From f17722dd860a10a748f6666aede4aa2da5b293a8 Mon Sep 17 00:00:00 2001
From: zhangzc <zhangzc@csdn.net>
Date: Mon, 29 Nov 2021 18:08:18 +0800
Subject: [PATCH] update dailycode pipline data

---
 .../1.cpp/10.exercises/solution.md"           |  60 ++++++-
 .../1.cpp/11.exercises/solution.md"           | 105 ++++++++----
 .../1.cpp/12.exercises/solution.md"           | 144 +++++++++++-----
 .../1.cpp/13.exercises/solution.md"           |  99 ++++++++---
 .../1.cpp/14.exercises/solution.md"           | 118 ++++++++++----
 .../1.cpp/15.exercises/solution.md"           | 136 ++++++++++++----
 .../1.cpp/16.exercises/solution.md"           |  87 +++++++---
 .../1.cpp/17.exercises/solution.md"           |  34 ++--
 .../1.cpp/18.exercises/solution.md"           | 130 +++++++++++----
 .../1.cpp/19.exercises/solution.md"           | 116 +++++++++----
 .../1.cpp/2.exercises/solution.md"            |  53 +++++-
 .../1.cpp/20.exercises/solution.md"           | 154 ++++++++++++++----
 .../1.cpp/21.exercises/solution.md"           |  69 ++++++--
 .../1.cpp/22.exercises/solution.md"           | 141 ++++++++++++----
 .../1.cpp/23.exercises/solution.md"           | 142 +++++++++++-----
 .../1.cpp/24.exercises/solution.md"           |  86 +++++++---
 .../1.cpp/25.exercises/solution.md"           |  47 +++++-
 .../1.cpp/26.exercises/solution.md"           |  58 +++++--
 .../1.cpp/27.exercises/solution.md"           |  90 +++++++---
 .../1.cpp/28.exercises/solution.md"           | 102 +++++++++---
 .../1.cpp/3.exercises/solution.md"            |  31 +++-
 .../1.cpp/4.exercises/solution.md"            |  41 ++++-
 .../1.cpp/5.exercises/solution.md"            |  64 +++++++-
 .../1.cpp/6.exercises/solution.md"            |  38 ++++-
 .../1.cpp/7.exercises/solution.md"            |  39 ++++-
 .../1.cpp/8.exercises/solution.md"            |  36 +++-
 .../1.cpp/9.exercises/solution.md"            |  50 +++++-
 27 files changed, 1741 insertions(+), 529 deletions(-)

diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/10.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/10.exercises/solution.md"
index ca502cbb7..0eec75f17 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/10.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/10.exercises/solution.md"
@@ -3,6 +3,42 @@
 <p>给定公司N名员工的工龄&#xff0c;要求按工龄增序输出每个工龄段有多少员工。输入首先给出正整数N&#xff0c;即员工总人数&#xff1b;
 随后给出N个整数&#xff0c;即每个员工的工龄&#xff0c;范围在[0, 99]。其中&#xff0c;0-9为第1个工龄段&#xff0c;10-19为第2个工龄段&#xff0c;…&#xff0c;90-99为第10个工龄段。按工龄的递增顺序输出每个工龄的员工个数&#xff0c;格式为&#xff1a;“工龄:人数”。每项占一行。如果人数为0则不输出该项。</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 i,n,a[11]= {0},x;
+	scanf("%d",&n);
+	for(i=0; i<n; i++)
+	{
+		scanf("%d",&x);
+		if(x<=9)
+			a[1]++;
+		else if(x>9&&x<=19)
+			a[2]++;
+		else if(x>19&&x<=29)
+			a[3]++;
+		else if(x>29&&x<=39)
+			a[4]++;
+		else if(x>39&&x<=49)
+			a[5]++;
+		else if(x>49&&x<=59)
+			a[6]++;
+		else if(x>59&&x<=69)
+			a[7]++;
+		else if(x>69&&x<=79)
+			a[8]++;
+		else if(x>79&&x<=89)
+			a[9]++;
+		else
+			a[10]++;
+	}
+	________________
+}
+```
+
 ## template
 
 ```cpp
@@ -46,7 +82,11 @@ int main()
 ## 答案
 
 ```cpp
-
+for(i=1;i<=10;i++){
+	if(a[i]>0){
+		printf("%d-%d:%d\n",i*10-10,i*10-1,a[i]);
+	}
+}
 ```
 
 ## 选项
@@ -54,17 +94,29 @@ int main()
 ### A
 
 ```cpp
-
+for(i=0;i<=10;i++){
+	if(a[i]>0){
+		printf("%d-%d:%d\n",i*10-10,i*10-1,a[i]);
+	}
+}
 ```
 
 ### B
 
 ```cpp
-
+for(i=1;i<=10;i++){
+	if(a[i]>0){
+		printf("%d-%d:%d\n",i*10,i-1,a[i]);
+	}
+}
 ```
 
 ### C
 
 ```cpp
-
+for(i=0;i<=10;i++){
+	if(a[i]>0){
+		printf("%d-%d:%d\n",i*10,i-1,a[i]);
+	}
+}
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/11.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/11.exercises/solution.md"
index de5427ef2..0bf09e871 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/11.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/11.exercises/solution.md"
@@ -2,40 +2,86 @@
 
 有一分数序列:2/1,-3/2,5/3,-8/5,13/8,-21/13,…, 由用户输入项目数N，求这个数列的前N 项之和
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```cpp
+#include <stdlib.h>
+#include <stdio.h>
+int main()
+{
+    int n;
+    scanf("%d", &n);
+    int i;
+    double a1 = 2, b1 = 1;
+    double a2 = 3, b2 = 2;
+    double sum = a1 / b1 - a2 / b2;
+    if (n == 1)
+        printf("%f\n", a1 / b1);
+    else if (n == 2)
+        printf("%f\n", sum);
+    else
+    {
+        for (i = 0; i < n - 2; i++)
+        {
+            double exp = a2 / b2;
+            _____________________
+            sum += exp;
+            double a = a1 + a2;
+            double b = b1 + b2;
+            a1 = a2;
+            b1 = b2;
+            a2 = a;
+            b2 = b;
+        }
+        printf("%f\n", sum);
+    }
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
-#include<stdlib.h> 
-#include<stdio.h>
-int main(){
-	int n;
-	scanf("%d",&n);
-	int i;
-	double a1 = 2, b1 = 1;
-	double a2 = 3, b2 = 2;
-	double sum = a1/b1 - a2/b2;
-	if(n==1) printf("%f\n",a1/b1);
-	else if (n==2) printf("%f\n",sum);
-	else{
-		for(i = 0;i<n-2;i++){
-			double exp = a2 / b2;
-			if(i%2==0) exp *= -1;
-			sum += exp;
-			double a = a1 + a2;
-			double b = b1 + b2;
-			a1 = a2;b1 = b2;
-			a2 = a; b2 = b;
-		}
-		printf("%f\n",sum);
-	}
-	return 0;
+#include <stdlib.h>
+#include <stdio.h>
+int main()
+{
+    int n;
+    scanf("%d", &n);
+    int i;
+    double a1 = 2, b1 = 1;
+    double a2 = 3, b2 = 2;
+    double sum = a1 / b1 - a2 / b2;
+    if (n == 1)
+        printf("%f\n", a1 / b1);
+    else if (n == 2)
+        printf("%f\n", sum);
+    else
+    {
+        for (i = 0; i < n - 2; i++)
+        {
+            double exp = a2 / b2;
+            if (i % 2 == 0)
+                exp *= -1;
+            sum += exp;
+            double a = a1 + a2;
+            double b = b1 + b2;
+            a1 = a2;
+            b1 = b2;
+            a2 = a;
+            b2 = b;
+        }
+        printf("%f\n", sum);
+    }
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+if (i % 2 == 0)
+    exp *= -1;
 ```
 
 ## 选项
@@ -43,17 +89,20 @@ int main(){
 ### A
 
 ```cpp
-
+if (i % 2 == 0)
+    exp = -1;
 ```
 
 ### B
 
 ```cpp
-
+if (i / 2 == 0)
+    exp *= -1;
 ```
 
 ### C
 
 ```cpp
-
+if (i / 2 == 0)
+    exp = -1;
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/12.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/12.exercises/solution.md"
index d97d0ae2b..60f309a0b 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/12.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/12.exercises/solution.md"
@@ -1,56 +1,106 @@
 # 输入一个正整数n（代表图形的行数），输出如样例形式的图形。
 
 输入：7
+
 输出：
-        D      D
-        CD    DC
-        BCD  DCB
-        ABCDDCBA
-        BCD  DCB
-        CD    DC
-        D      D
+
+```json
+    D      D
+    CD    DC
+    BCD  DCB
+    ABCDDCBA
+    BCD  DCB
+    CD    DC
+    D      D
+```
+
+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 <vector>
+#include <string>
+#include <iostream>
+#include <algorithm>
+using namespace std;
+int main()
+{
+    int n;
+    cin >> n;
+    vector<string> a(n, ""), b(n, "");
+    int m = (n + 1) / 2;
+    int p = 0;
+    for (int i = m - 1; i >= 0; i--)
+    {
+        for (int j = 0; j <= i; j++)
+            a[i].push_back('A' + j + p);
+        b[i] = a[i];
+        reverse(b[i].begin(), b[i].end());
+        ___________________
+        p++;
+    }
+    p = 0;
+    for (int i = n - 1; i >= m; i--)
+    {
+        a[i] = a[p];
+        b[i] = b[p++];
+    }
+    for (int i = 0; i < n; i++)
+        cout << a[i] << b[i] << endl;
+    return 0;
+}
+```
+
 
