diff --git a/TangDou/AcWing/NumTriangle/1015.md b/TangDou/AcWing_TiGao/T1/1015.md
similarity index 100%
rename from TangDou/AcWing/NumTriangle/1015.md
rename to TangDou/AcWing_TiGao/T1/1015.md
diff --git a/TangDou/AcWing/NumTriangle/1015_Dfs.cpp b/TangDou/AcWing_TiGao/T1/1015_Dfs.cpp
similarity index 100%
rename from TangDou/AcWing/NumTriangle/1015_Dfs.cpp
rename to TangDou/AcWing_TiGao/T1/1015_Dfs.cpp
diff --git a/TangDou/AcWing/NumTriangle/1015_Dfs_JiYIHua.cpp b/TangDou/AcWing_TiGao/T1/1015_Dfs_JiYIHua.cpp
similarity index 100%
rename from TangDou/AcWing/NumTriangle/1015_Dfs_JiYIHua.cpp
rename to TangDou/AcWing_TiGao/T1/1015_Dfs_JiYIHua.cpp
diff --git a/TangDou/AcWing/NumTriangle/1015_ErWei.cpp b/TangDou/AcWing_TiGao/T1/1015_ErWei.cpp
similarity index 100%
rename from TangDou/AcWing/NumTriangle/1015_ErWei.cpp
rename to TangDou/AcWing_TiGao/T1/1015_ErWei.cpp
diff --git a/TangDou/AcWing/NumTriangle/1015_YiWei.cpp b/TangDou/AcWing_TiGao/T1/1015_YiWei.cpp
similarity index 100%
rename from TangDou/AcWing/NumTriangle/1015_YiWei.cpp
rename to TangDou/AcWing_TiGao/T1/1015_YiWei.cpp
diff --git a/TangDou/AcWing/NumTriangle/1018.md b/TangDou/AcWing_TiGao/T1/1018.md
similarity index 100%
rename from TangDou/AcWing/NumTriangle/1018.md
rename to TangDou/AcWing_TiGao/T1/1018.md
diff --git a/TangDou/AcWing/NumTriangle/1018_ErWei.cpp b/TangDou/AcWing_TiGao/T1/1018_ErWei.cpp
similarity index 100%
rename from TangDou/AcWing/NumTriangle/1018_ErWei.cpp
rename to TangDou/AcWing_TiGao/T1/1018_ErWei.cpp
diff --git a/TangDou/AcWing/NumTriangle/1018_YiWei.cpp b/TangDou/AcWing_TiGao/T1/1018_YiWei.cpp
similarity index 100%
rename from TangDou/AcWing/NumTriangle/1018_YiWei.cpp
rename to TangDou/AcWing_TiGao/T1/1018_YiWei.cpp
diff --git a/TangDou/AcWing/NumTriangle/1027.md b/TangDou/AcWing_TiGao/T1/1027.md
similarity index 100%
rename from TangDou/AcWing/NumTriangle/1027.md
rename to TangDou/AcWing_TiGao/T1/1027.md
diff --git a/TangDou/AcWing/NumTriangle/1027_0.cpp b/TangDou/AcWing_TiGao/T1/1027_0.cpp
similarity index 100%
rename from TangDou/AcWing/NumTriangle/1027_0.cpp
rename to TangDou/AcWing_TiGao/T1/1027_0.cpp
diff --git a/TangDou/AcWing/NumTriangle/1027_1.cpp b/TangDou/AcWing_TiGao/T1/1027_1.cpp
similarity index 100%
rename from TangDou/AcWing/NumTriangle/1027_1.cpp
rename to TangDou/AcWing_TiGao/T1/1027_1.cpp
diff --git a/TangDou/AcWing/NumTriangle/1027_2.cpp b/TangDou/AcWing_TiGao/T1/1027_2.cpp
similarity index 100%
rename from TangDou/AcWing/NumTriangle/1027_2.cpp
rename to TangDou/AcWing_TiGao/T1/1027_2.cpp
diff --git a/TangDou/AcWing/NumTriangle/275.cpp b/TangDou/AcWing_TiGao/T1/275.cpp
