@@ -2,7 +2,7 @@
 
 例如：字符串中的内容为：a1Ab1D2，1<A,A<b 1<D 则调用该函数后，返回码为：3。 
 
-## template
+以下程序实现了这一功能，请你填补空白处内容：
 
 ```cpp
 #include <iostream>
@@ -25,6 +25,29 @@ int main()
 }
 ```
 
+## template
+
+```cpp
+#include <iostream>
+#include <string>
+using namespace std;
+int solve(string s)
+{
+	if (s.length() == 0) return 0;
+	int n = 0;
+	for (int i = 1; i < s.length(); i++)
+		if (s.c_str()[i] < s.c_str()[i - 1]) n++;
+	return n;
+}
+int main()
+{
+	string s = "a1Ab1D2";
+	int n = solve(s);
+	cout << n << endl;
+	return 0;
+}
+```
+
 ## 答案
 
 ```cpp
diff --git "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/1.exercises/solution.md" "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/1.exercises/solution.md"
index 98bd412192c3bf6c219ab104ce179036a2c7cd02..57521b8d48dcf0cded4221ee76c680454ea78006 100644
--- "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/1.exercises/solution.md"
+++ "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/1.exercises/solution.md"
@@ -1,15 +1,63 @@
 # 移动字符串
 
 给定一个字符串长度为 nn 的字符串 s1 (10<n<100) , 求出将字符串循环向左移动 k 位的字符串 s2 (1<k<n) , 例如：字符串 abcdefghijk , 循环向左移动 3 位就变成 defghijkabc
-输入描述
+
+**输入描述**
+
 输入仅两行，第一行为左移的位数 k , 第二行为字符串 s1 .
-输出描述
+
+**输出描述**
+
 输出仅一行，为将字符串 s1 左移 k 位得到的字符串 s2 .
-样例输入
+
+**样例输入**
+
+```json
 3
 abcdefghijk
-样例输出
+```
+
+**样例输出**
+
+```json
 defghijkabc
+```
+
+以下程序实现了这一功能，请你填补空白处内容：
+
+```cpp
+#include <iostream>
+#include <string.h>
+using namespace std;
+void reverse(char *a, int start, int end)
+{
+    int i, j, temp;
+    for (i = start, j = end; i < j; i++, j--)
+    {
+        temp = a[i];
+        a[i] = a[j];
+        a[j] = temp;
+    }
+}
+void turnleft(char *a, int i, int n)
+{
+    int left = i % n;
+    if (left == 0)
+        return;
+	______________;
+    return;
+}
+int main()
+{
+    char a[1024];
+    int i;
+    cin >> i;
+    cin >> a;
+    int n = strlen(a);
+    turnleft(a, i, n);
+    cout << a << endl;
+}
+```
 
 ## template
 
@@ -32,7 +80,9 @@ void turnleft(char *a, int i, int n)
     int left = i % n;
     if (left == 0)
         return;
-	______________
+	reverse(a, 0, left - 1);
+    reverse(a, left, n - 1);
+    reverse(a, 0, n - 1);
     return;
 }
 int main()
diff --git "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/10.exercises/solution.md" "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/10.exercises/solution.md"
index 4131997fc9abdfca5bdc55c1b9019d8e46949b7f..bbbac7efe6d4cd7ac06ddd9c9fc96a45d82b680e 100644
--- "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/10.exercises/solution.md"
+++ "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/10.exercises/solution.md"
@@ -2,39 +2,70 @@
 
 <pre>一个数如果恰好等于它的所有因子之和&#xff0c;这个数就称为“完数”。统计自然数 1 — 100 间完数的个数。</pre>
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```cpp
+#include <stdio.h>
+int perfect()
+{
+    int i, x, sum, cnt = 0;
+    for (i = 1; i <= 100; i++)
+    {
+        sum = 0;
+        for (x = 1; x < i; x++)
+        {
+            _____________;
+        }
+        if (i == sum)
+        {
+            cnt++;
+            printf("%d ", i);
+        }
+    }
+    return cnt;
+}
+int main()
+{
+    printf("\ncount=%d\n", perfect());
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
-#include<stdio.h>
+#include <stdio.h>
 int perfect()
 {
-	int i,x,sum,cnt=0;
-	for(i=1;i<=100;i++)
-	{
-		sum=0;
-		for(x=1;x<i;x++)
-		{
-			if(i%x==0)sum+=x;
-		}
-		if(i==sum)
-		{
-			cnt++;
-			printf("%d ",i);
-		}
-	}
-	return cnt;
+    int i, x, sum, cnt = 0;
+    for (i = 1; i <= 100; i++)
+    {
+        sum = 0;
+        for (x = 1; x < i; x++)
+        {
+            if (i % x == 0)
+                sum += x;
+        }
+        if (i == sum)
+        {
+            cnt++;
+            printf("%d ", i);
+        }
+    }
+    return cnt;
 }
 int main()
 {
-	printf("\ncount=%d\n",perfect());
-	return 0;
+    printf("\ncount=%d\n", perfect());
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+if (i / x == 0)
+    sum += x;
 ```
 
 ## 选项
@@ -42,17 +73,20 @@ int main()
 ### A
 
 ```cpp
-
+if (i - x == 0)
+    sum += x;
 ```
 
 ### B
 
 ```cpp
-
+if (i % x == 1)
+    sum += x;
 ```
 
 ### C
 
 ```cpp
-
+if (i / x == 1)
+    sum += x;
 ```
\ No newline at end of file
diff --git "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/11.exercises/solution.md" "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/11.exercises/solution.md"
index f37af3e254714a0b9c7a7d1b1c51bcfec24cfedb..f3dd039ad9fe9ccb6437d8f72ce29cf6dc5fd6b8 100644
--- "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/11.exercises/solution.md"
+++ "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/11.exercises/solution.md"
@@ -1,17 +1,56 @@
 # 好数对
 
