send topic success

data/3.dailycode高阶/1.cpp/10.exercises/solution.md

@@ -2,58 +2,130 @@
 
 <p>设有n种不同面值的硬币&#xff0c;第i种硬币的币值是Vi(其中V1&#61;1),重量是Wi,i&#61;1,2&#xff0c;...n且现在购买某种总币值为y的商品&#xff0c;需要用这些硬币付款&#xff0c;如果每种钱币使用的个数不限&#xff0c;那么如何选择付款的方法使得付出钱币的总重量最轻&#xff1f;使用动态规划设计策略设计一个求解该问题的算法。假设问题的输入实例是&#xff1a;</p><p style="margin-left:.38in">V1&#61;1&#xff0c; V2&#61;4&#xff0c; V3&#61;6&#xff0c; V4&#61;8</p><p style="margin-left:.38in">W1&#61;1&#xff0c; W2&#61;2&#xff0c;W3&#61;4&#xff0c;W4&#61;6</p><p style="margin-left:.38in">Y&#61;12</p><p>要求输出优化函数表和标记函数表、以及硬币支付方式。</p>
 
+以下程序实现了这一功能，请你填补空白处的内容：
+
+```cpp
+#include <stdio.h>
+void strcpy(int *a, int *b, int Y)
+{
+	for (int i = 0; i <= Y; i++)
+		*(a + i) = *(b + i);
+}
+void solve()
+{
+    int n;
+    scanf("%d", &n);
+    int type[n], weight[n], Y, i, j, k;
+    for (i = 0; i < n; i++)
+        scanf("%d", &type[i]);
+    for (i = 0; i < n; i++)
+        scanf("%d", &weight[i]);
+    scanf("%d", &Y);
+    int Min[Y + 1], Min_Path[Y + 1], path[n][Y + 1];
+    for (i = 0; i <= Y; i++)
+        Min[i] = 9999;
+    Min[0] = 0;
+    printf("\n");
+    for (j = 0; j < n; j++)
+    {
+		___________________
+        for (k = 1; k <= Y; k++)
+            printf("%-3d", Min[k]);
+        printf("\n");
+        strcpy(path[j], Min_Path, Y);
+    }
+    printf("\n");
+    for (i = 0; i < n; i++)
+    {
+        for (j = 1; j <= Y; j++)
+            printf("%-3d", path[i][j]);
+        printf("\n");
+    }
+    int y = Y;
+    printf("\n支付方式:");
+    while (y)
+    {
+        printf("%d ", Min_Path[y]);
+        y -= Min_Path[y];
+    }
+    printf("\n总重量:%d\n", Min[Y]);
+}
+int main()
+{
+    solve();
+    return 1;
+}
+```
 