 ## template
 
 ```cpp
-#include<stdio.h>
-#include<vector>
-#include<string>
-#include<iostream>
-#include<algorithm>
+#include <stdio.h>
+#include <vector>
+#include <string>
+#include <iostream>
+#include <algorithm>
 using namespace std;
-int main() {
-	int n;
-	cin>>n;
-	vector<string> a(n,""),b(n,"");
-	int m=(n+1)/2;
-	int p=0;
-	for(int i=m-1;i>=0;i--){
-		for(int j=0;j<=i;j++)
-			a[i].push_back('A'+j+p);
-		b[i]=a[i];
-		reverse(b[i].begin(),b[i].end());
-		for(int j=i+1;j<m;j++){
-			a[i]+=" ";
-			b[i]+=" ";
-		}
-		p++;
-	}
-	p=0;
-	for(int i=n-1;i>=m;i--){
-		a[i]=a[p];
-		b[i]=b[p++];
-	}
-	for(int i=0;i<n;i++)
-		cout<<a[i]<<b[i]<<endl;
-	return 0;
+int main()
+{
+    int n;
+    cin >> n;
+    vector<string> a(n, ""), b(n, "");
+    int m = (n + 1) / 2;
+    int p = 0;
+    for (int i = m - 1; i >= 0; i--)
+    {
+        for (int j = 0; j <= i; j++)
+            a[i].push_back('A' + j + p);
+        b[i] = a[i];
+        reverse(b[i].begin(), b[i].end());
+        for (int j = i + 1; j < m; j++)
+        {
+            a[i] += " ";
+            b[i] = " " + b[i];
+        }
+        p++;
+    }
+    p = 0;
+    for (int i = n - 1; i >= m; i--)
+    {
+        a[i] = a[p];
+        b[i] = b[p++];
+    }
+    for (int i = 0; i < n; i++)
+        cout << a[i] << b[i] << endl;
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+for (int j = i + 1; j < m; j++)
+{
+    a[i] += " ";
+    b[i] = " " + b[i];
+}
 ```
 
 ## 选项
@@ -58,17 +108,29 @@ int main() {
 ### A
 
 ```cpp
-
+for (int j = i + 1; j < m; j++)
+{
+    a[i] += " ";
+    b[i] += " ";
+}
 ```
 
 ### B
 
 ```cpp
-
+for (int j = i; j <= m; j++)
+{
+    a[i] += " ";
+    b[i] = " " + b[i];
+}
 ```
 
 ### C
 
 ```cpp
-
+for (int j = i; j <= m; j++)
+{
+    a[i] += " ";
+    b[i] += " ";
+}
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/13.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/13.exercises/solution.md"
index 52d1e3b1d..7ed51ed5b 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/13.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/13.exercises/solution.md"
@@ -6,43 +6,78 @@
 输出格式:
 在一行中输出循环右移M位以后的整数序列，之间用空格分隔，序列结尾不能有多余空格。
 输入样例:
-	6 2
-	1 2 3 4 5 6
+
+```json
+6 2
+1 2 3 4 5 6
+```
+
 输出样例:
-	5 6 1 2 3 4
 
+```json
+5 6 1 2 3 4
+```
+
+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, m, a[1000];
+    scanf("%d %d", &n, &m);
+    m = m % n;
+    int count = m;
+    ______________
+    for (int i = 0; i < count; i++)
+        scanf("%d", &a[i]);
+    int first = 1;
+    for (int i = 0; i < n; i++)
+    {
+        if (!first)
+            printf(" ");
+        printf("%d", a[i]);
+        first = 0;
+    }
+}
+```
 
 ## template
 
 ```cpp
-#include<stdio.h>
+#include <stdio.h>
 int main()
 {
-	int n,m,a[1000];
-	scanf("%d %d",&n,&m);
-	m = m % n;
-	int count=m;
-	while(m<n)
-	{
-		scanf("%d",&a[m]);
-		m++;
-	}
-	for(int i=0;i<count;i++)
-		scanf("%d",&a[i]);
-	int first=1;
-	for(int i=0;i<n;i++)
-	{
-		if(!first)printf(" ");
-		printf("%d",a[i]);
-		first=0;
-	}
+    int n, m, a[1000];
+    scanf("%d %d", &n, &m);
+    m = m % n;
+    int count = m;
+    while (m < n)
+    {
+        scanf("%d", &a[m]);
+        m++;
+    }
+    for (int i = 0; i < count; i++)
+        scanf("%d", &a[i]);
+    int first = 1;
+    for (int i = 0; i < n; i++)
+    {
+        if (!first)
+            printf(" ");
+        printf("%d", a[i]);
+        first = 0;
+    }
 }
 ```
 
 ## 答案
 
 ```cpp
-
+while (m < n)
+{
+	scanf("%d", &a[m]);
+	m++;
+}
 ```
 
 ## 选项
@@ -50,17 +85,29 @@ int main()
 ### A
 
 ```cpp
-
+while (m < n)
+{
+	scanf("%d", &a[m + 1]);
+	m++;
+}
 ```
 
 ### B
 
 ```cpp
-
+while (m < n)
+{
+	scanf("%d", &a[m]);
+	m--;
+}
 ```
 
 ### C
 
 ```cpp
-
+while (m < n)
+{
+	scanf("%d", &a[m + 1]);
+	m--;
+}
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/14.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/14.exercises/solution.md"
index 472fcec48..f5783816d 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/14.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/14.exercises/solution.md"
@@ -2,6 +2,45 @@
 
 <p>在一个数组A中存放100个数据&#xff0c;用子函数判断该数组中哪些是素数&#xff0c;并统计该素数的个数&#xff0c;在主函数中输出该素数的个数</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>
+#include <stdlib.h>
+int isPrime(int n)
+{
+    int i;
+    if (n < 2)
+        return 0;
+    _______________
+    return 1;
+}
+int CountPrime(int a[], int size)
+{
+    int i = 0, count = 0;
+    for (i = 0; i < size; i++)
+    {
+        if (isPrime(a[i]))
+        {
+            printf("%d ", a[i]);
+            count++;
+            if (count % 10 == 0)
+                printf("\n");
+        }
+    }
+    printf("\n");
+    return count;
+}
+int main()
+{
+    int a[100], i, count = 0;
+    for (i = 0; i < 100; i++)
+        a[i] = rand() % 1000;
+    printf("素数的个数:%d\n", CountPrime(a, 100));
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
@@ -9,45 +48,50 @@
 #include <stdlib.h>
 int isPrime(int n)
 {
-	int i = 2;
-	if(n<2) return 0;
-	for (i=2;i<n;i++)
-	{
-		if(n%i == 0)
-			return 0;
-	}
-	return 1;
+    int i;
+    if (n < 2)
+        return 0;
+    for (i = 2; i * i <= n; ++i)
+    {
+        if (n % i == 0)
+            return 0;
+    }
+    return 1;
 }
-int CountPrime(int a[],int size)
+int CountPrime(int a[], int size)
 {
-	int i =0,count = 0;
-	for (i = 0;i<size;i++)
-	{
-		if(isPrime(a[i]))
-		{
-			printf("%d ",a[i]);
-			count++;
-			if(count%10 == 0) 
-				printf("\n");
-		}
-	}
-	printf("\n");
-	return count;
+    int i = 0, count = 0;
+    for (i = 0; i < size; i++)
+    {
+        if (isPrime(a[i]))
+        {
+            printf("%d ", a[i]);
+            count++;
+            if (count % 10 == 0)
+                printf("\n");
+        }
+    }
+    printf("\n");
+    return count;
 }
 int main()
 {
-	int a[100],i,count = 0;
-	for(i = 0;i<100;i++)
-		a[i] = rand()%1000;
-	printf("素数的个数:%d\n",CountPrime(a,100));
-	return 0;
+    int a[100], i, count = 0;
+    for (i = 0; i < 100; i++)
+        a[i] = rand() % 1000;
+    printf("素数的个数:%d\n", CountPrime(a, 100));
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+for (i = 2; i * i <= n; ++i)
+{
+	if (n % i == 0)
+		return 0;
+}
 ```
 
 ## 选项
@@ -55,17 +99,29 @@ int main()
 ### A
 
 ```cpp
-
+for (i = 2; i <= n; ++i)
+{
+	if (n % i == 0)
+		return 0;
+}
 ```
 
 ### B
 
 ```cpp
-
+for (i = 2; i * i <= n; ++i)
+{
+	if (n % i == 0)
+		return 1;
+}
 ```
 
 ### C
 
 ```cpp
-
+for (i = 2; i * i <= n; ++i)
+{
+	if (n / i == 0)
+		return 0;
+}
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/15.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/15.exercises/solution.md"
index d05cdbd77..a29f44c02 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/15.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/15.exercises/solution.md"
@@ -1,11 +1,35 @@
 # 商品优惠计算器
 
 <p>商品优惠计算器
-使用if语句编程实现输入购货金额&#xff0c;输出实际付款金额。购货折扣率如下&#xff1a;
-购货金额≤500元 不打折
-500元&lt;购货金额≤1000元 9折
+使用if语句编程实现输入购货金额&#xff0c;输出实际付款金额。购货折扣率如下&#xff1a;<br />
+购货金额≤500元 不打折<br />
+500元&lt;购货金额≤1000元 9折<br />
 1000元&lt;购货金额 8折</p>
 