  已知一个集合A，对A中任意两个不同的元素求和，若求得的和仍在A内，则称其为好数对。例如，集合A={1 2 3 4}，1+2=3，1+3=4，则1,2和1,3 是两个好数对。编写程序求给定集合中好数对的个数。
+ 
  注：集合中最多有1000个元素，元素最大不超过10000
+ 
  程序运行示例1：
+
+ ```json
  4↙
  1 2 3 4↙
  2
+ ```
+
  程序运行示例2：
+
+ ```json
  7↙
  2456 3251 654 890 100 754 1234↙
  1
+ ```
+
  其中，“↙”表示输入
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```cpp
+#include <stdio.h>
+#include <stdlib.h>
+int main()
+{
+    int n, i, j, t;
+    scanf("%d", &n);
+    int *a = (int *)malloc(n * sizeof(int));
+    for (i = 0; i < n; i++)
+    {
+        scanf("%d", &a[i]);
+    }
+    int cout = 0;
+    for (i = 0; i < n; i++)
+    {
+        for (j = i + 1; j < n; j++)
+        {
+            for (t = 0; t < n; t++)
+                ____________;
+        }
+    }
+    printf("%d", cout);
+    free(a);
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
@@ -19,33 +58,34 @@
 #include <stdlib.h>
 int main()
 {
-	int n,i,j,t;
-	scanf("%d",&n);
-	int* a = (int*)malloc(n*sizeof(int));
-	for(i=0;i<n;i++)
-	{
-		scanf("%d",&a[i]);
-	}
-	int cout=0;
-	for( i=0;i<n;i++)
-	{ 
-		for( j=i+1;j<n;j++)
-		{
-			for(t = 0; t <n;t++)
-				if(a[i]+a[j]==a[t])
-					cout++;
-		}
-	}
-	printf("%d",cout);
-	free(a);
-	return 0;
+    int n, i, j, t;
+    scanf("%d", &n);
+    int *a = (int *)malloc(n * sizeof(int));
+    for (i = 0; i < n; i++)
+    {
+        scanf("%d", &a[i]);
+    }
+    int cout = 0;
+    for (i = 0; i < n; i++)
+    {
+        for (j = i + 1; j < n; j++)
+        {
+            for (t = 0; t < n; t++)
+                if (a[i] + a[j] == a[t])
+                    cout++;
+        }
+    }
+    printf("%d", cout);
+    free(a);
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+if (a[i] + a[j] == a[t])
+    cout++;
 ```
 
 ## 选项
@@ -53,17 +93,20 @@ int main()
 ### A
 
 ```cpp
-
+if (a[i] + a[j] <= a[t])
+    cout++;
 ```
 
 ### B
 
 ```cpp
-
+if (a[i] + a[j] >= a[t])
+    cout++;
 ```
 
 ### C
 
 ```cpp
-
+if (a[i] + a[j] < a[t])
+    cout++;
 ```
\ No newline at end of file
diff --git "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/12.exercises/solution.md" "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/12.exercises/solution.md"
index 7434c9c6a6407f6f5abce46c814357a54b8165c0..698eeb91d2d19d344c4b79ccaee8596ff1d83301 100644
--- "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/12.exercises/solution.md"
+++ "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/12.exercises/solution.md"
@@ -1,20 +1,34 @@
 # 难倒数万人的小学数学题
 
 汉堡包在大街上大摇大摆的走着,看着手机上一道难倒数万人的小学数学题:
+
+```json
 1 + 1 = 0
 1 + 6 = 1
 6 + 6 = 2
 8 + 1 = 2
 8 + 6 = 3
+```
+
 汉堡包看完之后发现上面这些加法的答案就是看1,6,8中圈圈的个数嘛！
+
 突然之间,所有大厦上的LED屏幕上的广告全部变成数字1,6,8三个数字的随机闪现。
+
 现给你一块n*m的LED屏幕，上面有且仅有一个数字(1,6,or 8)，请你输出你看见的那个字母。
-输入格式:
+
+**输入格式:**
+
 第一行输入两个整数n，m(2<= m, n <= 1000)；
+
 接下来n行，每行由m个数字0和1组成，其中1表示数字1，6，8的组成部分。
-输出格式:
+
+**输出格式:**
+
 输出一个整数，代表图形表示的数字。
-输入样例:
+
+**输入样例:**
+
+```json
 7 7
 0 0 0 0 0 0 0
 0 0 1 1 1 0 0
@@ -23,8 +37,56 @@
 0 0 1 0 1 0 0
 0 0 1 1 1 0 0
 0 0 0 0 0 0 0
-输出样例:
+```
+
+**输出样例:**
+
+```json
 8
+```
+
+以下程序实现了这一功能，请你填补空白处内容：
+
+```cpp
+#include <cstring>
+#include <algorithm>
+#include <iostream>
+#include <cstdio>
+#include <cmath>
+#include <stack>
+#include <queue>
+using namespace std;
+int main()
+{
+    int i, j, k = 1;
+    int n, m;
+    int num[1010] = {0};
+    int num_cmp = 0;
+    int flag = 1;
+    int led[1005][1005];
+    cin >> n >> m;
+    for (i = 1; i <= n; i++)
+    {
+        for (j = 1; j <= m; j++)
+        {
+            cin >> led[i][j];
+            if (led[i][j] == 1)
+                num[k]++;
+        }
+        if (num[k] != 0)
+            k++;
+    }
+    num_cmp = num[k - 1];
+    ______________________;
+    if (flag == 1)
+        cout << "1" << endl;
+    else if (flag == 2)
+        cout << "8" << endl;
+    else
+        cout << "6" << endl;
+    return 0;
+}
+```
 
 ## template
 
@@ -37,39 +99,56 @@
 #include <stack>
 #include <queue>
 using namespace std;
-int main(){
-	int i,j,k=1;
-	int n,m;
-	int num[1010]={0};
-	int num_cmp=0;
-	int flag=1;
-	int led[1005][1005];
-	cin >> n >> m;
-	for(i=1;i<=n;i++){
-		for(j=1;j<=m;j++){
-			cin >> led[i][j];
-			if(led[i][j]==1)  num[k]++;
-		}
-		if(num[k]!=0)k++;
-	}
-	num_cmp=num[k-1];
-	for(i=k-1;i>0;i--){
-		if(num[i]<num_cmp) {
-			flag++;
-			num_cmp=num[i];
-		}
-	}
-	if(flag==1) cout <<"1"<<endl;
-	else if(flag==2) cout <<"8"<<endl;
-	else  cout <<"6"<<endl;
-	return 0;
+int main()
+{
+    int i, j, k = 1;
+    int n, m;
+    int num[1010] = {0};
+    int num_cmp = 0;
+    int flag = 1;
+    int led[1005][1005];
+    cin >> n >> m;
+    for (i = 1; i <= n; i++)
+    {
+        for (j = 1; j <= m; j++)
+        {
+            cin >> led[i][j];
+            if (led[i][j] == 1)
+                num[k]++;
+        }
+        if (num[k] != 0)
+            k++;
+    }
+    num_cmp = num[k - 1];
+    for (i = k - 1; i > 0; i--)
+    {
+        if (num[i] < num_cmp)
+        {
+            flag++;
+            num_cmp = num[i];
+        }
+    }
+    if (flag == 1)
+        cout << "1" << endl;
+    else if (flag == 2)
+        cout << "8" << endl;
+    else
+        cout << "6" << endl;
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+for (i = k - 1; i > 0; i--)
+{
+	if (num[i] < num_cmp)
+	{
+		flag++;
+		num_cmp = num[i];
+	}
+}
 ```
 