 ## template
 
 ```cpp
-#include<stdio.h>
-void strcpy(int *a, int *b, int Y){
-	for(int i=0;i<=Y;i++) *(a+i) = *(b+i);
+#include <stdio.h>
+void strcpy(int *a, int *b, int Y)
+{
+    for (int i = 0; i <= Y; i++)
+        *(a + i) = *(b + i);
 }
-void solve(){
-	int n; scanf("%d",&n);
-	int type[n], weight[n], Y, i, j, k;
-	for(i=0;i<n;i++) scanf("%d",&type[i]);
-	for(i=0;i<n;i++) scanf("%d",&weight[i]);
-	scanf("%d",&Y);
-	int Min[Y+1], Min_Path[Y+1], path[n][Y+1];
-	for(i=0;i<=Y;i++) Min[i] = 9999;
-	Min[0] = 0;
-	printf("\n");
-	for(j=0;j<n;j++){
-		for(i=type[j]; i<=Y; i++)
-			if(Min[i] > Min[i-type[j]]+weight[j]){
-				Min_Path[i] = type[j];
-				Min[i] = Min[i-type[j]]+weight[j];
-			}
-		for(k=1;k<=Y;k++) printf("%-3d",Min[k]);
-		printf("\n");
-		strcpy(path[j],Min_Path,Y);
-	}
-	printf("\n");
-	for(i=0;i<n;i++){
-		for(j=1;j<=Y;j++)
-			printf("%-3d",path[i][j]);
-		printf("\n");
-	}
-	int y=Y;
-	printf("\n支付方式:");
-	while (y){
-		printf("%d ",Min_Path[y]);
-		y -= Min_Path[y];
-	}
-	printf("\n总重量:%d\n",Min[Y]);
+void solve()
+{
+    int n;
+    scanf("%d", &n);
+    int type[n], weight[n], Y, i, j, k;
+    for (i = 0; i < n; i++)
+        scanf("%d", &type[i]);
+    for (i = 0; i < n; i++)
+        scanf("%d", &weight[i]);
+    scanf("%d", &Y);
+    int Min[Y + 1], Min_Path[Y + 1], path[n][Y + 1];
+    for (i = 0; i <= Y; i++)
+        Min[i] = 9999;
+    Min[0] = 0;
+    printf("\n");
+    for (j = 0; j < n; j++)
+    {
+        for (i = type[j]; i <= Y; i++)
+            if (Min[i] > Min[i - type[j]] + weight[j])
+            {
+                Min_Path[i] = type[j];
+                Min[i] = Min[i - type[j]] + weight[j];
+            }
+        for (k = 1; k <= Y; k++)
+            printf("%-3d", Min[k]);
+        printf("\n");
+        strcpy(path[j], Min_Path, Y);
+    }
+    printf("\n");
+    for (i = 0; i < n; i++)
+    {
+        for (j = 1; j <= Y; j++)
+            printf("%-3d", path[i][j]);
+        printf("\n");
+    }
+    int y = Y;
+    printf("\n支付方式:");
+    while (y)
+    {
+        printf("%d ", Min_Path[y]);
+        y -= Min_Path[y];
+    }
+    printf("\n总重量:%d\n", Min[Y]);
 }
-int main(){
-	solve();
-	return 1;
+int main()
+{
+    solve();
+    return 1;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+for (i = type[j]; i <= Y; i++)
+	if (Min[i] > Min[i - type[j]] + weight[j])
+	{
+		Min_Path[i] = type[j];
+		Min[i] = Min[i - type[j]] + weight[j];
+	}
 ```
 
 ## 选项
@@ -61,17 +133,32 @@ int main(){
 ### A
 
 ```cpp
-
+for (i = type[j]; i <= Y; i++)
+	if (Min[i] < Min[i - type[j]] + weight[j])
+	{
+		Min_Path[i] = type[j];
+		Min[i] = Min[i - type[j]] + weight[j];
+	}
 ```
 
 ### B
 
 ```cpp
-
+for (i = type[j]; i <= Y; i++)
+	if (Min[i] > Min[i - type[j]] + weight[j])
+	{
+		Min_Path[i] = type[j];
+		Min[i] = Min[i - type[j]];
+	}
 ```
 
 ### C
 
 ```cpp
-
+for (i = type[j]; i <= Y; i++)
+	if (Min[i] < Min[i - type[j]] + weight[j])
+	{
+		Min_Path[i] = type[j];
+		Min[i] = Min[i - type[j]];
+	}
 ```
\ No newline at end of file

