diff --git a/20240826/learn.pdf b/20240826/learn.pdf
new file mode 100644
index 0000000..bceb52c
Binary files /dev/null and b/20240826/learn.pdf differ
diff --git a/20240827/preComp/image/learn/1724749114259.png b/20240827/preComp/image/learn/1724749114259.png
new file mode 100644
index 0000000..dddaf95
Binary files /dev/null and b/20240827/preComp/image/learn/1724749114259.png differ
diff --git a/20240827/preComp/learn.md b/20240827/preComp/learn.md
index 9fd9892..027d58f 100644
--- a/20240827/preComp/learn.md
+++ b/20240827/preComp/learn.md
@@ -1,3 +1,55 @@
+# 总览
+
+![1724749114259](image/learn/1724749114259.png)
+
+# 选择
+
+## 5
+
+> 算法
+
+n个数组成$Z_{1..n}$，建立数组$A_{1..{n\over2}}与B_{1..{n\over2}}$
+
+$$
+i=[1,{n\over2}],A_i=max(Z_{i},Z_{i+{n\over2}}),B_i=min(Z_i,Z_{i+{n\over2}})
+$$
+
+$$
+max\_number=A_1,i=[2,{n\over2}],max\_number=\max(max\_number,A_i)
+$$
+
+$$
+min\_number=B_1,i=[2,{n\over2}],min\_number=\min(min\_number,B_i)
+$$
+
+所以总次数是
+
+$$
+times={3\over2}n-2
+$$
+
+在题目中是2n个数所以是3n-2
+
+## 7
+
+### 完全图n个点时
+
+$$
+边数={n\times(n-1)\over2}
+$$
+
+### 非连通图+1即可
+
+> 解方程
+
+$$
+{n\times(n-1)\over2}\geq36
+$$
+
+> 最后记得非连通图要+1
+>
+> 答案是10
+
 # 阅读程序
 
 ## 1
@@ -85,3 +137,21 @@ int main() {
 - `cout.precision(n)`：设置浮点数输出的精度，即小数点后显示的位数。
 
 这两个函数常一起使用，控制浮点数的输出格式和精度。
+
+### 17.
+
+```cpp
+int/int=int
+int/double=double
+double/double=double
+```
+
+### 19
+
+```cpp
+//acos的逻辑，逻辑代码
+double acos(double num){
+	find t where{cos(t) == num;}
+	return t;//弧度制
+}
+```
diff --git a/20240827/preComp/learn.pdf b/20240827/preComp/learn.pdf
new file mode 100644
index 0000000..3eecc9c
Binary files /dev/null and b/20240827/preComp/learn.pdf differ