 ## 选项
@@ -77,17 +156,36 @@ int main(){
 ### A
 
 ```cpp
-
+for (i = k - 1; i > 0; i--)
+{
+	if (num[i] < num_cmp)
+	{
+		flag++;
+	}
+}
 ```
 
 ### B
 
 ```cpp
-
+for (i = k - 1; i > 0; i--)
+{
+	if (num[i] < num_cmp)
+	{
+		num_cmp = num[i];
+	}
+}
 ```
 
 ### C
 
 ```cpp
-
+for (i = k - 1; i > 0; i--)
+{
+	if (num[i] > num_cmp)
+	{
+		flag++;
+		num_cmp = num[i];
+	}
+}
 ```
\ No newline at end of file
diff --git "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/13.exercises/solution.md" "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/13.exercises/solution.md"
index d955dee7c03c7ab16c1a463c2e3e5f672c84a9bf..354b34b7538bdb8d1a5bf6ab537ec30d1ae5da1a 100644
--- "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/13.exercises/solution.md"
+++ "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/13.exercises/solution.md"
@@ -1,46 +1,85 @@
 # 约分
 
 编写程序，要求用户输入一个分数，然后将其约分为最简式。如：
+
 输入一个分数：8/12
+
 最简分式：2/3
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```cpp
+#include <stdio.h>
+#include <stdlib.h>
+int main()
+{
+    int a, b, x, y, c;
+    printf("输入一个分式：");
+    scanf("%d/%d", &a, &b);
+    _________________;
+    while (c)
+    {
+        x = y;
+        y = c;
+        c = x % y;
+    }
+    if (b / y != 1)
+        printf("最简分式为：%d/%d", a / y, b / y);
+    else
+        printf("最简分式为：%d", a / y);
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
-#include<stdio.h>
-#include<stdlib.h>
+#include <stdio.h>
+#include <stdlib.h>
 int main()
 {
-	int a,b,x,y,c;
-	printf("输入一个分式：");
-	scanf("%d/%d",&a,&b);
-	if(a<b)
-	{
-		x=b;y=a;
-	}
-	else
-	{
-		x=a;y=b;
-	}
-	c=x%y;
-	while(c)
-	{
-		x=y;
-		y=c;
-		c=x%y;
-	}
-	if(b/y!=1)
-		printf("最简分式为：%d/%d",a/y,b/y);
-	else
-		printf("最简分式为：%d",a/y);
-	return 0;
+    int a, b, x, y, c;
+    printf("输入一个分式：");
+    scanf("%d/%d", &a, &b);
+    if (a < b)
+    {
+        x = b;
+        y = a;
+    }
+    else
+    {
+        x = a;
+        y = b;
+    }
+    c = x % y;
+    while (c)
+    {
+        x = y;
+        y = c;
+        c = x % y;
+    }
+    if (b / y != 1)
+        printf("最简分式为：%d/%d", a / y, b / y);
+    else
+        printf("最简分式为：%d", a / y);
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+if (a < b)
+{
+	x = b;
+	y = a;
+}
+else
+{
+	x = a;
+	y = b;
+}
+c = x % y;
 ```
 
 ## 选项
@@ -48,17 +87,47 @@ int main()
 ### A
 
 ```cpp
-
+if (a < b)
+{
+	x = a;
+	y = b;
+}
+else
+{
+	x = b;
+	y = a;
+}
+c = x % y;
 ```
 
 ### B
 
 ```cpp
-
+if (a < b)
+{
+	x = a;
+	y = b;
+}
+else
+{
+	x = b;
+	y = a;
+}
+c = x / y;
 ```
 
 ### C
 
 ```cpp
-
+if (a < b)
+{
+	x = b;
+	y = a;
+}
+else
+{
+	x = a;
+	y = b;
+}
+c = x / y;
 ```
\ No newline at end of file
diff --git "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/14.exercises/solution.md" "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/14.exercises/solution.md"
index d41814508002fc5eefff42a11b5d1d62b06d5a6b..bf8482be990607f76f2a59b219e22ca07f1ddde3 100644
--- "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/14.exercises/solution.md"
+++ "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/14.exercises/solution.md"
@@ -2,48 +2,90 @@
 
 输入若干英文单词，将每个单词的首字母转换成大写字母，其他字母为小写，并按字典顺序排列
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```cpp
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+int cmp(const void *a, const void *b)
+{
+    return strcmp(*(char **)a, *(char **)b);
+}
+int main(int argc, char *argv[])
+{
+    int n = 0;
+    int i;
+    printf("how many words?\n");
+    scanf("%d", &n);
+    char **s = new char *[n];
+    for (i = 0; i < n; i++)
+    {
+        s[i] = new char[100];
+        scanf("%s", s[i]);
+        char *t = s[i];
+        while (*t != '\0')
+        {
+            _______________;
+        }
+    }
+    qsort(s, n, sizeof(char *), cmp);
+    for (i = 0; i < n; i++)
+    {
+        printf("%s\n", s[i]);
+    }
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
- #include <stdio.h>
+#include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
-int cmp(const void * a, const void * b)
+int cmp(const void *a, const void *b)
 {
-	return strcmp(*(char **)a, *(char **)b);
+    return strcmp(*(char **)a, *(char **)b);
 }
-int main(int argc, char* argv[])
+int main(int argc, char *argv[])
 {
-	int n = 0;
-	int i;
-	printf("how many words?\n");
-	scanf("%d", &n);
-	char ** s = new char *[n];
-	for (i = 0; i < n; i++)
-	{
-		s[i] = new char[100];
-		scanf("%s", s[i]);
-		char * t = s[i];
-		while (*t != '\0')
-		{
-			if (t == s[i] && (*t >= 'a' && *t <= 'z')) *t = *t - 'a' + 'A';
-			if (t > s[i] && (*t >= 'A' && *t <= 'Z')) *t = *t - 'A' + 'a';
-			t++;
-		}
-	}
-	qsort(s, n, sizeof(char *), cmp);
-	for (i = 0; i < n; i++)
-	{
-		printf("%s\n", s[i]);
-	}
-	return 0;
+    int n = 0;
+    int i;
+    printf("how many words?\n");
+    scanf("%d", &n);
+    char **s = new char *[n];
+    for (i = 0; i < n; i++)
+    {
+        s[i] = new char[100];
+        scanf("%s", s[i]);
+        char *t = s[i];
+        while (*t != '\0')
+        {
+            if (t == s[i] && (*t >= 'a' && *t <= 'z'))
+                *t = *t - 'a' + 'A';
+            if (t > s[i] && (*t >= 'A' && *t <= 'Z'))
+                *t = *t - 'A' + 'a';
+            t++;
+        }
+    }
+    qsort(s, n, sizeof(char *), cmp);
+    for (i = 0; i < n; i++)
+    {
+        printf("%s\n", s[i]);
+    }
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+if (t == s[i] && (*t >= 'a' && *t <= 'z'))
+	*t = *t - 'a' + 'A';
+if (t > s[i] && (*t >= 'A' && *t <= 'Z'))
+	*t = *t - 'A' + 'a';
+t++;
 ```
 