similarity index 100%
rename from TangDou/AcWing/NumTriangle/275.cpp
rename to TangDou/AcWing_TiGao/T1/275.cpp
diff --git a/TangDou/AcWing/NumTriangle/275.md b/TangDou/AcWing_TiGao/T1/275.md
similarity index 92%
rename from TangDou/AcWing/NumTriangle/275.md
rename to TangDou/AcWing_TiGao/T1/275.md
index 94cbbb2..6f7bdd0 100644
--- a/TangDou/AcWing/NumTriangle/275.md
+++ b/TangDou/AcWing_TiGao/T1/275.md
@@ -96,18 +96,13 @@ $1≤n,m≤50$
因此,我们可以通过微调其中的一条路线,使之不经过重合点 $C$,同时路线的总价值没有减少
-**得证**:最优解路线可以是不经过重复路线的 (部分证明方法参照了 [第一赞的题解](https://www.acwing.com/solution/content/12389/)了)
+**得证**:最优解路线可以是不经过重复路线的
接下来就是完全参照 $AcWing$ $1027$. 方格取数 的$DP$分析了
-关于 **重合格子** 判断一些条件都在这篇博客里详细写了
-
-这里我偷个懒,直接用我上篇博客的内容了 ~~(既然是我自己写的应该不算抄袭吧!)~~
-

-### Code(三重迭代写法)
**集合划分**
diff --git a/TangDou/LanQiaoBei/LanQiao14STEMA20230212/2.cpp b/TangDou/LanQiaoBei/LanQiao14STEMA20230212/2.cpp
new file mode 100644
index 0000000..5dd0d91
--- /dev/null
+++ b/TangDou/LanQiaoBei/LanQiao14STEMA20230212/2.cpp
@@ -0,0 +1,20 @@
+#include
+using namespace std;
+int main() {
+ int n, m;
+ cin >> n >> m;
+ int cnt = 0;
+ for (int i = n; i <= m; i++) {
+ int x = i;
+ bool flag = false;
+ while (x) {
+ int a = x % 10;
+ if (a % 2 == 0) flag = true;
+ x /= 10;
+ }
+ if (!flag) cnt++;
+ }
+ cout << cnt << endl;
+
+ return 0;
+}
diff --git a/TangDou/LanQiaoBei/LanQiao14STEMA20230212/8.cpp b/TangDou/LanQiaoBei/LanQiao14STEMA20230212/8.cpp
new file mode 100644
index 0000000..3ea813d
--- /dev/null
+++ b/TangDou/LanQiaoBei/LanQiao14STEMA20230212/8.cpp
@@ -0,0 +1,18 @@
+#include
+using namespace std;
+int main() {
+ int n, cnt = 0;
+ cin >> n;
+ cnt = n / 3;
+ if (n == 2)
+ cnt = -1;
+ else if (n == 4)
+ cnt = 4;
+ else if (n % 3 == 1)
+ cnt += 1;
+ else if (n % 3 == 2)
+ cnt += 2;
+
+ cout << cnt << endl;
+ return 0;
+}
\ No newline at end of file
diff --git a/TangDou/LanQiaoBei/LanQiao14STEMA20230212/题解.md b/TangDou/LanQiaoBei/LanQiao14STEMA20230212/题解.md
new file mode 100644
index 0000000..7bc7009
--- /dev/null
+++ b/TangDou/LanQiaoBei/LanQiao14STEMA20230212/题解.md
@@ -0,0 +1,3 @@
+https://blog.csdn.net/qq_36230375/article/details/134725732
+
+https://tiku.scratchor.com/paper/view/gjfvsjrgtpm30v3f
\ No newline at end of file
diff --git a/TangDou/LanQiaoBei/LanQiao15STEMA20231217/10.cpp b/TangDou/LanQiaoBei/LanQiao15STEMA20231217/10.cpp
new file mode 100644
index 0000000..df981a0
--- /dev/null
+++ b/TangDou/LanQiaoBei/LanQiao15STEMA20231217/10.cpp
@@ -0,0 +1,26 @@
+#include
+using namespace std;
+int k;
+string kill(string s) {
+ int p = 0;
+ for (int i = 0; i < s.size() - 1; i++)
+ if (s[i + 1] > s[i]) {
+ p = i;