+以下程序实现了这一功能，请你补全空白处内容：
+
+```cpp
+#include <stdio.h>
+#include <ctype.h>
+#include <stdbool.h>
+#include <stdlib.h>
+#include <string.h>
+#include <unistd.h>
+int main()
+{
+    float money = 0.0;
+    float pay = 0.0;
+    bool run = true;
+    while (run)
+    {
+        printf("\n请输入购货金额:\n");
+        scanf("%f", &money);
+        ____________________
+    }
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
@@ -17,36 +41,49 @@
 #include <unistd.h>
 int main()
 {
-	float money=0.0;
-	float pay=0.0;
-	bool run = true;
-	while(run)
-	{
-		printf("\n请输入购货金额:\n");
-		scanf("%f", &money);
-		if(money >1000)
-		{
-			pay=money*0.8;
-			printf("打八折，应付金额:%.2f\n",pay);
-		}
-		else if((money >500)&&(money <=1000))
-		{
-			pay=money*0.9;
-			printf("打九折，应付金额:%.2f\n",pay);
-		}
-		else if(money <=500)
-		{
-			printf("不打折，应付金额:%.2f\n",money);
-		}
-	}
-	return 0;
+    float money = 0.0;
+    float pay = 0.0;
+    bool run = true;
+    while (run)
+    {
+        printf("\n请输入购货金额:\n");
+        scanf("%f", &money);
+        if (money > 1000)
+        {
+            pay = money * 0.8;
+            printf("打八折，应付金额:%.2f\n", pay);
+        }
+        else if ((money > 500) && (money <= 1000))
+        {
+            pay = money * 0.9;
+            printf("打九折，应付金额:%.2f\n", pay);
+        }
+        else if (money <= 500)
+        {
+            printf("不打折，应付金额:%.2f\n", money);
+        }
+    }
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+if (money > 1000)
+{
+	pay = money * 0.8;
+	printf("打八折，应付金额:%.2f\n", pay);
+}
+else if ((money > 500) && (money <= 1000))
+{
+	pay = money * 0.9;
+	printf("打九折，应付金额:%.2f\n", pay);
+}
+else if (money <= 500)
+{
+	printf("不打折，应付金额:%.2f\n", money);
+}
 ```
 
 ## 选项
@@ -54,17 +91,56 @@ int main()
 ### A
 
 ```cpp
-
+if (money > 1000)
+{
+	pay = money * 0.8;
+	printf("打八折，应付金额:%.2f\n", pay);
+}
+elif ((money > 500) && (money <= 1000))
+{
+	pay = money * 0.9;
+	printf("打九折，应付金额:%.2f\n", pay);
+}
+elif (money <= 500)
+{
+	printf("不打折，应付金额:%.2f\n", money);
+}
 ```
 
 ### B
 
 ```cpp
-
+if (money > 1000)
+{
+	pay = money * 0.8;
+	printf("打八折，应付金额:%.2f\n", pay);
+}
+else ((money > 500) && (money <= 1000))
+{
+	pay = money * 0.9;
+	printf("打九折，应付金额:%.2f\n", pay);
+}
+else (money <= 500)
+{
+	printf("不打折，应付金额:%.2f\n", money);
+}
 ```
 
 ### C
 
 ```cpp
-
+if (money > 1000):
+{
+	pay = money * 0.8;
+	printf("打八折，应付金额:%.2f\n", pay);
+}
+if ((money > 500) && (money <= 1000)):
+{
+	pay = money * 0.9;
+	printf("打九折，应付金额:%.2f\n", pay);
+}
+if (money <= 500):
+{
+	printf("不打折，应付金额:%.2f\n", money);
+}
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/16.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/16.exercises/solution.md"
index d0066de96..2ef04c3ce 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/16.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/16.exercises/solution.md"
@@ -1,45 +1,79 @@
 # 移动数组中的元素
 
 题目描述
+
 将一维数组中的元素循环左移 k 个位置
+
 输入描述
+
 第 1 行是一维数组元素的个数 n (数组大小)
+
 第 2 行是一个整数 k , 表示移动的位置
+
 下面 n 行为数组的元素个数
+
 输出描述
+
 输出 n 行，表示移动后的数字
 
+以下程序实现了这一功能，请你补全空白处内容：
+
+```cpp
+#include <stdio.h>
+#define N 10000
+int main()
+{
+    int k, a[N], b[N], n, t, w, i;
+    scanf("%d", &n);
+    scanf("%d", &k);
+    for (i = 0; i < n; i++)
+        scanf("%d", &a[i]);
+    for (i = 0; i < k % n; i++)
+        b[i] = a[i];
+    for (i = 0; i < n; i++)
+    {
+        __________________
+    }
+    for (i = 0; i < n; i++)
+        printf("%d\n", a[i]);
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
-#include<stdio.h>
+#include <stdio.h>
 #define N 10000
 int main()
 {
-	int k,a[N],b[N],n,t,w,i;
-	scanf("%d",&n);
-	scanf("%d",&k);
-	for(i=0;i<n;i++)
-		scanf("%d",&a[i]);
-	for (i = 0; i < k % n; i++)
-		b[i] = a[i];
-	for (i = 0; i < n; i++)
-	{
-		if (i < n - k % n)
-			a[i] = a[i + k % n];
-		else
-			a[i] = b[i - n + k % n];
-	}
-	for(i=0;i<n;i++)
-		printf("%d\n",a[i]);
-	return 0;
+    int k, a[N], b[N], n, t, w, i;
+    scanf("%d", &n);
+    scanf("%d", &k);
+    for (i = 0; i < n; i++)
+        scanf("%d", &a[i]);
+    for (i = 0; i < k % n; i++)
+        b[i] = a[i];
+    for (i = 0; i < n; i++)
+    {
+        if (i < n - k % n)
+            a[i] = a[i + k % n];
+        else
+            a[i] = b[i - n + k % n];
+    }
+    for (i = 0; i < n; i++)
+        printf("%d\n", a[i]);
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+if (i < n - k % n)
+	a[i] = a[i + k % n];
+else
+	a[i] = b[i - n + k % n];
 ```
 
 ## 选项
@@ -47,17 +81,26 @@ int main()
 ### A
 
 ```cpp
-
+if (i < n - k % n)
+	a[i] = a[i + k + n];
+else
+	a[i] = b[i + k - n];
 ```
 
 ### B
 
 ```cpp
-
+if (i < n - k % n)
+	a[i] = a[i + k % n];
+else
+	a[i] = b[i - n + k];
 ```
 
 ### C
 
 ```cpp
-
+if (i < n - k % n)
+	a[i] = a[i + k - n];
+else
+	a[i] = b[i - n + k % n];
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/17.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/17.exercises/solution.md"
index c62c5d930..d1ac78259 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/17.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/17.exercises/solution.md"
@@ -1,35 +1,33 @@
 # 个位数是6，且能被3整除的五位数共有多少个？
 
-<p>要求：
-1、必须包含循环结构、顺序结构、选择分支结构。
-2、必须包含数组
-3、必须包含一个以上函数
-4、可以包含方针</p>
+请你直接选择出正确答案
 
 ## template
 
 ```cpp
 #include <iostream>
 using namespace std;
-void search1(){
-	int i,t=0;
-	for(i=10000;i<=99999;i++){
-		if(i%3==0&&i%10==6)
-			t++;
-	}
-	cout<<t;
+void search1()
+{
+    int i, t = 0;
+    for (i = 10000; i <= 99999; i++)
+    {
+        if (i % 3 == 0 && i % 10 == 6)
+            t++;
+    }
+    cout << t;
 }
 int main()
 {
-	search1();
-	return 0;
+    search1();
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+3000
 ```
 
 ## 选项
@@ -37,17 +35,17 @@ int main()
 ### A
 
 ```cpp
-
+2000
 ```
 
 ### B
 
 ```cpp
-
+3096
 ```
 
 ### C
 
 ```cpp
-
+2048
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/18.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/18.exercises/solution.md"
index b49dcde10..a0badb839 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/18.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/18.exercises/solution.md"
@@ -1,47 +1,103 @@
-# 一个整数的序列，要求对其重新排序
+# 冒泡排序法排序
 
-<p>一个整数的序列&#xff0c;要求对其重新排序。排序要求:
-1.奇数在前&#xff0c;偶数在后&#xff1b;
-2.奇数按从大到小排序&#xff1b;
-3.偶数按从小到大排序。<br />
-输入一行&#xff0c;包含整数个数n&#xff0c; n个整数值&#xff0c;彼此以一个空格分开。
-按照要求排序后输出一行&#xff0c;包含排序后的n 个整数。
- </p>
+输入n（1≤n≤10）个整数，用冒泡排序法对其从小到大排序，共进行n-1趟，要求输出每一趟的排序情况
 
-## template
+输入格式:
+
+先输入个数n，再输入n个整数。
+
+输出格式:
+
+第1趟结果
+
+第2趟结果
+
+......
+
+第n-1趟结果
+
+每个数后面有一个空格，每个序列占一行。
+
+输入样例:
+
+```json
+5
+4 2 3 2 1
+```
+
+输出样例:
+
+```json
+2 3 2 1 4
+2 2 1 3 4
+2 1 2 3 4
+1 2 2 3 4
+```
+
+以下程序实现了这一功能，请你补全空白处内容：
 