 ## 选项
@@ -51,17 +93,29 @@ int main(int argc, char* argv[])
 ### A
 
 ```cpp
-
+if (t == s[i] && (*t >= 'a' && *t <= 'z'))
+	t = t - 'a' + 'A';
+if (t > s[i] && (*t >= 'A' && *t <= 'Z'))
+	t = t - 'A' + 'a';
+t++;
 ```
 
 ### B
 
 ```cpp
-
+if (t == s[i] && (*t >= 'a' && *t <= 'z'))
+	t = t - 'A' + 'a';
+if (t > s[i] && (*t >= 'A' && *t <= 'Z'))
+	t = t - 'a' + 'A';
+t++;
 ```
 
 ### C
 
 ```cpp
-
+if (t == s[i] && (*t >= 'a' && *t <= 'z'))
+	*t = *t - 'A' + 'a';
+if (t > s[i] && (*t >= 'A' && *t <= 'Z'))
+	*t = *t - 'a' + 'A';
+t++;
 ```
\ No newline at end of file
diff --git "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/15.exercises/solution.md" "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/15.exercises/solution.md"
index fa91f80c84a58eefba35b07c645e03749f026d3d..dea7371ba5d385f3cb32296dc217544f74b4b1e0 100644
--- "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/15.exercises/solution.md"
+++ "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/15.exercises/solution.md"
@@ -1,22 +1,62 @@
 # 国名排序
 
-【字符数组】国名排序
-Description:
+**题目描述:**
+
 小李在准备明天的广交会，明天有来自世界各国的客房跟他们谈生意，小李要尽快的整理出名单给经理，你能帮他把客户来自的国家按英文字典次序排好吗？
 例如小李手上有来自加拿大，美国，中国的名单，排好的名单应是美国，加拿大，中国
-Input
-　　第一行为一个n（n<=100）表示n个国家，第２行到第n+1行分别为n个国家的名字．
-Output
-　　输出共计n行，为n个国家按字典顺序的排列，每行为一个国家
-Sample Input:
-	3
-	China
-	Canada
-	America
-Sample Output:
-	America
-	Canada
-	China
+
+**输入**
+
+第一行为一个n（n<=100）表示n个国家，第２行到第n+1行分别为n个国家的名字．
+
+**输出**
+
+输出共计n行，为n个国家按字典顺序的排列，每行为一个国家
+
+**输入样例:**
+
+```json
+3
+China
+Canada
+America
+```
+
+**输出样例:**
+
+```json
+America
+Canada
+China
+```
+
+以下程序实现了这一功能，请你填补空白处内容：
+
+```cpp
+#include <iostream>
+#include <string>
+using namespace std;
+string a[1000];
+int main()
+{
+    int i, n;
+    cin >> n;
+    for (i = 1; i <= n; i++)
+    {
+        cin >> a[i];
+    }
+    for (i = 1; i <= n; i++)
+    {
+        for (int j = i + 1; j <= n; j++)
+        {
+            _________;
+        }
+    }
+    for (int i = 1; i <= n; i++)
+        cout << a[i] << endl;
+    return 0;
+}
+```
 
 ## template
 
@@ -27,30 +67,31 @@ using namespace std;
 string a[1000];
 int main()
 {
-	int i,n;
-	cin>>n;
-	for(i=1; i<=n; i++)
-	{
-		cin>>a[i];
-	}
-	for(i=1;i<=n;i++)
-	{
-		for(int j=i+1;j<=n;j++)
-		{
-			if(a[i]>a[j])
-				swap(a[i],a[j]);
-		}
-	}
-	for(int i=1;i<=n;i++)
-		cout<<a[i]<<endl;
-	return 0;
+    int i, n;
+    cin >> n;
+    for (i = 1; i <= n; i++)
+    {
+        cin >> a[i];
+    }
+    for (i = 1; i <= n; i++)
+    {
+        for (int j = i + 1; j <= n; j++)
+        {
+            if (a[i] > a[j])
+                swap(a[i], a[j]);
+        }
+    }
+    for (int i = 1; i <= n; i++)
+        cout << a[i] << endl;
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+if (a[i] > a[j])
+    swap(a[i], a[j]);
 ```
 
 ## 选项
@@ -58,17 +99,20 @@ int main()
 ### A
 
 ```cpp
-
+if (a[i] < a[j])
+    swap(a[i], a[j]);
 ```
 
 ### B
 
 ```cpp
-
+if (a[i] <= a[j])
+    swap(a[i], a[j]);
 ```
 
 ### C
 
 ```cpp
-
+if (a[i] == a[j])
+    swap(a[i], a[j]);
 ```
\ No newline at end of file
diff --git "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/16.exercises/solution.md" "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/16.exercises/solution.md"
index 72625aa4e00d3b8e46cfaa86c4a010fe0596d9fe..445a43553dcac54bf459104f2a87ed0fb7c29488 100644
--- "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/16.exercises/solution.md"
+++ "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/16.exercises/solution.md"
@@ -2,34 +2,58 @@
 
 目前有一个长度为 n 的木棍，当做直角三角形的斜边。A,B,C要从许多整数长度的木棍中选出三根，分别长为 a, b, c。 现在，蒜头君和花椰妹的木棍组成一条直角边长度为 a + b，白菜君组成另外一条直角边 c，并且要求 a + b ≤ c。请问一共可以有多少种取木棍的方案。 提示：a = 3, b = 4 与 a = 4, b = 3 算作同一种方案。
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```cpp
+#include <stdio.h>
+int main()
+{
+    int n;
+    int cnt = 0;
+    scanf("%d", &n);
+    for (int a = 1; a < n; a++)
+        _______________;
+    printf("一共有%d种方案", cnt);
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
 #include <stdio.h>
 int main()
 {
-	int n;
-	int cnt = 0;
-	scanf("%d", &n);
-	for (int a = 1; a < n; a++)
-		for (int b = a; b < n - a; b++)
-			for (int c = 1; c < n; c++)
-			{
-				if ((a+b)*(a+b)+c*c==n*n)
-				{
-					printf("a=%d b=%d c=%d\n", a, b, c);
-					cnt++;
-				}
-			}
-	printf("一共有%d种方案", cnt);
-	return 0;
+    int n;
+    int cnt = 0;
+    scanf("%d", &n);
+    for (int a = 1; a < n; a++)
+        for (int b = a; b < n - a; b++)
+            for (int c = 1; c < n; c++)
+            {
+                if ((a + b) * (a + b) + c * c == n * n)
+                {
+                    printf("a=%d b=%d c=%d\n", a, b, c);
+                    cnt++;
+                }
+            }
+    printf("一共有%d种方案", cnt);
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+for (int b = a; b < n - a; b++)
+	for (int c = 1; c < n; c++)
+	{
+		if ((a + b) * (a + b) + c * c == n * n)
+		{
+			printf("a=%d b=%d c=%d\n", a, b, c);
+			cnt++;
+		}
+	}
 ```
 