+ break;
+ }
+
+ string res;
+ for (int i = 0; i < s.size(); i++)
+ if (i != p) res += s[i];
+ return res;
+}
+
+int main() {
+ string s;
+ cin >> s;
+
+ cin >> k;
+ for (int i = 1; i <= k; i++) s = kill(s);
+ cout << s << endl;
+ return 0;
+}
\ No newline at end of file
diff --git a/TangDou/LanQiaoBei/LanQiao15STEMA20231217/11.cpp b/TangDou/LanQiaoBei/LanQiao15STEMA20231217/11.cpp
new file mode 100644
index 0000000..7c08cb7
--- /dev/null
+++ b/TangDou/LanQiaoBei/LanQiao15STEMA20231217/11.cpp
@@ -0,0 +1,49 @@
+#include
+using namespace std;
+const int N = 100010;
+vector g[N]; // 邻接表,一个正图,一个反图
+int c[N]; // 每个树洞中松鼠的数量
+/*
+测试用例
+4
+5
+3
+6
+1
+1 2
+1 3
+2 4
+2
+*/
+int dfs(int u, int fa, int k) {
+ if (k == 0) return c[u];
+ int sum = c[u];
+ for (int i = 0; i < g[u].size(); i++) {
+ if (g[u][i] == fa) continue;
+ sum += dfs(g[u][i], u, k - 1);
+ }
+ return sum;
+}
+
+int main() {
+#ifndef ONLINE_JUDGE
+ freopen("11.in", "r", stdin);
+#endif
+
+ int n; // 树洞的数量
+ cin >> n;
+
+ for (int i = 1; i <= n; i++) cin >> c[i]; // 每个树洞中松鼠的数量
+
+ for (int i = 1; i < n; i++) {
+ int a, b; // 表示两个树洞相连接
+ cin >> a >> b;
+ g[a].push_back(b); // 用邻接表保存一下某个树洞与哪两个其它树洞相连接
+ g[b].push_back(a);
+ }
+
+ int k;
+ cin >> k;
+ for (int i = 1; i <= n; i++) cout << dfs(i, -1, k) << endl;
+ return 0;
+}
\ No newline at end of file
diff --git a/TangDou/LanQiaoBei/LanQiao15STEMA20231217/11.in b/TangDou/LanQiaoBei/LanQiao15STEMA20231217/11.in
new file mode 100644
index 0000000..14224a9
--- /dev/null
+++ b/TangDou/LanQiaoBei/LanQiao15STEMA20231217/11.in
@@ -0,0 +1,9 @@
+4
+5
+3
+6
+1
+1 2
+1 3
+2 4
+2
\ No newline at end of file
diff --git a/TangDou/LanQiaoBei/LanQiao15STEMA20231217/5.cpp b/TangDou/LanQiaoBei/LanQiao15STEMA20231217/5.cpp
new file mode 100644
index 0000000..8bae870
--- /dev/null
+++ b/TangDou/LanQiaoBei/LanQiao15STEMA20231217/5.cpp
@@ -0,0 +1,14 @@
+#include
+using namespace std;
+int func(int x, int y, int z) {
+ if (x == 1 || y == 1 || z == 1) return 1;
+ if (x < y && x < z)
+ return func(x, y - 1, z) + func(x, y, z - 1);
+ if (y < x && y < z)
+ return func(x - 1, y, z) + func(x, y, z - 1);
+ return func(x - 1, y, z) + func(x, y - 1, z);
+}
+int main() {
+ cout << func(3, 3, 2);
+ return 0;
+}
\ No newline at end of file
diff --git a/TangDou/LanQiaoBei/LanQiao15STEMA20231217/5.md b/TangDou/LanQiaoBei/LanQiao15STEMA20231217/5.md
new file mode 100644
index 0000000..a7f5536
--- /dev/null
+++ b/TangDou/LanQiaoBei/LanQiao15STEMA20231217/5.md
@@ -0,0 +1,36 @@
+```cpp {.line-numbers}
+int func( int x, int y, int z )
+{
+ if( x == 1 || y == 1 || z == 1 ) return 1;
+ if( x < y && x < z )