 ```cpp
-#include <iostream>
-#include <algorithm>
-using namespace std;
-bool cmp(int a, int b){
-    int x = a % 2;
-    int y = b % 2;
-    if(x == y)
-        if(x == 0)
-            return a < b;
-        else
-            return a > b;
-    else
-        return x > y;
+#include "stdio.h"
+int main()
+{
+    int arr[10];
+    int n;
+    scanf("%d", &n);
+    for (int i = 0; i < n; i++)
+        scanf("%d", &arr[i]);
+    for (int i = 0; i < n - 1; i++)
+    {
+        for (int j = 0; j < n - i - 1; j++)
+        {
+            if (arr[j] > arr[j + 1])
+            {
+                ____________________
+            }
+        }
+        for (int j = 0; j < n; j++)
+            printf("%d ", arr[j]);
+        printf("\n");
+    }
+    return 0;
 }
+```
+
+## template
+
+```cpp
+#include "stdio.h"
 int main()
 {
-    int n, i;
-    cin >> n;
-    int a[n];
-    for(i = 0; i < n; i++)
-        cin >> a[i];
-    sort(a, a+n, cmp);
-    for(i = 0; i < n; i++)
-        cout << a[i] << " ";
+    int arr[10];
+    int n;
+    scanf("%d", &n);
+    for (int i = 0; i < n; i++)
+        scanf("%d", &arr[i]);
+    for (int i = 0; i < n - 1; i++)
+    {
+        for (int j = 0; j < n - i - 1; j++)
+        {
+            if (arr[j] > arr[j + 1])
+            {
+                int t = arr[j];
+                arr[j] = arr[j + 1];
+                arr[j + 1] = t;
+            }
+        }
+        for (int j = 0; j < n; j++)
+            printf("%d ", arr[j]);
+        printf("\n");
+    }
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+int t = arr[j];
+arr[j] = arr[j + 1];
+arr[j + 1] = t;
 ```
 
 ## 选项
@@ -49,17 +105,23 @@ int main()
 ### A
 
 ```cpp
-
+int t = arr[j];
+arr[j + 1] = t;
+arr[j] = arr[j + 1];
 ```
 
 ### B
 
 ```cpp
-
+arr[j] = arr[j + 1];
+int t = arr[j];
+arr[j + 1] = t;
 ```
 
 ### C
 
 ```cpp
-
+int t = arr[j];
+arr[j + 1] = arr[j];
+arr[j] = t;
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/19.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/19.exercises/solution.md"
index 3fe84a7be..b2d4d445a 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/19.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/19.exercises/solution.md"
@@ -2,48 +2,90 @@
 
 给出一个由O和X组成的串，长度为1~80，统计得分，每个O的得分为目前连续出现的O的个数，X的得分为0，例如OOXXO的得分为1+2+0+0+1.
 
+以下程序实现了这一功能，请你补全空白处内容：
+
+```cpp
+#include "stdlib.h"
+#include "string.h"
+#include "stdio.h"
+int main()
+{
+    char s[85];
+    int score = 0, x;
+    scanf("%s", &s);
+    for (int i = 0; s[i]; i++)
+    {
+        if (s[i] == 'X')
+            score += 0;
+        if (s[i] == 'O')
+        {
+            int temp = 1;
+            if (i == 0)
+                score += 1;
+            else
+            {
+                x = i;
+                while (x > 0)
+                {
+                    ________________
+                }
+                score += temp;
+            }
+        }
+    }
+    printf("%d\n", score);
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
-#include"stdlib.h"
-#include"string.h"
-#include"stdio.h"
+#include "stdlib.h"
+#include "string.h"
+#include "stdio.h"
 int main()
 {
-	char s[85];
-	int score = 0,x;
-	scanf("%s",&s);
-	for (int i = 0;s[i];i++)
-	{
-		if (s[i] == 'X')
-		score += 0;
-		if (s[i] == 'O')
-		{
-			int temp = 1;
-			if (i == 0)score += 1;
-			else
-			{
-				x=i;
-				while (x > 0)
-				{
-					if (s[x - 1] == 'O')temp++;
-					else break; 
-					x--;
-				}
-				score += temp;
-			}
-		}
-	}
-	printf("%d\n", score);
-	system("pause");
-	return 0;
+    char s[85];
+    int score = 0, x;
+    scanf("%s", &s);
+    for (int i = 0; s[i]; i++)
+    {
+        if (s[i] == 'X')
+            score += 0;
+        if (s[i] == 'O')
+        {
+            int temp = 1;
+            if (i == 0)
+                score += 1;
+            else
+            {
+                x = i;
+                while (x > 0)
+                {
+                    if (s[x - 1] == 'O')
+                        temp++;
+                    else
+                        break;
+                    x--;
+                }
+                score += temp;
+            }
+        }
+    }
+    printf("%d\n", score);
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+if (s[x - 1] == 'O')
+	temp++;
+else
+	break;
+x--;
 ```
 
 ## 选项
@@ -51,17 +93,23 @@ int main()
 ### A
 
 ```cpp
-
+if (s[x - 1] == 'O')
+	temp++;
+else
+	break;
 ```
 
 ### B
 
 ```cpp
-
+if (s[x - 1] == 'O')
+	temp++;
 ```
 
 ### C
 
 ```cpp
-
+if (s[x - 1] == 'O')
+	temp++;
+x--;
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/2.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/2.exercises/solution.md"
index 0a542e0f0..310f99ce1 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/2.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/2.exercises/solution.md"
@@ -1,12 +1,37 @@
 # 一个班有10个同学，通过键盘输入成绩，并打印输出，每行输出5个同学的成绩。并求出平均成绩，最高分、最低分并输出。
 
 <p>一个班有10个同学&#xff0c;通过键盘输入成绩&#xff0c;
-并打印输出&#xff0c;每行输出5个同学的成绩。并求出平均成绩&#xff0c;最高分、最低分并输出。算法分析&#xff1a;
-(1)定义一个数组用来存放10个成绩数据。
-(2)用循环结构实现成绩输入&#xff1b;
-(3)用循环结构实现成绩输出,并控制换行&#xff1b;
+并打印输出&#xff0c;每行输出5个同学的成绩。并求出平均成绩&#xff0c;最高分、最低分并输出。算法分析&#xff1a;<br />
+(1)定义一个数组用来存放10个成绩数据。<br />
+(2)用循环结构实现成绩输入&#xff1b;<br />
+(3)用循环结构实现成绩输出,并控制换行&#xff1b;<br />
 (4)使用循环结构求平均成绩、最高分、最低分并输出。</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 x,i,max=0,min=0;
+	double sum=0,ave=0;
+	int a[10];
+	for(i=0;i<10;i++){
+		scanf("%d",&a[i]);
+		if(i==0)
+			min = a[i];
+		sum+=a[i];
+		____________
+	}
+	ave=sum/10;
+	for(i=0;i<5;i++)
+		printf("%d ",a[i]);
+	 printf("\n");
+	for(i=5;i<10;i++)
+		printf("%d ",a[i]);
+	printf("平均成绩%f,最高分%d,最低分%d ",ave,max,min);
+}
+```
+
 ## template
 
 ```cpp
@@ -38,7 +63,10 @@ int main(){
 ## 答案
 
 ```cpp
-
+if(max<a[i])
+	max=a[i];
+if(min>a[i])
+	min=a[i];
 ```
 
 ## 选项
@@ -46,17 +74,26 @@ int main(){
 ### A
 
 ```cpp
-
+if(max>a[i])
+	max=a[i];
+if(min<a[i])
+	min=a[i];
 ```
 
 ### B
 
 ```cpp
-
+if(max<a[i])
+	max=a[i + 1];
+if(min>a[i])
+	min=a[i - 1];
 ```
 
 ### C
 
 ```cpp
-
+if(max<a[i])
+	min=a[i];
+if(min>a[i])
+	max=a[i];
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/20.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/20.exercises/solution.md"
index bf7e69104..b1a078a9a 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/20.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/20.exercises/solution.md"
@@ -23,49 +23,106 @@ a,b均为自然数
 1
 WA</code></pre>
 