data/3.dailycode高阶/1.cpp/4.exercises/solution.md

 # 计算出现次数最多的整数及其出现次数
 
 <p>【问题描述】
-输入一组无序的整数&#xff0c;编程输出其中出现次数最多的整数及其出现次数。
-【输入形式】
-先从标准输入读入整数的个数&#xff08;大于等于1&#xff0c;小于等于100&#xff09;&#xff0c;然后在下一行输入这些整数&#xff0c;各整数之间以一个空格分隔。
-【输出形式】
-在标准输出上输出出现次数最多的整数及其出现次数&#xff0c;两者以一个空格分隔&#xff1b;若出现次数最多的整数有多个&#xff0c;则按照整数升序分行输出。
-【样例输入】
-10
-0 -50 0 632 5813 -50 9 -50 0 632
-【样例输出】
--50 3
-0 3
-【样例说明】
+输入一组无序的整数&#xff0c;编程输出其中出现次数最多的整数及其出现次数。</br>
+【输入形式】</br>
+先从标准输入读入整数的个数&#xff08;大于等于1&#xff0c;小于等于100&#xff09;&#xff0c;然后在下一行输入这些整数&#xff0c;各整数之间以一个空格分隔。</br>
+【输出形式】</br>
+在标准输出上输出出现次数最多的整数及其出现次数&#xff0c;两者以一个空格分隔&#xff1b;若出现次数最多的整数有多个&#xff0c;则按照整数升序分行输出。</br>
+【样例输入】</br>
+10</br>
+0 -50 0 632 5813 -50 9 -50 0 632</br>
+【样例输出】</br>
+-50 3</br>
+0 3</br>
+【样例说明】</br>
 输入了10个整数&#xff0c;其中出现次数最多的是-50和0&#xff0c;都是出现3次。</p>
 
+以下程序实现了这一功能，请你填补空白处的内容：
+
+```cpp
+#include <stdio.h>
+int main()
+{
+    int a[50], b[50], c[50], n, i, j, t, max;
+    scanf("%d", &n);
+    for (i = 0; i < n; i++)
+    {
+        scanf("%d", &a[i]);
+    }
+    for (i = 1; i < n; i++)
+        for (j = 0; j < n - 1; j++)
+        {
+            if (a[j] > a[j + 1])
+            {
+                t = a[j];
+                a[j] = a[j + 1];
+                a[j + 1] = t;
+            }
+        }
+    j = 0;
+    t = -1;
+    for (i = 0; i < n - 1; i++)
+    {
+		____________________
+    }
+    b[j] = n - 1 - t;
+    c[j] = n - 1;
+    max = b[0];
+    for (i = 1; i <= j; i++)
+    {
+        if (max < b[i])
+        {
+            max = b[i];
+        }
+    }
+    for (i = 0; i <= j; i++)
+        if (b[i] == max)
+        {
+            t = c[i];
+            printf("%d %d\n", a[t], b[i]);
+        }
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
 #include <stdio.h>
 int main()
 {
-	int a[50],b[50],c[50],n,i,j,t,max;
-	scanf("%d",&n);
-	for(i=0;i<n;i++)
-	   {
-		scanf("%d",&a[i]);
-	   }
-	for(i=1;i<n;i++)
-		for(j=0;j<n-1;j++)
-		{
-		  if(a[j]>a[j+1])
-		  {
-			  t=a[j];
-			  a[j]=a[j+1];
-			  a[j+1]=t;
-		  }
-		}
-		j=0;
-		t=-1;
-	for(i=0;i<n-1;i++)
-		{
-			if(a[i]!=a[i+1])
-			{
-				b[j]=i-t;
-				c[j]=i;
-				t=i;
-				j++;
-			}
-		}
-		   b[j]=n-1-t;
-		   c[j]=n-1;
-		max=b[0];
-	for(i=1;i<=j;i++)
-	{
-		if(max<b[i])
-			{
-				max=b[i];
-			}
-	}
-	for(i=0;i<=j;i++)
-		if(b[i]==max)
-	{
-		t=c[i];
-		printf("%d %d\n",a[t],b[i]);
-	}
-	return 0;
+    int a[50], b[50], c[50], n, i, j, t, max;
+    scanf("%d", &n);
+    for (i = 0; i < n; i++)
+    {
+        scanf("%d", &a[i]);
+    }
+    for (i = 1; i < n; i++)
+        for (j = 0; j < n - 1; j++)
+        {
+            if (a[j] > a[j + 1])
+            {
+                t = a[j];
+                a[j] = a[j + 1];
+                a[j + 1] = t;
+            }
+        }
+    j = 0;
+    t = -1;
+    for (i = 0; i < n - 1; i++)
+    {
+        if (a[i] != a[i + 1])
+        {
+            b[j] = i - t;
+            c[j] = i;
+            t = i;
+            j++;
+        }
+    }
+    b[j] = n - 1 - t;
+    c[j] = n - 1;
+    max = b[0];
+    for (i = 1; i <= j; i++)
+    {
+        if (max < b[i])
+        {
+            max = b[i];
+        }
+    }
+    for (i = 0; i <= j; i++)
+        if (b[i] == max)
+        {
+            t = c[i];
+            printf("%d %d\n", a[t], b[i]);
+        }
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+if (a[i] != a[i + 1])
+{
+	b[j] = i - t;
+	c[j] = i;
+	t = i;
+	j++;
+}
 ```
 
 ## 选项
@@ -80,17 +134,35 @@ int main()
 ### A
 