 ## 选项
@@ -37,17 +61,41 @@ int main()
 ### A
 
 ```cpp
-
+for (int b = a; b < n - a; b++)
+	for (int c = 1; c < n; c++)
+	{
+		if ((a + b) * (a + b) + c * c >= n * n)
+		{
+			printf("a=%d b=%d c=%d\n", a, b, c);
+			cnt++;
+		}
+	}
 ```
 
 ### B
 
 ```cpp
-
+for (int b = a; b < n - a; b++)
+	for (int c = 1; c < n; c++)
+	{
+		if ((a + b) * (a + b) + c * c <= n * n)
+		{
+			printf("a=%d b=%d c=%d\n", a, b, c);
+			cnt++;
+		}
+	}
 ```
 
 ### C
 
 ```cpp
-
+for (int b = a; b < n - a; b++)
+	for (int c = 1; c < n; c++)
+	{
+		if ((a + b) * (a + b) + c * c > n * n)
+		{
+			printf("a=%d b=%d c=%d\n", a, b, c);
+			cnt++;
+		}
+	}
 ```
\ No newline at end of file
diff --git "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/17.exercises/solution.md" "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/17.exercises/solution.md"
index f4c518701c47f2f6208020e687d0f9c95335df5f..66218540b7994a4b840d41fed3fc3d8d10afcc25 100644
--- "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/17.exercises/solution.md"
+++ "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/17.exercises/solution.md"
@@ -10,42 +10,88 @@
 <strong>输出</strong><br />
 输出共一行 m 个整数&#xff0c;表示每回合报出的编号&#xff0c;每两个整数之间一个空格&#xff0c;最后一个数后面没有空格。</p>
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```cpp
+#include <iostream>
+#include <cstring>
+#include <vector>
+#include <algorithm>
+using namespace std;
+int ai[100010], qi[100010];
+int main()
+{
+    int a, q;
+    while (cin >> a >> q)
+    {
+        for (int i = 0; i < a; i++)
+            cin >> ai[i];
+        for (int i = 0; i < q; i++)
+            cin >> qi[i];
+        for (int i = 0; i < q; i++)
+        {
+            int left = 0, right = a - 1, mid;
+            while (left < right)
+            {
+                mid = (left + right) >> 1;
+                if (ai[mid] <= qi[i])
+                    left = mid + 1;
+                else
+                    right = mid;
+            }
+            _______________________;
+            i ? cout << " " << ai[left - 1] : cout << ai[left - 1];
+        }
+        cout << endl;
+    }
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
-#include<iostream>
-#include<cstring>
-#include<vector>
-#include<algorithm>
+#include <iostream>
+#include <cstring>
+#include <vector>
+#include <algorithm>
 using namespace std;
 int ai[100010], qi[100010];
 int main()
 {
-	int a, q;
-	while (cin >> a >> q)
-	{
-		for (int i = 0; i < a; i++)cin >> ai[i];
-		for (int i = 0; i < q; i++)cin >> qi[i];
-		for (int i = 0; i < q; i++) {
-			int left = 0, right = a - 1, mid;
-			while (left < right) {
-				mid = (left + right) >> 1;
-				if (ai[mid] <= qi[i])left = mid + 1;
-				else right = mid;
-			}
-			if (left - 1 < 0 || ai[left] < qi[i])left++;
-			i ? cout << " " << ai[left - 1] : cout << ai[left - 1];			
-		}
-		cout << endl;
-	}
-	return 0;
+    int a, q;
+    while (cin >> a >> q)
+    {
+        for (int i = 0; i < a; i++)
+            cin >> ai[i];
+        for (int i = 0; i < q; i++)
+            cin >> qi[i];
+        for (int i = 0; i < q; i++)
+        {
+            int left = 0, right = a - 1, mid;
+            while (left < right)
+            {
+                mid = (left + right) >> 1;
+                if (ai[mid] <= qi[i])
+                    left = mid + 1;
+                else
+                    right = mid;
+            }
+            if (left - 1 < 0 || ai[left] < qi[i])
+                left++;
+            i ? cout << " " << ai[left - 1] : cout << ai[left - 1];
+        }
+        cout << endl;
+    }
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+if (left - 1 < 0 || ai[left] < qi[i])
+    left++;
 ```
 
 ## 选项
@@ -53,17 +99,20 @@ int main()
 ### A
 
 ```cpp
-
+if (left - 1 < 0)
+    left++;
 ```
 
 ### B
 
 ```cpp
-
+if (ai[left] < qi[i])
+    left++;
 ```
 
 ### C
 
 ```cpp
-
+if (left - 1 < 0 && ai[left] < qi[i])
+    left++;
 ```
\ No newline at end of file
diff --git "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/18.exercises/solution.md" "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/18.exercises/solution.md"
index d99ab4dff7c0ddac05696a9a73b2478cfc0d38a4..14ba51dd34010e76c6e1c9470aec7d746229e2a8 100644
--- "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/18.exercises/solution.md"
+++ "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/18.exercises/solution.md"
@@ -1,17 +1,76 @@
 # 不喜欢带钱的小C
 
-题目描述：
+**题目描述：**
+
 小C不喜欢带钱，有一次竟被他碰上了一家不能使用移动支付（也不能找钱）的神秘商店。请问小C至少准备多少张RMB才能恰好支付n元。RMB的面额有100元，50元，20元，10元，5元，1元。
-输入格式：
+
+**输入格式：**
+
 输入一个整数n
-输出格式：
+
+**输出格式：**
+
 最少带几张。
-样例输入1：
+
+**样例输入1：**
+
+```json
 50
-样例输出1：
+```
+
+**样例输出1：**
+
+```json
 1
-约定：
+```
+
+**约定：**
+
+```json
 1<=n<=100
+```
+
+以下程序实现了这一功能，请你填补空白处内容：
+
+```cpp
+#include <iostream>
+using namespace std;
+int solve(int tar, int * meta, int metan, int * seed = NULL, int seedn = 0)
+{
+	if (tar == 0)
+	{
+		return seedn;
+	}
+	int min = -1;
+	int m;
+	int * seed1 = new int[seedn + 1];
+	if (seed)
+		memcpy(seed1, seed, sizeof(int) * seedn);
+	for (int i = 0; i < metan; i++)
+	{
+		if (meta[i] <= tar)
+		{
+			seed1[seedn] = meta[i];
+			_____________________;
+			if (m != -1 && (min == -1 || min > m))
+				min = m;
+			break;
+		}
+	}
+	delete[] seed1;
+	return min;
+}
+int main()
+{
+	int arr[] = { 100, 50, 20, 10, 5, 1 };
+	int n = 6;
+	int total;
+	cin >> total;
+	int result = solve(total, arr, n);
+	cout << result << endl;
+	return 0;
+}
+```
 
 ## template
 
@@ -58,7 +117,7 @@ int main()
 ## 答案
 
