'commit'

2025-08-30 18:35:01 +08:00
commit 1f397eca87
6258 changed files with 1487241 additions and 0 deletions
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/.clang-format
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/.clang-format
@@ -0,0 +1,213 @@
+# 语言: None, Cpp, Java, JavaScript, ObjC, Proto, TableGen, TextProto
+Language: Cpp
+BasedOnStyle:	LLVM
+
+# 访问说明符(public、private等)的偏移
+AccessModifierOffset: -4
+
+# 开括号(开圆括号、开尖括号、开方括号)后的对齐: Align, DontAlign, AlwaysBreak(总是在开括号后换行)
+AlignAfterOpenBracket: Align
+
+# 连续赋值时，对齐所有等号
+AlignConsecutiveAssignments: false
+
+# 连续声明时，对齐所有声明的变量名
+AlignConsecutiveDeclarations: false
+
+# 右对齐逃脱换行(使用反斜杠换行)的反斜杠
+AlignEscapedNewlines: Right
+
+# 水平对齐二元和三元表达式的操作数
+AlignOperands: true
+
+# 对齐连续的尾随的注释
+AlignTrailingComments: true
+
+# 不允许函数声明的所有参数在放在下一行
+AllowAllParametersOfDeclarationOnNextLine: false
+
+# 不允许短的块放在同一行
+AllowShortBlocksOnASingleLine: true
+
+# 允许短的case标签放在同一行
+AllowShortCaseLabelsOnASingleLine: true
+
+# 允许短的函数放在同一行: None, InlineOnly(定义在类中), Empty(空函数), Inline(定义在类中，空函数), All
+AllowShortFunctionsOnASingleLine: None
+
+# 允许短的if语句保持在同一行
+AllowShortIfStatementsOnASingleLine: true
+
+# 允许短的循环保持在同一行
+AllowShortLoopsOnASingleLine: true
+
+# 总是在返回类型后换行: None, All, TopLevel(顶级函数，不包括在类中的函数), 
+# AllDefinitions(所有的定义，不包括声明), TopLevelDefinitions(所有的顶级函数的定义)
+AlwaysBreakAfterReturnType: None
+
+# 总是在多行string字面量前换行
+AlwaysBreakBeforeMultilineStrings: false
+
+# 总是在template声明后换行
+AlwaysBreakTemplateDeclarations: true
+
+# false表示函数实参要么都在同一行，要么都各自一行
+BinPackArguments: true
+
+# false表示所有形参要么都在同一行，要么都各自一行
+BinPackParameters: true
+
+# 大括号换行，只有当BreakBeforeBraces设置为Custom时才有效
+BraceWrapping:
+  # class定义后面
+  AfterClass: false
+  # 控制语句后面
+  AfterControlStatement: false
+  # enum定义后面
+  AfterEnum: false
+  # 函数定义后面
+  AfterFunction: false
+  # 命名空间定义后面
+  AfterNamespace: false
+  # struct定义后面
+  AfterStruct: false
+  # union定义后面
+  AfterUnion: false
+  # extern之后
+  AfterExternBlock: false
+  # catch之前
+  BeforeCatch: false
+  # else之前
+  BeforeElse: false
+  # 缩进大括号
+  IndentBraces: false
+  # 分离空函数
+  SplitEmptyFunction: false
+  # 分离空语句
+  SplitEmptyRecord: false
+  # 分离空命名空间
+  SplitEmptyNamespace: false
+
+# 在二元运算符前换行: None(在操作符后换行), NonAssignment(在非赋值的操作符前换行), All(在操作符前换行)
+BreakBeforeBinaryOperators: NonAssignment
+
+# 在大括号前换行: Attach(始终将大括号附加到周围的上下文), Linux(除函数、命名空间和类定义，与Attach类似), 
+#   Mozilla(除枚举、函数、记录定义，与Attach类似), Stroustrup(除函数定义、catch、else，与Attach类似), 
+#   Allman(总是在大括号前换行), GNU(总是在大括号前换行，并对于控制语句的大括号增加额外的缩进), WebKit(在函数前换行), Custom
+#   注：这里认为语句块也属于函数
+BreakBeforeBraces: Custom
+
+# 在三元运算符前换行
+BreakBeforeTernaryOperators: false
+
+# 在构造函数的初始化列表的冒号后换行
+BreakConstructorInitializers: AfterColon
+
+#BreakInheritanceList: AfterColon
+
+BreakStringLiterals: false
+
+# 每行字符的限制，0表示没有限制
+ColumnLimit: 0
+
+CompactNamespaces: true
+
+# 构造函数的初始化列表要么都在同一行，要么都各自一行
+ConstructorInitializerAllOnOneLineOrOnePerLine: false
+
+# 构造函数的初始化列表的缩进宽度
+ConstructorInitializerIndentWidth: 4
+
+# 延续的行的缩进宽度
+ContinuationIndentWidth: 4
+
+# 去除C++11的列表初始化的大括号{后和}前的空格
+Cpp11BracedListStyle: true
+
+# 继承最常用的指针和引用的对齐方式
+DerivePointerAlignment: false
+
+# 固定命名空间注释
+FixNamespaceComments: true
+
+# 缩进case标签
+IndentCaseLabels: false
+
+IndentPPDirectives: None
+
+# 缩进宽度
+IndentWidth: 4
+
+# 函数返回类型换行时，缩进函数声明或函数定义的函数名