 ```cpp
-
+if (a[i] != a[i + 1])
+{
+	b[j] = i + t;
+	c[j] = i - 1;
+	t = i;
+	j++;
+}
 ```
 
 ### B
 
 ```cpp
-
+if (a[i] != a[i + 1])
+{
+	b[j] = i + t;
+	c[j] = i + 1;
+	t = i;
+	j++;
+}
 ```
 
 ### C
 
 ```cpp
-
+if (a[i] != a[i + 1])
+{
+	b[j] = i + t;
+	c[j] = i;
+	t = i;
+	j++;
+}
 ```
\ No newline at end of file

data/3.dailycode高阶/1.cpp/5.exercises/solution.md

 # 六角填数
 
-<p>![图片说明](https://img-ask.csdn.net/upload/202006/14/1592104139_240178.png)
-六角填数
-题目描述
+![图片说明](https://img-ask.csdn.net/upload/202006/14/1592104139_240178.png)
+
+**题目描述**
+
 如下图所示六角形中，有12个点，依次填入1~12的数字，使得每条直线上的数字之和都相同。其中，已经替你填好了点1，2，3的数字，请你计算其他位置所代表的数字是多少？
-输入
+
+**输入**
+
 输入仅一行，以空格隔开，分别表示已经填好的点1,2,3的数字。
-输出
+
+**输出**
+
 输出仅一行，以空格隔开，分别表示所有位置所代表的数字。
-样例输入：
+
+**样例输入：**
+
+```json
 1 8 2
-样例输出：
-1 8 2 9 7 11 10 12 3 5 6 4</p>
+```
+
+**样例输出：**
+
+```json
+1 8 2 9 7 11 10 12 3 5 6 4
+```
+
+以下程序实现了这一功能，请你填补空白处的内容：
+
+```cpp
+#include <iostream>
+#include <cmath>
+#include <cstdio>
+#include <cstring>
+using namespace std;
+#define eps 10e-10
+#define N 15
+int a[N];
+bool vis[N];
+void dfs(int x)
+{
+    if (x == 1 || x == 2 || x == 3)
+    {
+        dfs(x + 1);
+        return;
+    }
+    if (x > 12)
+    {
+        int t[6];
+        t[0] = a[1] + a[3] + a[6] + a[8];
+        t[1] = a[1] + a[4] + a[7] + a[11];
+        t[2] = a[2] + a[3] + a[4] + a[5];
+        t[3] = a[2] + a[6] + a[9] + a[12];
+        t[4] = a[8] + a[9] + a[10] + a[11];
+        t[5] = a[12] + a[10] + a[7] + a[5];
+        for (int i = 1; i < 6; ++i)
+        {
+            if (t[i] != t[i - 1])
+                return;
+        }
+        for (int i = 1; i <= 12; i++)
+            cout << a[i] << " ";
+        return;
+    }
+    for (int i = 1; i < 13; ++i)
+    {
+		________________
+    }
+}
+int main()
+{
+    memset(vis, 0, sizeof(vis));
+    cin >> a[1];
+    vis[a[1]] = 1;
+    cin >> a[2];
+    vis[a[2]] = 1;
+    cin >> a[3];
+    vis[a[3]] = 1;
+    dfs(1);
+    return 0;
+}
+```
 
 ## template
 
@@ -25,51 +94,66 @@ using namespace std;
 #define N 15
 int a[N];
 bool vis[N];
-void dfs(int x){
-	if(x == 1 || x == 2 || x == 3){
-		dfs(x+1);
-		return ;
-	}
-	if(x > 12){
-		int t[6];
-		t[0] = a[1] + a[3] + a[6] + a[8];
-		t[1] = a[1] + a[4] + a[7] + a[11];