 ```cpp
-
+m = solve(tar - meta[i], meta, metan, seed1, seedn + 1);
 ```
 
 ## 选项
@@ -66,17 +125,17 @@ int main()
 ### A
 
 ```cpp
-
+m = solve(tar - meta[i], meta, metan, seed1, seedn);
 ```
 
 ### B
 
 ```cpp
-
+m = solve(tar + meta[i], meta, metan, seed1, seedn);
 ```
 
 ### C
 
 ```cpp
-
+m = solve(tar + meta[i], meta, metan, seed1, seedn + 1);
 ```
\ No newline at end of file
diff --git "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/2.exercises/solution.md" "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/2.exercises/solution.md"
index ebe26578b0364a25a0aa46b1e5fa9bb1e9ebe5e8..ab21c82353df2222eaf22cb3efdf759ca5ef3278 100644
--- "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/2.exercises/solution.md"
+++ "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/2.exercises/solution.md"
@@ -1,10 +1,18 @@
 # 数字归类
 
-题目描述
+**题目描述**
+
 一个数里面若含有数字1，则归类到1字类，含有数字2，则归类到2字类，所以一个数可能同时归类到不同的数字类。对于0、1、2、3、4、5、6、7、8、9这十个数字类，因研究需要，急于想知道某一堆数中，究竟归类到这些数字类的个数。
-样例输入
+
+**样例输入**
+
+```json
 123 456 175 2 61 9998 12 5053 382
-样例输出
+```
+
+**样例输出**
+
+```json
 0: 1
 1: 4
 2: 4
@@ -15,9 +23,41 @@
 7: 1
 8: 2
 9: 1
+```
+
 提示
+
 注意：输出结果中冒号后面有空格
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```cpp
+#include <stdio.h>
+#include <string.h>
+int result[10];
+int main(void)
+{
+    memset(result, 0, sizeof(int) * 10);
+    int n;
+    int arr[10];
+    while (scanf("%d", &n) != EOF)
+    {
+        memset(arr, 0, sizeof(int) * 10);
+        if (n == 0)
+            arr[0] = 1;
+        while (n > 0)
+        {
+			____________;
+        }
+        for (int i = 0; i < 10; i++)
+            result[i] += arr[i];
+    }
+    for (int i = 0; i < 10; i++)
+        printf("%d: %d\n", i, result[i]);
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
@@ -36,7 +76,8 @@ int main(void)
             arr[0] = 1;
         while (n > 0)
         {
-			____________
+			arr[n % 10] = 1;
+            n = n / 10;
         }
         for (int i = 0; i < 10; i++)
             result[i] += arr[i];
diff --git "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/3.exercises/solution.md" "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/3.exercises/solution.md"
index 5391420f6594e80213e2baf1009357ab4939d321..d5112dbb987e780aa2cded0b5f658f6b180becf4 100644
--- "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/3.exercises/solution.md"
+++ "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/3.exercises/solution.md"
@@ -1,14 +1,41 @@
 # 利用字母组成图形
 
 利用字母可以组成一些美丽的图形，下面给出了一个例子：
+
+```json
 ABCDEFG
 BABCDEF
 CBABCDE
 DCBABCD
 EDCBABC
+```
+
 这是一个5行7列的图形，请找出这个图形的规律，并输出一个n行m列的图形。
+
 输入一行，包含两个整数n和m，分别表示你要输出的图形的行数的列数。
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```cpp
+#include <stdio.h>
+#include <math.h>
+int main()
+{
+    int m, n;
+    scanf("%d%d", &n, &m);
+    int i, j;
+    for (i = 0; i < n; i++)
+    {
+        for (j = 0; j < m; j++)
+        {
+            ______________;
+        }
+        printf("\n");
+    }
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
@@ -23,7 +50,7 @@ int main()
     {
         for (j = 0; j < m; j++)
         {
-            ______________
+            printf("%c", 65 + abs(i - j));
         }
         printf("\n");
     }
diff --git "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/4.exercises/solution.md" "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/4.exercises/solution.md"
index 4e58868c8ae196ace419086e3a15eb04353f18f8..eb9b298b4ddfc7233b40ec808008874a80d47374 100644
--- "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/4.exercises/solution.md"
+++ "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/4.exercises/solution.md"
@@ -5,6 +5,31 @@
 <p style="text-align:center"><img alt="" src="https://img-ask.csdnimg.cn/upload/1624519041254.jpg" />
  </p>
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```cpp
+#include <stdio.h>
+int main()
+{
+    int a[10], i, n;
+    int isfind = 0;
+    printf("please set array values:");
+    for (i = 0; i < 10; i++)
+        scanf("%d", &a[i]);
+    printf("please enter one num:");
+    scanf("%d", &n);
+    for (i = 0; i < 10; i++)
+    {
+		______________;
+    }
+    if (isfind)
+        printf("\n");
+    else
+        printf("not find\n");
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
@@ -20,7 +45,11 @@ int main()
     scanf("%d", &n);
     for (i = 0; i < 10; i++)
     {
-		______________
+		if (a[i] < n)
+        {
+            isfind = 1;
+            printf("%d ", a[i]);
+        }
     }
     if (isfind)
         printf("\n");
diff --git "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/5.exercises/solution.md" "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/5.exercises/solution.md"
index 15b259953d1fc700a43ca958d97e060971a27fa3..ce6c18047bd1099f794707fb6213911ea900b2c1 100644
--- "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/5.exercises/solution.md"
+++ "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/5.exercises/solution.md"
@@ -15,6 +15,40 @@
 <pre>
 <code>9,1,1</code></pre>
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```cpp
+#include <stdio.h>
+int main()
+{
+    int a[6][6];
+    int m, n;
+    int i, j;
+    int max;
+    int indexx = 0, indexy = 0;
+    scanf("%d %d", &m, &n);
+    for (i = 0; i < m; i++)
+    {
+        for (j = 0; j < n; j++)
+        {
+            scanf("%d", &a[i][j]);
+            if (i == 0 && j == 0)
+            {
+                max = a[i][j];
+                indexx = 0;
+                indexy = 0;
+            }
+            else
+            {
+                _____________;
+            }
+        }
+    }
+    printf("%d,%d,%d\n", max, indexx, indexy);
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
@@ -40,7 +74,12 @@ int main()
             }
             else
             {
-                _____________
+                if (a[i][j] > max)
+                {
+                    max = a[i][j];
+                    indexx = i;
+                    indexy = j;
+                }
             }
         }
     }
diff --git "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/6.exercises/solution.md" "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/6.exercises/solution.md"
index 937e7b9176739a85514391a76046d780dd2e72dc..cb044bf52361abe6098305ea2bb0cb25be09ca34 100644
--- "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/6.exercises/solution.md"
+++ "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/6.exercises/solution.md"
@@ -2,8 +2,39 @@
 
 已知一个浮点数A（0<A<5），求它由哪两个整数B/C相除的值最接近，有相同值时要求B最小
 例如：
+
+```json
 A=0.2
 B=1 C=5
+```
+
+以下程序实现了这一功能，请你填补空白处内容：
+
+```cpp
+#include <stdio.h>
+#include <math.h>
+int main()
+{
+    float A = 0.2f;
+    int x = 0;
+    if (A < 0)
+    {
+        x = 1;
+        A = 1 / A;
+    }
+    float delta = 1;
+    int B = 1, C = 1;
+    do
+    {
+        ____________;
+    } while (delta > 0.000001);
+    if (x == 0)
+        printf("%d / %d", C, B - 1);
+    else
+        printf("%d / %d", B - 1, C);
+    return 0;
+}
+```
 
 ## template
 
@@ -23,7 +54,9 @@ int main()
     int B = 1, C = 1;
     do
     {
-        ____________
+        C = (int)(B * A);
+        delta = fabs(C / (float)B - A);
+        B++;
     } while (delta > 0.000001);
     if (x == 0)
         printf("%d / %d", C, B - 1);
diff --git "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/7.exercises/solution.md" "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/7.exercises/solution.md"
index 1de9536903a9f1dc64abcff231c579664a3415bb..d8245a81fe46fb16a27279e0a5be60e9eaee09f3 100644
--- "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/7.exercises/solution.md"
+++ "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/7.exercises/solution.md"
@@ -1,47 +1,107 @@
 # 矩阵问题
 
-题目描述
+**题目描述**
+
 编写以下函数：
+
 （1）在一个二维数组中形成以下形式的n阶矩阵：
 
-```json
-[1 1 1 1 1
-2 1 1 1 1
-3 2 1 1 1
-4 3 2 1 1
-5 4 3 2 1]
-```
+ ```json
+ [1 1 1 1 1
+ 2 1 1 1 1
+ 3 2 1 1 1
+ 4 3 2 1 1
+ 5 4 3 2 1]
+ ```
 
 （2）去掉靠边的元素，生成新的n-2阶矩阵；
+
 （3）求生成的n阶矩阵主对角线上的元素之和；
+
 （4）以方阵形式输出数组。
+
 在main函数中调用以上函数进行测试。
-输入
+
+**输入**
+
 输入生成矩阵的阶数(n>=2)
-输出
+
+**输出**
+
 以方阵形式输出生成的n阶矩阵、去掉靠边的元素生成的新的n-2阶矩阵、以及生成的n阶矩阵主对角线上的元素之和，最后一行要回车
-样例输入
+
+**样例输入**
+
+```json
 5
-样例输出
-Generated matrix:
+```
+
+**样例输出**
 