+以下程序实现了这一功能，请你补全空白处内容：
+
+```cpp
+#include <stdio.h>
+int main()
+{
+    int a;
+    int b;
+    char operation;
+    int num;
+    while (scanf("%d", &a) != EOF)
+    {
+        scanf("%c", &operation);
+        scanf("%d", &b);
+        if (operation == '+')
+        {
+            num = a + b;
+            printf("%d\n", num);
+        }
+        else if (operation == '-')
+        {
+            num = a - b;
+            printf("%d\n", num);
+        }
+        else if (operation == '*')
+        {
+            num = a * b;
+            printf("%d\n", num);
+        }
+        _________________________
+        else
+        {
+            printf("%s\n", "WA");
+        }
+    }
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
-#include<stdio.h>
+#include <stdio.h>
 int main()
 {
-	int a;
-	int b;
-	char operation;
-	int num;
-	while (scanf("%d", &a) != EOF)
-	{
-		scanf("%c", &operation);
-		scanf("%d", &b);
-		if (operation == '+')
-		{
-			num = a + b;
-			printf("%d\n", num);
-		}
-		else if (operation == '-') {
-			num = a - b;
-			printf("%d\n", num);
-		}
-		else if (operation == '*') {
-			num = a * b;
-			printf("%d\n", num);
-		}
-		else if (operation == '/' && b != 0) {
-			num = a / b;
-			printf("%d\n", num);
-		}
-		else {
-			printf("%s\n", "WA");
-		}
-	}
-	return 0;
+    int a;
+    int b;
+    char operation;
+    int num;
+    while (scanf("%d", &a) != EOF)
+    {
+        scanf("%c", &operation);
+        scanf("%d", &b);
+        if (operation == '+')
+        {
+            num = a + b;
+            printf("%d\n", num);
+        }
+        else if (operation == '-')
+        {
+            num = a - b;
+            printf("%d\n", num);
+        }
+        else if (operation == '*')
+        {
+            num = a * b;
+            printf("%d\n", num);
+        }
+        else if (operation == '/' && b != 0)
+        {
+            num = a / b;
+            printf("%d\n", num);
+        }
+        else if (operation == '%' && b != 0)
+        {
+            num = a % b;
+            printf("%d\n", num);
+        }
+        else
+        {
+            printf("%s\n", "WA");
+        }
+    }
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+else if (operation == '/' && b != 0)
+{
+	num = a / b;
+	printf("%d\n", num);
+}
+else if (operation == '%' && b != 0)
+{
+	num = a % b;
+	printf("%d\n", num);
+}
 ```
 
 ## 选项
@@ -73,17 +130,44 @@ int main()
 ### A
 
 ```cpp
-
+else if (operation == '/')
+{
+	num = a / b;
+	printf("%d\n", num);
+}
+else if (operation == '%')
+{
+	num = a % b;
+	printf("%d\n", num);
+}
 ```
 
 ### B
 
 ```cpp
-
+else if (operation == '/' && b != 0)
+{
+	num = a / b;
+	printf("%d\n", num);
+}
+else if (operation == '%')
+{
+	num = a % b;
+	printf("%d\n", num);
+}
 ```
 
 ### C
 
 ```cpp
-
+else if (operation == '/')
+{
+	num = a / b;
+	printf("%d\n", num);
+}
+else if (operation == '%' && b != 0)
+{
+	num = a % b;
+	printf("%d\n", num);
+}
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/21.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/21.exercises/solution.md"
index a6f926b06..8a8a22f45 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/21.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/21.exercises/solution.md"
@@ -1,15 +1,47 @@
 # 输入两个小写英语字母后， 输出两个字母的差
 
 两个小写字母的差用整数输出。
+
 但是，输入小写以外的文字就结束程序。
+
 比较两个字母的部分用diff这个名字的函数书写。执行结果
+
 请输入两个小写：a b
+
 字母a和字母b的区别是1。
+
 请输入两个小写：f b
-字母f和字母b的差异是4。
+
+字母f和字母b的差是4。
+
 请输入两个小写：0 a
+
 退出程序。
 
+以下程序实现了这一功能，请你补全空白处内容：
+
+```cpp
+#include <iostream>
+using namespace std;
+int diff(char ch1, char ch2)
+{
+    __________________
+}
+int main()
+{
+    char a, b;
+    while (1)
+    {
+        cout << "请输入两个小写：";
+        cin >> a >> b;
+        if (a > 'z' || a < 'a' || b > 'z' || b < 'a')
+            break;
+        cout << "文字" << a << "和文字" << b << "的差是" << diff(a, b) << "。\n";
+    }
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
@@ -17,26 +49,30 @@
 using namespace std;
 int diff(char ch1, char ch2)
 {
-	if (ch1 > ch2) return ch1 - ch2; else return ch2 - ch1;
+    if (ch1 > ch2)
+        return ch1 - ch2;
+    else
+        return ch2 - ch1;
 }
 int main()
 {
-	char a, b;
-	while (1)
-	{
-		cout << "请输入两个小写：";
-		cin >> a >> b;
-		if (a > 'z' || a < 'a' || b > 'z' || b < 'a') break;
-		cout << "文字" << a << "和文字" << b << "的差异是" << diff(a, b) << "。\n";
-	}
-	return 0;
+    char a, b;
+    while (1)
+    {
+        cout << "请输入两个小写：";
+        cin >> a >> b;
+        if (a > 'z' || a < 'a' || b > 'z' || b < 'a')
+            break;
+        cout << "文字" << a << "和文字" << b << "的差是" << diff(a, b) << "。\n";
+    }
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+return abs(ch1 - ch2);
 ```
 
 ## 选项
@@ -44,17 +80,20 @@ int main()
 ### A
 
 ```cpp
-
+return ch1 - ch2;
 ```
 
 ### B
 
 ```cpp
-
+return ch2 - ch1;
 ```
 
 ### C
 
 ```cpp
-
+if (ch1 < ch2)
+	return ch1 - ch2;
+else
+	return ch2 - ch1;
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/22.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/22.exercises/solution.md"
index 36646df5b..1db4bcfdf 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/22.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/22.exercises/solution.md"
@@ -1,54 +1,110 @@
 # 排序
 
 试题描述
-   由键盘上输入n个整数，请将这些数从大到小排序，然后输出排序后的数列。
+
+由键盘上输入n个整数，请将这些数从大到小排序，然后输出排序后的数列。
+
 输入
-   输入包含两行：
-   第一行是n（1 <= n <= 1000）。
-   第二行是n个整数，邻近两数之间用一个空格隔开。
+
+输入包含两行：
+
+第一行是n（1 <= n <= 1000）。
+
+第二行是n个整数，邻近两数之间用一个空格隔开。
+
 输出
-   输出排序后的n个整数，邻近两数之间用一个空格隔开。
+
+输出排序后的n个整数，邻近两数之间用一个空格隔开。
+
 输入示例
-   5
-   8 2 5 1 2
+
+```json
+5
+8 2 5 1 2
+```
+
 输出示例
-   8 5 2 2 1
+
+```json
+8 5 2 2 1
+```
+
 数据范围
-   输入和输出均为int范围的整数
+
+输入和输出均为int范围的整数
+
+以下程序实现了这一功能，请你补全空白处内容：
+
+```cpp
+#include <iostream>
+using namespace std;
+int main()
+{
+    int n, tmp;
+    cin >> n;
+    int *a = new int[n];
+    for (int i = 0; i < n; i++)
+        cin >> a[i];
+    for (int i = 0; i < n - 1; i++)
+    {
+        for (int j = i + 1; j < n; j++)
+        {
+            __________________
+        }
+    }
+    for (int i = 0; i < n; i++)
+    {
+        cout << a[i];
+        if (i != n - 1)
+            cout << " ";
+    }
+    return 0;
+}
+```
 
 ## template
 
 ```cpp
- #include<iostream>
+#include <iostream>
 using namespace std;
-int main() {
-	int n,tmp;
-	cin >> n;
-	int *a = new int[n];
-	for (int i = 0; i < n; i++)
-		cin >> a[i];
-	for (int i = 0; i < n - 1; i++) {		
-		for (int j = i + 1; j < n; j++) {
-			if (a[i] < a[j]) {
-				tmp = a[i];
-				a[i] = a[j];
-				a[j] = tmp;
-			}
-		}
-	}
-	for (int i = 0; i < n; i++) {
-		cout << a[i];
-		if (i != n - 1)
-			cout << " ";
-	}
-	return 0;
+int main()
+{
+    int n, tmp;
+    cin >> n;
+    int *a = new int[n];
+    for (int i = 0; i < n; i++)
+        cin >> a[i];
+    for (int i = 0; i < n - 1; i++)
+    {
+        for (int j = i + 1; j < n; j++)
+        {
+            if (a[i] < a[j])
+            {
+                tmp = a[i];
+                a[i] = a[j];
+                a[j] = tmp;
+            }
+        }
+    }
+    for (int i = 0; i < n; i++)
+    {
+        cout << a[i];
+        if (i != n - 1)
+            cout << " ";
+    }
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+if (a[i] < a[j])
+{
+	tmp = a[i];
+	a[i] = a[j];
+	a[j] = tmp;
+}
 ```
 
 ## 选项
@@ -56,17 +112,32 @@ int main() {
 ### A
 
 ```cpp
-
+if (a[i] > a[j])
+{
+	tmp = a[i];
+	a[i] = a[j];
+	a[j] = tmp;
+}
 ```
 