-		t[2] = a[2] + a[3] + a[4] + a[5];
-		t[3] = a[2] + a[6] + a[9] + a[12];
-		t[4] = a[8] + a[9] + a[10] + a[11];
-		t[5] = a[12] + a[10] + a[7] + a[5];
-		for(int i = 1; i < 6; ++i){
-			if(t[i] != t[i-1])return ;
-		}
-		for (int i = 1; i <= 12; i++) cout << a[i] << " ";
-		return ;
-	}
-	for(int i = 1;i < 13; ++i){
-		if(!vis[i]){
-			vis[i] = 1;
-			a[x] = i;
-			dfs(x+1);
-			vis[i] = 0;
-		}
-	}
+void dfs(int x)
+{
+    if (x == 1 || x == 2 || x == 3)
+    {
+        dfs(x + 1);
+        return;
+    }
+    if (x > 12)
+    {
+        int t[6];
+        t[0] = a[1] + a[3] + a[6] + a[8];
+        t[1] = a[1] + a[4] + a[7] + a[11];
+        t[2] = a[2] + a[3] + a[4] + a[5];
+        t[3] = a[2] + a[6] + a[9] + a[12];
+        t[4] = a[8] + a[9] + a[10] + a[11];
+        t[5] = a[12] + a[10] + a[7] + a[5];
+        for (int i = 1; i < 6; ++i)
+        {
+            if (t[i] != t[i - 1])
+                return;
+        }
+        for (int i = 1; i <= 12; i++)
+            cout << a[i] << " ";
+        return;
+    }
+    for (int i = 1; i < 13; ++i)
+    {
+        if (!vis[i])
+        {
+            vis[i] = 1;
+            a[x] = i;
+            dfs(x + 1);
+            vis[i] = 0;
+        }
+    }
 }
-int main(){
-	memset(vis,0,sizeof(vis));
-	cin >> a[1];
-	vis[a[1]] = 1;
-	cin >> a[2];
-	vis[a[2]] = 1;
-	cin >> a[3];
-	vis[a[3]] = 1;
-	dfs(1);
-	return 0;
+int main()
+{
+    memset(vis, 0, sizeof(vis));
+    cin >> a[1];
+    vis[a[1]] = 1;
+    cin >> a[2];
+    vis[a[2]] = 1;
+    cin >> a[3];
+    vis[a[3]] = 1;
+    dfs(1);
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+if (!vis[i])
+{
+	vis[i] = 1;
+	a[x] = i;
+	dfs(x + 1);
+	vis[i] = 0;
+}
 ```
 
 ## 选项
@@ -77,17 +161,35 @@ int main(){
 ### A
 
 ```cpp
-
+if (!vis[i])
+{
+	vis[i] = 1;
+	a[x] = i;
+	dfs(x - 1);
+	vis[i] = 0;
+}
 ```
 
 ### B
 
 ```cpp
-
+if (!vis[i])
+{
+	vis[i] = 0;
+	a[x] = i;
+	dfs(x + 1);
+	vis[i] = 1;
+}
 ```
 
 ### C
 
 ```cpp
-
+if (!vis[i])
+{
+	vis[i] = 0;
+	a[x] = i + 1;
+	dfs(x + 1);
+	vis[i] = 1;
+}
 ```
\ No newline at end of file

data/3.dailycode高阶/1.cpp/6.exercises/solution.md

 # 阶乘和数
 
 一个正整数如果等于组成它的各位数字的阶乘之和，该整数称为阶乘和数。
+
 例如，145=1!+4!+5!，则145是一个三位阶乘和数。
+
 请输出所有阶乘数（不会超过十万）
 