 ```json
+Generated matrix:
+
 1 1 1 1 1
 2 1 1 1 1
 3 2 1 1 1
 4 3 2 1 1
 5 4 3 2 1
-```
+
 del the elements on the side:
 
-```json
 1 1 1
 2 1 1
 3 2 1
-```
 
 The sum of the diagonal:5
+```
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```cpp
+#include <iostream>
+using namespace std;
+int main()
+{
+    while (1)
+    {
+        int a;
+        cin >> a;
+        int array[a][a];
+        for (int i = 0; i < a; i++)
+            for (int j = 0; j < a; j++)
+            {
+                if (j < i)
+                    array[i][j] = i + 1 - j;
+                else
+                    array[i][j] = 1;
+            }
+        cout << "Generated matrix:" << endl;
+        for (int i = 0; i < a; i++)
+        {
+            for (int j = 0; j < a; j++)
+            {
+                cout << array[i][j];
+            }
+            cout << endl;
+        }
+        cout << "del the elements on the side:" << endl;
+        for (int i = 1; i < a - 1; i++)
+        {
+            ________________;
+            cout << endl;
+        }
+        int sum = 0;
+        int i, j;
+        for (i = a - 2, j = 1; i >= 1; i--, j++)
+        {
+            sum += array[i][j];
+        }
+        cout << "The sum of the diagonal:" << sum << endl;
+    }
+    return 0;
+}
+```
 
 ## template
 
@@ -96,7 +156,10 @@ int main()
 ## 答案
 
 ```cpp
-
+for (int j = 1; j < a - 1; j++)
+{
+    cout << array[i][j];
+}
 ```
 
 ## 选项
@@ -104,17 +167,26 @@ int main()
 ### A
 
 ```cpp
-
+for (int j = 1; j < a - 1; j++)
+{
+    cout << array[i - 1][j - 1];
+}
 ```
 
 ### B
 
 ```cpp
-
+for (int j = 1; j < a - 1; j++)
+{
+    cout << array[i + 1][j + 1];
+}
 ```
 
 ### C
 
 ```cpp
-
+for (int j = 0; j < a; j++)
+{
+    cout << array[i][j];
+}
 ```
\ No newline at end of file
diff --git "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/8.exercises/solution.md" "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/8.exercises/solution.md"
index 888e255ef29142f9a43928683e4f36f6870cc9c8..a9db8c2c84ddf684b271136ceb26a3cc7c0a4c79 100644
--- "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/8.exercises/solution.md"
+++ "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/8.exercises/solution.md"
@@ -1,64 +1,133 @@
 # 偶数 or 奇数
 