 ### B
 
 ```cpp
-
+if (a[i] < a[j])
+{
+	a[i] = a[j];
+	tmp = a[i];
+	a[j] = tmp;
+}
 ```
 
 ### C
 
 ```cpp
-
+if (a[i] > a[j])
+{
+	a[i] = a[j];
+	tmp = a[i];
+	a[j] = tmp;
+}
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/23.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/23.exercises/solution.md"
index ad04f88c6..3ee1c04c3 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/23.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/23.exercises/solution.md"
@@ -1,11 +1,57 @@
 # 字符数组
 
-<p>编写一个以两个字符数组作为输入的函数。
-如果第二个数组包含在第一个数组中&#xff0c;则函数返回第一个数组中第二个数组开始的第一个索引。
-如果第二个数组不被包含在第一个数组&#xff0c;然后函数应该return -1
-输入 [’c’,’a’,’l’,’l’,’i’,’n’,’g’] 和 [’a’,’l’,’l’]  就 return 1.
+<p>编写一个以两个字符数组作为输入的函数。<br />
+如果第二个数组包含在第一个数组中&#xff0c;则函数返回第一个数组中第二个数组开始的第一个索引。<br />
+如果第二个数组不被包含在第一个数组&#xff0c;然后函数应该return -1<br />
+输入 [’c’,’a’,’l’,’l’,’i’,’n’,’g’] 和 [’a’,’l’,’l’] 就 return 1.<br />
 输入 [’c’,’a’,’l’,’l’,’i’,’n’,’g’] 和 [’a’,’n’] 就 return -1.</p>
 
+以下程序实现了这一功能，请你补全空白处内容：
+
+```cpp
+#include <iostream>
+#include <string>
+using namespace std;
+int main()
+{
+    char a[128], b[128];
+    int numA, numB;
+    cout << "请输入第一个数组元素个数：";
+    cin >> numA;
+    cout << "请输入第一个数组元素：";
+    for (int i = 0; i < numA; ++i)
+        cin >> a[i];
+    cin.clear();
+    cin.sync();
+    cout << "请输入第二个数组元素个数：";
+    cin >> numB;
+    cout << "请输入第二个数组元素：";
+    for (int i = 0; i < numB; ++i)
+        cin >> b[i];
+    int num = 0;
+    string index;
+    for (int j = 0; j < numB; j++)
+    {
+        for (int k = 0; k < numA; k++)
+        {
+            if (b[j] == a[k])
+            {
+                __________________
+            }
+        }
+    }
+    if (num == numB)
+    {
+        cout << "第二个数组包含在第一个数组中" << endl;
+        cout << "第一个数组中第二个数组开始的第一个索引为:" << index.substr(0, 1) << endl;
+    }
+    else
+        cout << "第二个数组不被包含在第一个数组";
+    system("pause");
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
@@ -14,50 +60,52 @@
 using namespace std;
 int main()
 {
-	char a[128],b[128];
-	int numA, numB;
-	cout << "请输入第一个数组元素个数：";
-	cin >> numA;
-	cout << "请输入第一个数组元素：";
-	for (int i = 0; i < numA; ++i)
-		cin >> a[i];
-	cin.clear();
-	cin.sync();
-	cout << "请输入第二个数组元素个数：";
-	cin >> numB;
-	cout << "请输入第二个数组元素：";
-	for (int i = 0; i < numB; ++i)
-		cin >> b[i];
-	int num = 0;		
-	string index;		
-	for (int j = 0; j < numB; j++)		
-	{
-		for (int k = 0; k < numA; k++)
-		{
-			if (b[j] == a[k])
-			{
-				index += to_string(k);
-				num++;
-				break;
-			}
-		}
-	}
-	if (num == numB)	
-	{
-		cout << "第二个数组包含在第一个数组中" << endl;
-		cout << "第一个数组中第二个数组开始的第一个索引为:" << index.substr(0,1) << endl;
-	}
-	else
-		cout << "第二个数组不被包含在第一个数组";
-	system("pause");
-	return 0;
+    char a[128], b[128];
+    int numA, numB;
+    cout << "请输入第一个数组元素个数：";
+    cin >> numA;
+    cout << "请输入第一个数组元素：";
+    for (int i = 0; i < numA; ++i)
+        cin >> a[i];
+    cin.clear();
+    cin.sync();
+    cout << "请输入第二个数组元素个数：";
+    cin >> numB;
+    cout << "请输入第二个数组元素：";
+    for (int i = 0; i < numB; ++i)
+        cin >> b[i];
+    int num = 0;
+    string index;
+    for (int j = 0; j < numB; j++)
+    {
+        for (int k = 0; k < numA; k++)
+        {
+            if (b[j] == a[k])
+            {
+                index += to_string(k);
+                num++;
+                break;
+            }
+        }
+    }
+    if (num == numB)
+    {
+        cout << "第二个数组包含在第一个数组中" << endl;
+        cout << "第一个数组中第二个数组开始的第一个索引为:" << index.substr(0, 1) << endl;
+    }
+    else
+        cout << "第二个数组不被包含在第一个数组";
+    system("pause");
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+index += to_string(k);
+num++;
+break;
 ```
 
 ## 选项
@@ -65,17 +113,21 @@ int main()
 ### A
 
 ```cpp
-
+index += to_string(k);
+num++;
 ```
 
 ### B
 
 ```cpp
-
+index = to_string(k);
+num++;
 ```
 
 ### C
 
 ```cpp
-
+index = to_string(k);
+num++;
+break;
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/24.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/24.exercises/solution.md"
index af1ad2842..82c56cf99 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/24.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/24.exercises/solution.md"
@@ -4,36 +4,64 @@
 输入：B<回车>。
 输出：B在第2个位置，其后面第四个字母是F
 
+以下程序实现了这一功能，请你补全空白处内容：
+
+```cpp
+#include <stdio.h>
+int main()
+{
+    char c, c2;
+    printf("输入:");
+    c = getchar();
+    int m = 0, n = 0;
+    if (c >= 'A' && c <= 'z')
+    {
+        m = c - 'A' + 1;
+        _____________________
+    }
+    if (n > 0)
+        printf("%c在第%d个位置,其后面第四个字母是%c\n", c, m, c2);
+    else
+        printf("%c在第%d个位置,其后面没有第四个字母\n", c, m);
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
 #include <stdio.h>
-int main(){
-	char c,c2;
-	printf("输入:");
-	c = getchar();
-	int m=0,n=0;
-	if(c>='A'&& c<='z')
-	{
-		m = c - 'A' + 1;
-		if(m < 23)
-		{
-			c2 = c + 4;
-			n = m+4;
-		}
-	}
-	if(n > 0)
-		printf("%c在第%d个位置,其后面第四个字母是%c\n",c,m,c2);
-	else
-		printf("%c在第%d个位置,其后面没有第四个字母\n",c,m);
-	return 0;
+int main()
+{
+    char c, c2;
+    printf("输入:");
+    c = getchar();
+    int m = 0, n = 0;
+    if (c >= 'A' && c <= 'z')
+    {
+        m = c - 'A' + 1;
+        if (m < 23)
+        {
+            c2 = c + 4;
+            n = m + 4;
+        }
+    }
+    if (n > 0)
+        printf("%c在第%d个位置,其后面第四个字母是%c\n", c, m, c2);
+    else
+        printf("%c在第%d个位置,其后面没有第四个字母\n", c, m);
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+if (m < 23)
+{
+	c2 = c + 4;
+	n = m + 4;
+}
 ```
 
 ## 选项
@@ -41,17 +69,29 @@ int main(){
 ### A
 
 ```cpp
-
+if (m <= 23)
+{
+	c2 = c + 4;
+	n = m + 4;
+}
 ```
 
 ### B
 
 ```cpp
-
+if (m > 23)
+{
+	c2 = c + 4;
+	n = m + 4;
+}
 ```
 
 ### C
 
 ```cpp
-
+if (m >= 23)
+{
+	c2 = c + 4;
+	n = m + 4;
+}
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/25.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/25.exercises/solution.md"
index 8c41f83c1..dcad0a4dd 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/25.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/25.exercises/solution.md"
@@ -2,6 +2,41 @@
 
 <p>&nbsp;给你一个整数数组 arr 。请你将数组中的元素按照其二进制表示中，数字 1 的数目升序排序。 如果存在多个数字二进制中 1 的数目相同，则必须将它们按照数值大小升序排列。请你返回排序后的数组。</p>
 
+以下程序实现了这一功能，请你补全空白处内容：
+
+```cpp
+#include<stdio.h>
+#include<stdlib.h>
+int oneNum(int x)
+{
+	int cnt = 0;
+	while(x)
+	{
+		____________
+	}
+	return cnt;
+}
+int cmp(const void *a,const void*b)
+{
+	int al,bl;
+	int ret;
+	al = *(int*)a;
+	bl = *(int*)b;
+	ret =  oneNum(al) - oneNum(bl);
+	return  ret ? ret : al - bl;
+}
+int main()
+{
+	int s[]={1,2,3,5,6,7,8},i;
+	int len = sizeof(s)/sizeof(*s);
+	qsort(s,len,4,cmp);
+	for(i = 0 ; i < len ; i++)
+	{
+		printf("%d\n",s[i]);
+	}
+}
+```
+
 ## template
 
 ```cpp
@@ -41,7 +76,8 @@ int main()
 ## 答案
 
 ```cpp
-
+cnt++;
+x = x & (x - 1);
 ```
 
 ## 选项
@@ -49,17 +85,20 @@ int main()
 ### A
 
 ```cpp
-
+cnt++;
+x = x | (x - 1);
 ```
 
 ### B
 
 ```cpp
-
+cnt++;
+x = x & (x + 1);
 ```
 
 ### C
 
 ```cpp
-
+cnt++;
+x = x & x - 1;
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/26.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/26.exercises/solution.md"
index b74b7ba2c..d6ca549ce 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/26.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/26.exercises/solution.md"
@@ -1,35 +1,61 @@
 # 检查一个3位数是否是水仙花数
 
 检查一个3位数是否是水仙花数。
+
 输入：一个数字，比如 371，输出：x是水仙花数，
+
 如果不是，则输出：x不是水仙花数。
+
 注：x为输入的数字
 
+以下程序实现了这一功能，请你补全空白处内容：
+
+```cpp
+#include <iostream>
+using namespace std;
+int main()
+{
+    int a, b, c, y, n = 0;
+    cout << "请输入三位数字：" << endl;
+    cin >> n;
+    a = n % 1000 / 100;
+    b = n % 100 / 10;
+    c = n % 10 / 1;
+    ___________________
+    if (y == n)
+        cout << n << "是水仙花数" << endl;
+    else
+        cout << n << "不是水仙花数" << endl;
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
-#include <iostream> 
+#include <iostream>
 using namespace std;
 int main()
 {
-	int a, b, c, y, n = 0;
-	cout << "请输入三位数字：" << endl;
-	cin >> n;
-	a = n % 1000 / 100;  
-	b = n % 100 / 10;	
-	c = n % 10 / 1;	  
-	y = a*a*a + b*b*b + c*c*c;
-	if (y == n) cout << n << "是水仙花数" << endl;
-	else cout << n << "不是水仙花数" << endl;
-	system("pause");
-	return 0;
+    int a, b, c, y, n = 0;
+    cout << "请输入三位数字：" << endl;
+    cin >> n;
+    a = n % 1000 / 100;
+    b = n % 100 / 10;
+    c = n % 10 / 1;
+    y = a * a * a + b * b * b + c * c * c;
+    if (y == n)
+        cout << n << "是水仙花数" << endl;
+    else
+        cout << n << "不是水仙花数" << endl;
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+y = a * a * a + b * b * b + c * c * c;
 ```
 