+以下程序实现了这一功能，请你填补空白处的内容：
+
+```cpp
+#include <stdio.h>
+int b[10] = {0, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880};
+int main()
+{
+    int x, sum = 0, a;
+    for (int j = 1; j <= 100000; j++)
+    {
+        sum = 0;
+        a = j;
+        while (j != 0)
+        {
+            ______________
+        }
+        j = a;
+        if (sum == a)
+        {
+            printf("%d ", j);
+        }
+    }
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
 #include <stdio.h>
 int b[10] = {0, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880};
-int main(){
-    int x, sum = 0, a;
-    for(int j = 1; j <= 100000; j++){
-        sum = 0; 
-        a = j;
-        while(j != 0){
-            x = j % 10;
-            j /= 10;
-            sum += b[x];
-        }
-        j = a;
-        if(sum == a){
-            printf("%d ", j);
-        }
-    }
-    return 0;
-} 
+int main()
+{
+    int x, sum = 0, a;
+    for (int j = 1; j <= 100000; j++)
+    {
+        sum = 0;
+        a = j;
+        while (j != 0)
+        {
+            x = j % 10;
+            j /= 10;
+            sum += b[x];
+        }
+        j = a;
+        if (sum == a)
+        {
+            printf("%d ", j);
+        }
+    }
+    return 0;
+}
 ```
 
 ## 答案
 
 ```cpp
-
+x = j % 10;
+j /= 10;
+sum += b[x];
 ```
 
 ## 选项
@@ -39,17 +73,23 @@ int main(){
 ### A
 
 ```cpp
-
+x = j / 10;
+j %= 10;
+sum += b[x];
 ```
 
 ### B
 
 ```cpp
-
+x = j % 10;
+j %= 10;
+sum += b[x];
 ```
 
 ### C
 
 ```cpp
-
+x = j / 10;
+j /= 10;
+sum += b[x];
 ```
\ No newline at end of file

data/3.dailycode高阶/1.cpp/7.exercises/solution.md

@@ -2,6 +2,52 @@
 
 <p>在主函数中随机生成一有n个元素的一维数组(元素的取值范围载10-90之间),调用排序函数对该数组进行排序,并输出排序结果。</p>
 
+以下程序实现了这一功能，请你填补空白处的内容：
+
+```cpp
+#include <stdio.h>
+#include <random>
+#include <iostream>
+#include <string>
+#include <time.h>
+using namespace std;
+void InsertSort(int a[], int n)
+{
+    for (int i = 1; i < n; i++)
+    {
+        if (a[i] < a[i - 1])
+        {
+            int j = i - 1;
+            int x = a[i];
+			________________
+        }
+    }
+}
+int main()
+{
+    cout << "请输入产生多个随机数：";
+    int ranNum[100], num = 0;
+    srand((unsigned)time(NULL));
+    cin >> num;
+    cout << "随机数组：";
+    for (int i = 0; i < num; i++)
+    {
+        ranNum[i] = 10 + (rand() % 80);
+        cout << " " << ranNum[i];
+    }
+    cout << endl;
+    InsertSort(ranNum, num);
+    cout << "排序后整数序列：";
+    for (int j = num - 1; j >= 0; j--)
+    {
+        cout << ranNum[j] << " ";
+    }
+    cout << endl;
+    system("pause");
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
@@ -13,47 +59,57 @@
 using namespace std;
 void InsertSort(int a[], int n)
 {
-	for (int i = 1; i<n; i++) {
-		if (a[i] < a[i - 1]) {			   
-			int j = i - 1;
-			int x = a[i];		
-			a[i] = a[i - 1];		   
-			while (x < a[j]) {  
+    for (int i = 1; i < n; i++)
+    {
+        if (a[i] < a[i - 1])
+        {
+            int j = i - 1;
+            int x = a[i];
+			a[i] = a[i - 1];
+			while (x < a[j])
+			{
 				a[j + 1] = a[j];
-				j--;		 
+				j--;
 			}
-			a[j + 1] = x;	  
-		}
-	}
+			a[j + 1] = x;
+        }
+    }
 }
 int main()
 {
-	cout << "请输入产生多个随机数：";
-	int ranNum[100],num = 0;
-	srand((unsigned)time(NULL));
-	cin >> num;
-	cout << "随机数组：";
-	for (int i = 0; i < num; i++)
-	{
-		ranNum[i] = 10 + (rand() % 80);
-		cout << " " << ranNum[i];
-	}
-	cout << endl;
-	InsertSort(ranNum, num);
-	cout << "排序后整数序列：";
-	for (int j = num-1; j >= 0; j--) {
-		cout << ranNum[j] << " ";
-	}
-	cout << endl;
-	system("pause");
-	return 0;
+    cout << "请输入产生多个随机数：";
+    int ranNum[100], num = 0;
+    srand((unsigned)time(NULL));
+    cin >> num;
+    cout << "随机数组：";
+    for (int i = 0; i < num; i++)
+    {
+        ranNum[i] = 10 + (rand() % 80);
+        cout << " " << ranNum[i];
+    }
+    cout << endl;
+    InsertSort(ranNum, num);
+    cout << "排序后整数序列：";
+    for (int j = num - 1; j >= 0; j--)
+    {
+        cout << ranNum[j] << " ";
+    }
+    cout << endl;
+    system("pause");
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+a[i] = a[i - 1];
+while (x < a[j])
+{
+	a[j + 1] = a[j];
+	j--;
+}
+a[j + 1] = x;
 ```
 