-偶数 or 奇数
-时间限制: 1 Sec  内存限制: 128 MB
-题目描述
+**题目描述**
+
 编程班老师搞了一个有 N (1 <= N <= 100) 个正整数 I (1 <= I <= 10^60) 的表，叫 同学们去统计每个数里面数字（0，1，2，3，4，5，6，7，8，9）（注 0 为偶数）的奇偶数 字个数。写一个程序读入 N 个整数，统计每个整数的数字奇偶个数。
-输入
+
+**输入**
+
 第 1 行: 一个单独的整数: N 
+
 第 2 到第 N+1 行: 每行一个长长（小于等于 60 位）的整数，需要统计数字奇偶个数。
-输出
- 1..N 行: 第 j 行根据第 j 个整数输出奇数个数与偶数个数。
-样例输入
+
+**输出**
+
+1..N 行: 第 j 行根据第 j 个整数输出奇数个数与偶数个数。
+
+**样例输入**
+
+```json
 2
 1024
 5931
-样例输出
+```
+
+**样例输出**
+
+```json
 1 3
 4 0
+```
+
+以下程序实现了这一功能，请你填补空白处内容：
+
+```cpp
+#include <iostream>
+#include <string>
+using namespace std;
+string a[105];
+int b[105], c[105];
+int qiujishu(string x)
+{
+    int jishu = 0;
+    for (int i = 0; x.c_str()[i]; i++)
+    {
+        (1)____________;
+    }
+    return jishu;
+}
+int qiuoushu(string x)
+{
+    int oushu = 0;
+    for (int i = 0; x.c_str()[i]; i++)
+    {
+        (2)____________;
+    }
+    return oushu;
+}
+int main()
+{
+    int n;
+    cin >> n;
+    for (int i = 1; i <= n; i++)
+    {
+        cin >> a[i];
+        b[i] = qiujishu(a[i]);
+        c[i] = qiuoushu(a[i]);
+    }
+    for (int i = 1; i <= n; i++)
+    {
+        cout << b[i] << " " << c[i] << endl;
+    }
+    return 0;
+}
+```
 
 ## template
 
 ```cpp
-#include<iostream>
-#include<string>
+#include <iostream>
+#include <string>
 using namespace std;
 string a[105];
-int b[105],c[105];
-int qiujishu(string x){
-	int jishu=0;
-	for(int i = 0; x.c_str()[i]; i++){
-		if ((x.c_str()[i] - '0') % 2 == 1) jishu++;
-	}
-	return jishu;
+int b[105], c[105];
+int qiujishu(string x)
+{
+    int jishu = 0;
+    for (int i = 0; x.c_str()[i]; i++)
+    {
+        if ((x.c_str()[i] - '0') % 2 == 1)
+    		jishu++;
+    }
+    return jishu;
 }
-int qiuoushu(string x){
-	int oushu=0;
-	for(int i = 0; x.c_str()[i]; i++){
-		if ((x.c_str()[i] - '0') % 2 == 0) oushu++;
-	}
-	return oushu;
+int qiuoushu(string x)
+{
+    int oushu = 0;
+    for (int i = 0; x.c_str()[i]; i++)
+    {
+        if ((x.c_str()[i] - '0') % 2 == 0)
+    		oushu++;
+    }
+    return oushu;
 }
 int main()
 {
-	int n;
-	cin>>n;
-	for(int i=1;i<=n;i++){
-		cin>>a[i];
-		b[i]=qiujishu(a[i]);
-		c[i]=qiuoushu(a[i]);
-	}
-	for(int i=1;i<=n;i++){
-		cout<<b[i]<<" "<<c[i]<<endl;
-	}
-	return 0;
+    int n;
+    cin >> n;
+    for (int i = 1; i <= n; i++)
+    {
+        cin >> a[i];
+        b[i] = qiujishu(a[i]);
+        c[i] = qiuoushu(a[i]);
+    }
+    for (int i = 1; i <= n; i++)
+    {
+        cout << b[i] << " " << c[i] << endl;
+    }
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+(1)
+if ((x.c_str()[i] - '0') % 2 == 1)
+    jishu++;
+(2)
+if ((x.c_str()[i] - '0') % 2 == 0)
+    oushu++;
 ```
 
 ## 选项
@@ -66,17 +135,32 @@ int main()
 ### A
 
 ```cpp
-
+(1)
+if ((x.c_str()[i]) % 2 == 1)
+    jishu++;
+(2)
+if ((x.c_str()[i]) % 2 == 0)
+    oushu++;
 ```
 
 ### B
 
 ```cpp
-
+(1)
+if ((x.c_str()[i]) % 2 == 0)
+    jishu++;
+(2)
+if ((x.c_str()[i]) % 2 == 1)
+    oushu++;
 ```
 
 ### C
 
 ```cpp
-
+(1)
+if ((x.c_str()[i]) / 2 == 1)
+    jishu++;
+(2)
+if ((x.c_str()[i]) / 2 == 0)
+    oushu++;
 ```
\ No newline at end of file
diff --git "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/9.exercises/solution.md" "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/9.exercises/solution.md"
index 1c793212e30ad948565e0c6e8f4ad270565c3b60..b5c2cd3b1483cf88860c51845bda7c9fdf380982 100644
--- "a/data/2.dailycode\344\270\255\351\230\266/1.cpp/9.exercises/solution.md"
+++ "b/data/2.dailycode\344\270\255\351\230\266/1.cpp/9.exercises/solution.md"
@@ -19,6 +19,32 @@
 <pre>
 B 252</pre>
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```cpp
+#include <iostream>
+using namespace std;
+int main() {
+	double n, A, B;	
+	cin >> n;
+	B = 12 * 0.7 * n;
+	if (n <= 10) {
+		A = 10 * n;
+	}
+	else if (n > 10 && n <= 20) {
+		A = 10 * 10 + (n - 10) * 8.2;
+	}
+	___________________________;
+	if (A < B) {
+		cout << "A " << A;
+	}
+	else {
+		cout << "B " << B;
+	}
+	return 0;
+}
+```
+
 ## template
 
 ```cpp
@@ -50,7 +76,9 @@ int main() {
 ## 答案
 
 ```cpp
-
+else {
+	A = 10 * 8.2 + (n - 20) * 7.5 + 10 * 10;
+}
 ```
 
 ## 选项
@@ -58,17 +86,23 @@ int main() {
 ### A
 
 ```cpp
-
+else {
+	A = 20 * 7.5;
+}
 ```
 
 ### B
 
 ```cpp
-
+else {
+	A = (n - 10) * 8.2 + (n - 20) * 7.5 + 10 * 10;
+}
 ```
 
 ### C
 
 ```cpp
-
+else {
+	A = (n - 10) * 8.2 + (n - 20) * 7.5;
+}
 ```
\ No newline at end of file