+ return func( x, y - 1, z ) + func( x, y, z - 1 );
+ if( y < x && y < z )
+ return func( x - 1, y, z ) + func( x, y, z - 1 );
+ return func( x - 1, y, z ) + func( x, y - 1, z );
+}
+int main()
+{
+ cout << func( 3, 3, 2 );
+ return 0;
+}
+```
+将$func()$简写成$f()$
+
+人脑模拟电脑
+
+$f(3,3,2)=f(2,3,2)+f(3,2,2)$
+
+其中
+$f(2,3,2)=f(1,3,2)+f(2,2,2)$ ①
+$f(3,2,2)=f(2,2,2)+f(3,1,2)$ ②
+
+最终$res=①+②$
+
+$f(1,3,2)=1$
+$f(2,2,2)=f(1,2,2)+f(2,1,2)=1+1=2$
+
+$\therefore$ ①=3
+
+②=3
+
+①+②=6
diff --git a/TangDou/LanQiaoBei/LanQiao15STEMA20231217/9.cpp b/TangDou/LanQiaoBei/LanQiao15STEMA20231217/9.cpp
new file mode 100644
index 0000000..0b2e4de
--- /dev/null
+++ b/TangDou/LanQiaoBei/LanQiao15STEMA20231217/9.cpp
@@ -0,0 +1,26 @@
+#include
+using namespace std;
+
+int main() {
+ string s;
+ cin >> s;
+
+ int m;
+ cin >> m;
+ m--;
+
+ while (m--) {
+ string t;
+ for (int i = 0; i < s.size(); i++) {
+ int c = 1;
+ while (i + 1 < s.size() && s[i + 1] == s[i]) {
+ c++;
+ i++;
+ }
+ t = t + to_string(c) + s[i];
+ }
+ s = t;
+ }
+ cout << s << endl;
+ return 0;
+}
diff --git a/TangDou/LanQiaoBei/LanQiao15STEMA20231217/试题.txt b/TangDou/LanQiaoBei/LanQiao15STEMA20231217/试题.txt
new file mode 100644
index 0000000..30c6074
--- /dev/null
+++ b/TangDou/LanQiaoBei/LanQiao15STEMA20231217/试题.txt
@@ -0,0 +1 @@
+https://tiku.scratchor.com/paper/view/z8k4y0xsf0ue8xlh
diff --git a/TangDou/Test/TestA.cpp b/TangDou/Test/TestA.cpp
index 1b6071b..9f73825 100644
--- a/TangDou/Test/TestA.cpp
+++ b/TangDou/Test/TestA.cpp
@@ -1,27 +1,26 @@
#include
-
using namespace std;
-/**
- 8
- 4 2
- 1
- 2 2
- 3 3 3
- 4 4 4 4
-*/
-int n, sum;
int main() {
- cin >> n;
- for (int i = 1;; i++) {
- n -= i;
- if (n <= 0) {
- // cout << i << " " << i + n << endl;
- sum += i * (i + n); // 最后一行
- for (int j = 1; j < i; j++) sum += j * j;
- cout << sum << endl;
- break;
+ string s;
+ cin >> s;
+
+ int m;
+ cin >> m;
+ m--;
+
+ while (m--) {
+ string t;
+ for (int i = 0; i < s.size(); i++) {
+ int c = 1;
+ while (i + 1 < s.size() && s[i + 1] == s[i]) {
+ c++;
+ i++;
+ }
+ t = t + to_string(c) + s[i];
}
+ s = t;
}
+ cout << s << endl;
return 0;
-}
\ No newline at end of file
+}
diff --git a/小乔证件/蓝桥杯报名.txt b/小乔证件/蓝桥杯报名.txt
index 35fc9a2..b2b7f9f 100644
--- a/小乔证件/蓝桥杯报名.txt
+++ b/小乔证件/蓝桥杯报名.txt
@@ -5,19 +5,7 @@ https://www.lanqiaoqingshao.cn/home
黄琬乔
220105201212060427
-办法
-1、使用qq截图功能,(随意找一个聊天框,然后点击小剪刀功能),然后在图片的下方有一个识别文字的功能,很好用。
-截图快捷键: ctrl+alt+a
-
-2、复制粘贴到ChatGPT,让它来帮助给出答案,但似乎有时它的回复不正确。
-
https://tongyi.aliyun.com/
kgdxpr@163.com
DsideaL4r5t6y7u!@#
-
-
-(1) 用加速猫翻墙
-(2) https://chat.forefront.ai/
-superhuanghai@gmail.com
-mdcija780522