 ## 选项
@@ -61,17 +117,35 @@ int main()
 ### A
 
 ```cpp
-
+a[i] = a[i - 1];
+while (x < a[j])
+{
+	a[j + 1] = a[j];
+	j++;
+}
+a[j + 1] = x;
 ```
 
 ### B
 
 ```cpp
-
+a[i] = a[i - 1];
+while (x < a[j])
+{
+	a[j - 1] = a[j];
+	j--;
+}
+a[j - 1] = x;
 ```
 
 ### C
 
 ```cpp
-
+a[i] = a[i - 1];
+while (x < a[j])
+{
+	a[j - 1] = a[j];
+	j++;
+}
+a[j + 1] = x;
 ```
\ No newline at end of file

data/3.dailycode高阶/1.cpp/8.exercises/solution.md

 # 结合两个字符串
 
 写一个结合两个字符串的方法，从第一个字符串中取出一个字符，然后从第二个字符串中取出一个字符，以此类推。一旦一个字符串没有字符，它就应该继续使用另一个字符串
+
 输入：两个字符串，如s1="day"和s2="time"
+
 输出：一个结果字符串，对于上面的输入情况，它将是“dtaiyme”。
 
+以下程序实现了这一功能，请你填补空白处的内容：
+
+```cpp
+#include <iostream>
+#include <string>
+using namespace std;
+string StrCon(const string& a, const string& b)
+{
+	string c;
+	int n = a.size(), m = b.size();
+	if (0 == n)	return a;
+	if (0 == m) return b;
+	int i, j;
+	for (i = 0, j = 0; i < n && j < m; ++i, ++j)
+	{
+		c += a[i];
+		c += b[i];
+	}
+	__________________
+	return c;
+}
+int main()
+{
+	string s = "day", t = "time";
+	cout << StrCon(s, t) << endl;
+	system("pause");
+	return 0;
+}
+```
+
 ## template
 
 ```cpp
@@ -40,7 +72,10 @@ int main()
 ## 答案
 
 ```cpp
-
+while (i < n)
+	c += a[i++];
+while (j < m)
+	c += b[j++];
 ```
 
 ## 选项
@@ -48,17 +83,26 @@ int main()
 ### A
 