 ## 选项
@@ -37,17 +63,17 @@ int main()
 ### A
 
 ```cpp
-
+y = a * a + b * b + c * c;
 ```
 
 ### B
 
 ```cpp
-
+y = a * a * a * a + b * b * b * b + c * c * c * c;
 ```
 
 ### C
 
 ```cpp
-
+y = a * b * c;
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/27.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/27.exercises/solution.md"
index 808a7de3f..d846e2e26 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/27.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/27.exercises/solution.md"
@@ -1,41 +1,81 @@
 # 秒数转换
 
 输入一个秒数，转换成HH：MM：SS的格式输出。
+
 输入样例  
+
 365
+
 输出样例 
+
 00:06:05
 
+以下程序实现了这一功能，请你补全空白处内容：
+
+```cpp
+#include <cstdio>
+int n;
+void print(int x)
+{
+    if (x == 0)
+        printf("00");
+    else if (x < 10)
+        printf("0%d", x);
+    else
+        printf("%d", x);
+    return;
+}
+int main()
+{
+    scanf("%d", &n);
+    int s, f, m;
+	____________
+    print(s);
+    printf(":");
+    print(f);
+    printf(":");
+    print(m);
+    return 0;
+}
+```
+
 ## template
 
 ```cpp
-#include<cstdio>
+#include <cstdio>
 int n;
-void print(int x){
-	if(x==0)printf("00");
-	else if(x<10)printf("0%d",x);
-	else printf("%d",x);
-	return;
+void print(int x)
+{
+    if (x == 0)
+        printf("00");
+    else if (x < 10)
+        printf("0%d", x);
+    else
+        printf("%d", x);
+    return;
 }
-int main(){
-	scanf("%d",&n);
-	int s,f,m;
-	s=n/3600;
-	f=n/60%60;
-	m=n%60;
-	print(s);
-	printf(":");
-	print(f);
-	printf(":");
-	print(m);
-	return 0;
+int main()
+{
+    scanf("%d", &n);
+    int s, f, m;
+    s = n / 3600;
+    f = n / 60 % 60;
+    m = n % 60;
+    print(s);
+    printf(":");
+    print(f);
+    printf(":");
+    print(m);
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+s = n / 3600;
+f = n / 60 % 60;
+m = n % 60;
 ```
 
 ## 选项
@@ -43,17 +83,23 @@ int main(){
 ### A
 
 ```cpp
-
+s = n / 3600;
+f = n / 60 / 60;
+m = n / 60;
 ```
 
 ### B
 
 ```cpp
-
+s = n / 3600;
+f = n % 60 % 60;
+m = n % 60;
 ```
 
 ### C
 
 ```cpp
-
+s = n / 3600;
+f = n / 60 % 60;
+m = n / 60;
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/28.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/28.exercises/solution.md"
index 711137778..4164a0838 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/28.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/28.exercises/solution.md"
@@ -1,7 +1,57 @@
-# 找出string中只出现过一次的字符
+# 找出字符串中第一个只出现一次的字符
 
-找出string中只出现过一次的字符
-例如"abcdef abcd"中需要得到ef
+输入描述：
+
+输入一个非空字符串
+
+输出描述：
+
+输出第一个只出现一次的字符，如果不存在输出-1
+
+示例1
+
+输入：
+
+```json
+asdfasdfo
+```
+
+输出：
+
+```json
+o
+```
+
+以下程序实现了这一功能，请你补全空白处内容：
+
+```cpp
+#include <iostream>
+#include <string>
+using namespace std;
+int main()
+{
+    string str;
+    while (getline(cin, str))
+    {
+        int a[256] = {0};
+        bool flag = false;
+        for (int i = 0; i < str.size(); ++i)
+            ____________________
+        for (int i = 0; i < str.size(); ++i)
+        {
+            if (a[str[i]] == 1)
+            {
+                cout << str[i] << endl;
+                flag = true;
+                break;
+            }
+        }
+        if (flag == false)
+            cout << "-1" << endl;
+    }
+    return 0;
+}
+```
 
 ## template
 
@@ -11,31 +61,33 @@
 using namespace std;
 int main()
 {
-    string temp = "";
-    cout << "请输入字符串：";
-    cin >> temp;
-    string str = "";        
-    string str1 = "";        
-    for (int i = 0; i < temp.length(); i++)
-    {
-        string tempSub = temp.substr(i, 1);    
-        int b = temp.rfind(tempSub);                
-        if (i == b && str1.find(tempSub) == -1)        
-            str +=temp.substr(i, 1);
-        else if (str1.find(tempSub) == -1)
-            str1 += temp.substr(i, 1);
-    }
-    cout << "只出现一次的字符：" << str << endl;
-    cout << "重复出现的字符的字符：" << str1 << endl;
-    system("pause");
-    return 0;
+    string str;
+    while (getline(cin, str))
+    {
+        int a[256] = {0};
+        bool flag = false;
+        for (int i = 0; i < str.size(); ++i)
+            ++a[str[i]];
+        for (int i = 0; i < str.size(); ++i)
+        {
+            if (a[str[i]] == 1)
+            {
+                cout << str[i] << endl;
+                flag = true;
+                break;
+            }
+        }
+        if (flag == false)
+            cout << "-1" << endl;
+    }
+    return 0;
 }
 ```
 
 ## 答案
 
 ```cpp
-
+++a[str[i]];
 ```
 
 ## 选项
@@ -43,17 +95,17 @@ int main()
 ### A
 
 ```cpp
-
+a[str[i]]++;
 ```
 
 ### B
 
 ```cpp
-
+a[str[i]++];
 ```
 
 ### C
 
 ```cpp
-
+a[str[i]--];
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/3.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/3.exercises/solution.md"
index 627c85b9f..3c5891795 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/3.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/3.exercises/solution.md"
@@ -2,6 +2,29 @@
 
 <p>本关任务&#xff1a;编写程序&#xff0c;有五种水果&#xff0c;apple、banana、orage、strawberry、pear&#xff0c;每一种有一个价格&#xff08;浮点小数&#xff09;&#xff0c;由老板输入&#xff0c;请提示用户选择什么水果&#xff0c;购买数量&#xff08;按照斤两&#xff09;&#xff0c;然后将总价显示出来。</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(){
+	typedef enum {apple, banana, orange, strawberry, pear} fruits;
+	double prices[5];			 
+	fruits purchase;			 
+	int fruit;				   
+	double amount;			   
+	for(int i = 0; i < 5; i++)
+		scanf("%lf", &prices[i]);
+	printf("水果编号：1.苹果  2.香蕉  3.橘子  4.草莓  5.梨\n");
+	printf("请输入购买的水果(1~5)，以及购买数量（按照斤两）：\n");
+	scanf("%d %lf", &fruit, &amount);
+	purchase = (fruits)fruit;	 
+	
+	________________________
+
+	return 0; 
+}
+```
+
 ## template
 
 ```cpp
@@ -26,7 +49,7 @@ int main(){
 ## 答案
 
 ```cpp
-
+printf("总价为：%.3lf", prices[purchase-1]*amount);
 ```
 
 ## 选项
@@ -34,17 +57,17 @@ int main(){
 ### A
 
 ```cpp
-
+printf("总价为：%.3lf", prices[purchase]*amount);
 ```
 
 ### B
 
 ```cpp
-
+printf("总价为：%.3lf", prices(purchase-1)*amount);
 ```
 
 ### C
 
 ```cpp
-
+printf("总价为：%.3lf", prices(purchase)*amount);
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/4.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/4.exercises/solution.md"
index 0e611af67..0b74e41a4 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/4.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/4.exercises/solution.md"
@@ -2,6 +2,35 @@
 
 求10-100之间个位数为7的质数
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```cpp
+#include <stdio.h>
+int isp(int n)
+{
+	int i;
+	if (n<2)
+		return 0;
+	for (i=2;i*i<=n;++i)
+	{
+		_______________
+	}
+	return 1;
+}
+int main()
+{
+	int i=17;
+	while (i<=100)
+	{
+		if (isp(i))
+			printf("%d  ",i);
+		i+=10;   
+	}
+	printf("\n");  
+	return 0;
+}
+```
+
 ## template
 
 ```cpp
@@ -35,7 +64,8 @@ int main()
 ## 答案
 
 ```cpp
-
+if (n%i==0)
+	return 0;
 ```
 
 ## 选项
@@ -43,17 +73,20 @@ int main()
 ### A
 
