update dailycode pipline data

data/1.dailycode初阶/1.cpp/29.exercises/solution.md

 # 找出小于平均值的数。
 
-<p>从键盘输入一个正整数存入变量n中&#xff0c;再输入n个整数&#xff0c;然后找出所有小于平均值的数&#xff0c;并按输入顺序输出。
- 
- </p>
+<p>从键盘输入一个正整数存入变量n中&#xff0c;再输入n个整数&#xff0c;然后找出所有小于平均值的数&#xff0c;并按输入顺序输出。</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>
+int main()
+{
+    int i, n, sum = 0, a[100];
+    float ave;
+    scanf("%d", &n);
+    for (i = 0; i < n; i++)
+    {
+        scanf("%d", &a[i]);
+        sum += a[i];
+    }
+    ave = sum * 1.0 / n;
+    for (i = 0; i < n; i++)
+    {
+        __________________
+    }
+    return 0;
+}
+```
 
 ## template
 
 ```cpp
-#include<stdio.h>
+#include <stdio.h>
 int main()
 {
-	int i,n,sum=0,a[100];
+    int i, n, sum = 0, a[100];
     float ave;
-	scanf("%d",&n);
-	for(i=0;i<n;i++){
-		scanf("%d",&a[i]);
-		sum+=a[i];
+    scanf("%d", &n);
+    for (i = 0; i < n; i++)
+    {
+        scanf("%d", &a[i]);
+        sum += a[i];
     }
-	ave=sum*1.0/n;
-	for(i=0;i<n;i++){
-		if(a[i]<ave)
-			printf("%d",a[i]);
+    ave = sum * 1.0 / n;
+    for (i = 0; i < n; i++)
+    {
+        if (a[i] < ave)
+            printf("%d", a[i]);
     }
     return 0;
 }
@@ -29,7 +52,8 @@ int main()
 ## 答案
 
 ```cpp
-
+if (a[i] < ave)
+	printf("%d", a[i]);
 ```
 
 ## 选项
@@ -37,17 +61,20 @@ int main()
 ### A
 
 ```cpp
-
+if (a[i] <= ave)
+	printf("%d", a[i]);
 ```
 
 ### B
 
 ```cpp
-
+if (a[i] > ave)
+	printf("%d", a[i]);
 ```
 
 ### C
 
 ```cpp
-
+if (a[i] >= ave)
+	printf("%d", a[i]);
 ```
\ No newline at end of file

data/1.dailycode初阶/1.cpp/30.exercises/solution.md

@@ -6,49 +6,51 @@
 
 ```cpp
 #include <stdio.h>
-int fun(int n,int *x,int *y)
+int fun(int n, int *x, int *y)
 {
-	int sum = 0,i;
+    int sum = 0, i;
     int size = 0;
     int dd = 1;
-	for (i = 1; i <= (*x);i++)
+    for (i = 1; i <= (*x); i++)
     {
-		sum += (2*i-1);
+        sum += (2 * i - 1);
     }
     if (sum == n)
     {
-		*y = 2*(*x) -1;
+        *y = 2 * (*x) - 1;
         return (*x);
-	}else if (sum > n)
+    }
+    else if (sum > n)
     {
-		(*y) = n - (sum - (2 * (*x) -1));
-		size = 2* (*x) -1;
+        (*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))
+            if (i <= (*x))
                 dd *= 2;
             else
                 dd /= 2;
         }
         return dd;
-	}else
+    }
+    else
     {
         (*x)++;
-		return fun(n,x,y);
+        return fun(n, x, y);
     }
 }
 int main()
 {
     int n;
-	int row = 1,col = 0;
+    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);
+    scanf("%d", &n);
+    val = fun(n, &row, &col);
+    printf("第%d项是：%d\n", n, val);
     return 0;
 }
 ```