 ```cpp
-
+while (i > n)
+	c += a[i++];
+while (j < m)
+	c += b[j++];
 ```
 
 ### B
 
 ```cpp
-
+while (i > n)
+	c += a[i++];
+while (j > m)
+	c += b[j++];
 ```
 
 ### C
 
 ```cpp
-
+while (i < n)
+	c += a[i++];
+while (j > m)
+	c += b[j++];
 ```
\ No newline at end of file

data/3.dailycode高阶/1.cpp/9.exercises/solution.md

@@ -4,6 +4,30 @@
 例如给定序列： 99 35 83 38 ， 处理完成后得到：38 83 99 35
 给定序列&#xff1a; 6 7 3 4 &#xff0c;处理完成后得到&#xff1a;3 4 6 7</p>
 
+以下程序实现了这一功能，请你填补空白处的内容：
+
+```cpp
+#include <stdio.h>
+#include <math.h>
+int main()
+{
+    int i, j, n, a[10005], t;
+    scanf("%d", &n);
+    for (i = 0; i < n; i++)
+    {
+        scanf("%d", &a[i]);
+        if (i % 2 == 1)
+        {
+			__________________
+        }
+    }
+    for (i = n - 1; i >= 0; i--)
+    {
+        printf("%d ", a[i]);
+    }
+}
+```
+
 ## template
 
 ```cpp
@@ -11,28 +35,37 @@
 #include <math.h>
 int main()
 {
-	int i,j,n,a[10005],t;
-	scanf("%d",&n);
-	for(i=0;i<n;i++){
-		scanf("%d",&a[i]);
-		if(i%2==1){
-			if((a[i]%3==0)||(a[i]%5==0)||(a[i-1]%3==0)||(a[i-1]%5==0)){
-				t=a[i];
-				a[i]=a[i-1];
-				a[i-1]=t;
-			}
-		}
-	}
-	for(i=n-1;i>=0;i--){
-		printf("%d ",a[i]);
-	}
+    int i, j, n, a[10005], t;
+    scanf("%d", &n);
+    for (i = 0; i < n; i++)
+    {
+        scanf("%d", &a[i]);
+        if (i % 2 == 1)
+        {
+            if ((a[i] % 3 == 0) || (a[i] % 5 == 0) || (a[i - 1] % 3 == 0) || (a[i - 1] % 5 == 0))
+            {
+                t = a[i];
+                a[i] = a[i - 1];
+                a[i - 1] = t;
+            }
+        }
+    }
+    for (i = n - 1; i >= 0; i--)
+    {
+        printf("%d ", a[i]);
+    }
 }
 ```
 
 ## 答案
 
 ```cpp
-
+if ((a[i] % 3 == 0) || (a[i] % 5 == 0) || (a[i - 1] % 3 == 0) || (a[i - 1] % 5 == 0))
+{
+	t = a[i];
+	a[i] = a[i - 1];
+	a[i - 1] = t;
+}
 ```
 
 ## 选项
@@ -40,17 +73,32 @@ int main()
 ### A
 
 ```cpp
-
+if ((a[i] % 3 == 0) || (a[i] % 5 == 0))
+{
+	t = a[i];
+	a[i] = a[i - 1];
+	a[i - 1] = t;
+}
 ```
 
 ### B
 
 ```cpp
-
+if (((a[i - 1] % 3 == 0) || (a[i - 1] % 5 == 0))
+{
+	t = a[i];
+	a[i] = a[i - 1];
+	a[i - 1] = t;
+}
 ```
 
 ### C
 
 ```cpp
-
+if ((a[i] % 3 == 0) && (a[i] % 5 == 0) && (a[i - 1] % 3 == 0) && (a[i - 1] % 5 == 0))
+{
+	t = a[i];
+	a[i] = a[i - 1];
+	a[i - 1] = t;
+}
 ```
\ No newline at end of file