 ```cpp
-
+if (n%i==0)
+	return 1;
 ```
 
 ### B
 
 ```cpp
-
+if (n/i==0)
+	return 1;
 ```
 
 ### C
 
 ```cpp
-
+if (n/i==0)
+	return 0;
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/5.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/5.exercises/solution.md"
index ada2dbaf7..5a7c433f7 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/5.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/5.exercises/solution.md"
@@ -1,22 +1,70 @@
 # 数据合并
 
 题目描述
+
 将两个从小到大排列的一维数组 (维长分别为 m,n , 其中 m,n≤100) 仍按从小到大的排列顺序合并到一个新的一维数组中，输出新的数组.
+
 输入描述
-第 1行一个正整数 m , 表示第一个要合并的一维数组中的元素个数
+
+第 1 行一个正整数 m , 表示第一个要合并的一维数组中的元素个数
+
 第 2 行一个正整数 n , 表示第二个要合并的一维数组中的元素个数
+
 第 3 行输入 m 个整数 (每个数用空格分开) , 表示第一个数组元素的值.
+
 第 4 行输入 n 个整数 (每个数用空格分开) , 表示第二个数组元素的值.
+
 输出描述
+
 一行，表示合并后的数据，共 m +n 个数
+
 样例输入
+
+```json
 3
 4
 1 3 5
 2 4 6 8
+```
+
 样例输出
+
+```json
 1 2 3 4 5 6 8
-要多组输入输出
+```
+
+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;
+void merge(int * a1, int m, int * a2, int n)
+{
+	int m1 = m - 1;
+	int n1 = n - 1;
+	for (int i = m + n - 1; i >= 0; i--)
+	{
+		if (m1 < 0) a1[i] = a2[n1--];
+		else if (n1 < 0) a1[i] = a1[m1--];
+		__________________
+	}
+}
+int main()
+{
+	int m;
+	int n;
+	cin >> m;
+	cin >> n;
+	int a1[201];
+	int a2[101];
+	for (int i = 0; i < m; i++) cin >> a1[i];
+	for (int i = 0; i < n; i++) cin >> a2[i];
+	merge(a1, m, a2, n);
+	for (int i = 0; i < m + n; i++) cout << a1[i] << " ";
+	return 0;
+}
+```
+
 
 ## template
 
@@ -54,7 +102,8 @@ int main()
 ## 答案
 
 ```cpp
-
+else if (a1[m1] < a2[n1]) a1[i] = a2[n1--];
+else a1[i] = a1[m1--];
 ```
 
 ## 选项
@@ -62,17 +111,20 @@ int main()
 ### A
 
 ```cpp
-
+else if (a1[m1] < a2[n1]) a1[i] = a2[n1];
+else a1[i] = a1[m1];
 ```
 
 ### B
 
 ```cpp
-
+else if (a1[m1] > a2[n1]) a1[i] = a2[n1--];
+else a1[i] = a1[m1--];
 ```
 
 ### C
 
 ```cpp
-
+else if (a1[m1] > a2[n1]) a1[i] = a2[n1];
+else a1[i] = a1[m1];
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/6.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/6.exercises/solution.md"
index 6c51a4497..9486cc243 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/6.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/6.exercises/solution.md"
@@ -1,10 +1,34 @@
 # 计算所有4位正整数中同时能被13和20整除的数的和
 
-<p>计算所有4位正整数中同时能被13和20整除的数的和&#xff0c; 并同时做到如下显示:
-①显示这些数:
-②显示这些数的个数:
+<p>计算所有4位正整数中同时能被13和20整除的数的和&#xff0c; 并同时做到如下显示:<br />
+①显示这些数:<br />
+②显示这些数的个数:<br />
 ③显示这些数的和。</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 i = 1000;
+	int count = 0;
+	int sum = 0;
+	printf("所有4位正整数中同时能被13和20整除的数:\n");
+	for(i = 1000;i<10000;i++)
+	{
+		if(_________________)
+		{
+			count++;
+			sum = sum + i;
+			printf("%d、",i);
+		}
+	}
+	printf("\n这些数一共有%d个\n",count);
+	printf("这些数的和是：%d\n",sum);
+}
+```
+
 ## template
 
 ```cpp
@@ -32,7 +56,7 @@ int main()
 ## 答案
 
 ```cpp
-
+i % 13 == 0 && i % 20 == 0
 ```
 
 ## 选项
@@ -40,17 +64,17 @@ int main()
 ### A
 
 ```cpp
-
+i / 13 == 0 && i / 20 == 0
 ```
 
 ### B
 
 ```cpp
-
+i % 13 == 0 & i % 20 == 0
 ```
 
 ### C
 
 ```cpp
-
+(i % 13 == 0) & (i % 20 == 0)
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/7.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/7.exercises/solution.md"
index 5d47adbde..93b28fdf0 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/7.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/7.exercises/solution.md"
@@ -2,6 +2,33 @@
 
 <p>实现从键盘输入一个字符时&#xff0c;如果该字符为小写字母&#xff0c;则转换为大写字母输出&#xff1b;如果该字符为大写字母&#xff0c;则转换为小写字母输出&#xff1b;如果为其他字符&#xff0c;则原样输出。</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()
+{
+	char x,y;
+	printf("请输入一个字符：");
+	scanf("%c",&x);
+	if(x>='A'&&x<='Z')
+	{
+		(1)___________
+		printf("此字母是一个大写字母,转换后的小写字母是：%c",y);
+	}
+	else if(x>='a'&&x<='z')
+	{
+		(2)___________
+		printf("此字母是一个小写字母,转换后的大写字母是%c",y);
+	}
+	else
+	{
+		printf("%c",x);
+	}
+	return 0;
+}
+```
+
 ## template
 
 ```cpp
@@ -32,7 +59,8 @@ int main()
 ## 答案
 
 ```cpp
-
+(1):y=x+32;
+(2):y=x-32;
 ```
 
 ## 选项
@@ -40,17 +68,20 @@ int main()
 ### A
 
 ```cpp
-
+(1):y=x-32;
+(2):y=x+32;
 ```
 
 ### B
 
 ```cpp
-
+(1):y=x+31;
+(2):y=x-31;
 ```
 
 ### C
 
 ```cpp
-
+(1):y=x-31;
+(2):y=x+31;
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/8.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/8.exercises/solution.md"
index 0e448bec7..e270c9264 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/8.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/8.exercises/solution.md"
@@ -9,6 +9,34 @@
 输出格式：
 输出一个正整数&#xff0c;是上述序列中所有素数之和。</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 isprime( int n);
+int main()
+{
+	int i,n,m,s=0;
+	scanf("%d", &n);
+	for( i=0; i<n; i++ ) {
+		scanf("%d", &m);
+		if(isprime(m)){
+			s+=m;
+		}
+	}
+	printf("%d", s);
+	return 0;
+}
+int isprime(int n)
+{
+	int i;
+	if(n<=1) return 0;
+	for(i=2;i*i<=n;i++)
+		_______________
+	return 1;
+}
+```
+
 ## template
 
 ```cpp
@@ -40,7 +68,7 @@ int isprime(int n)
 ## 答案
 
 ```cpp
-
+if(n%i==0) return 0;
 ```
 
 ## 选项
@@ -48,17 +76,17 @@ int isprime(int n)
 ### A
 
 ```cpp
-
+if(n/i==0) return 0;
 ```
 
 ### B
 
 ```cpp
-
+if(n%i==0) return 1;
 ```
 
 ### C
 
 ```cpp
-
+if(n%i==1) return 0;
 ```
\ No newline at end of file
diff --git "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/9.exercises/solution.md" "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/9.exercises/solution.md"
index 73780adfa..3d869510d 100644
--- "a/data/1.dailycode\345\210\235\351\230\266/1.cpp/9.exercises/solution.md"
+++ "b/data/1.dailycode\345\210\235\351\230\266/1.cpp/9.exercises/solution.md"
@@ -7,6 +7,32 @@
 输出
 只有一行且只有一个整数&#xff0c;乐乐班级的总分数。</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[10005];
+	long long t=0;
+	int n,i,j,x;
+	scanf("%d",&n);
+	for(i=0;i<n;i++)
+		scanf("%d",&a[i]);
+	for(i=0;i<n;i++){
+		for(j=0;j<n;j++){
+			_____________
+		}
+	}
+	for(i=0;i<n;i++)
+	{
+		t+=a[i]*(i+1);
+	}
+	printf("%lld",t);
+	return 0;
+}
+```
+
 ## template
 
 ```cpp
@@ -40,7 +66,11 @@ int main()
 ## 答案
 
 ```cpp
-
+if(a[i]>a[j]){
+	x=a[i];
+	a[i]=a[j];
+	a[j]=x;
+}
 ```
 
 ## 选项
@@ -48,17 +78,29 @@ int main()
 ### A
 
 ```cpp
-
+if(a[i]<a[j]){
+	x=a[i];
+	a[i]=a[j];
+	a[j]=x;
+}
 ```
 
 ### B
 
 ```cpp
-
+if(a[i]>a[j]){
+	x=a[i];
+	a[j]=x;
+	a[i]=a[j];
+}
 ```
 
 ### C
 
 ```cpp
-
+if(a[i]<a[j]){
+	x=a[i];
+	a[j]=x;
+	a[i]=a[j];
+}
 ```
\ No newline at end of file
-- 
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