+IndentWrappedFunctionNames: false
+
+# 保留在块开始处的空行
+KeepEmptyLinesAtTheStartOfBlocks: false
+
+# 连续空行的最大数量
+MaxEmptyLinesToKeep: 1
+
+# 命名空间的缩进: None, Inner(缩进嵌套的命名空间中的内容), All
+NamespaceIndentation: None
+
+# 指针和引用的对齐: Left, Right, Middle
+PointerAlignment: Right
+
+# 允许重新排版注释
+ReflowComments: true
+
+# 允许排序#include
+SortIncludes: false
+
+# 允许排序 using 声明
+SortUsingDeclarations: false
+
+# 在C风格类型转换后添加空格
+SpaceAfterCStyleCast: false
+
+# 在Template 关键字后面添加空格
+SpaceAfterTemplateKeyword: true
+
+# 在赋值运算符之前添加空格
+SpaceBeforeAssignmentOperators: true
+
+# SpaceBeforeCpp11BracedList: true
+
+# SpaceBeforeCtorInitializerColon: true
+
+# SpaceBeforeInheritanceColon: true
+
+# 开圆括号之前添加一个空格: Never, ControlStatements, Always
+SpaceBeforeParens: ControlStatements
+
+# SpaceBeforeRangeBasedForLoopColon: true
+
+# 在空的圆括号中添加空格
+SpaceInEmptyParentheses: false
+
+# 在尾随的评论前添加的空格数(只适用于//)
+SpacesBeforeTrailingComments: 1
+
+# 在尖括号的<后和>前添加空格
+SpacesInAngles: false
+
+# 在C风格类型转换的括号中添加空格
+SpacesInCStyleCastParentheses: false
+
+# 在容器(ObjC和JavaScript的数组和字典等)字面量中添加空格
+SpacesInContainerLiterals: true
+
+# 在圆括号的(后和)前添加空格
+SpacesInParentheses: false
+
+# 在方括号的[后和]前添加空格，lamda表达式和未指明大小的数组的声明不受影响
+SpacesInSquareBrackets: false
+
+# 标准: Cpp03, Cpp11, Auto
+Standard: Cpp11
+
+# tab宽度
+TabWidth: 4
+
+# 使用tab字符: Never, ForIndentation, ForContinuationAndIndentation, Always
+UseTab: Never
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/.vscode/c_cpp_properties.json
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/.vscode/c_cpp_properties.json
@@ -0,0 +1,25 @@
+{
+    "configurations": [
+      {
+        "name": "Win32",
+        "compilerArgs": [
+          "-fexec-charset=GBK"
+        ],
+        "includePath": [
+          "${workspaceFolder}/**",
+                "C:/TDM-GCC-64/lib/gcc/x86_64-w64-mingw32/10.3.0/**",
+                "C:/TDM-GCC-64/x86_64-w64-mingw32/include/**"
+              ],
+              "defines": [
+                "_DEBUG",
+                "UNICODE",
+                "_UNICODE"
+              ],
+              "compilerPath": "C:/TDM-GCC-64/bin/gcc.exe",
+              "cStandard": "c11",
+              "cppStandard": "c++17",
+              "intelliSenseMode": "gcc-x64"
+            }
+          ],
+          "version": 4
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/.vscode/extensions.json
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/.vscode/extensions.json
@@ -0,0 +1,5 @@
+{
+  "recommendations": [
+    "MS-CEINTL.vscode-language-pack-zh-hans"
+  ]
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/.vscode/launch.json
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/.vscode/launch.json
@@ -0,0 +1,28 @@
+// https://github.com/Microsoft/vscode-cpptools/blob/master/launch.md
+{
+  "version": "0.2.0",
+  "configurations": [
+    {
+      "name": "(gdb) Launch", // 配置名称，将会在启动配置的下拉菜单中显示
+      "type": "cppdbg", // 配置类型，cppdbg对应cpptools提供的调试功能；可以认为此处只能是cppdbg
+      "request": "launch", // 请求配置类型，可以为launch（启动）或attach（附加）
+      "program": "${fileDirname}\\${fileBasenameNoExtension}.exe", // 将要进行调试的程序的路径
+      "args": [], // 程序调试时传递给程序的命令行参数，一般设为空即可
+      "stopAtEntry": false, // 设为true时程序将暂停在程序入口处，相当于在main上打断点
+      "externalConsole": false,
+      "cwd": "${fileDirname}\\", // 调试程序时的工作目录，此为工作区文件夹；改成${fileDirname}可变为文件所在目录
+      "environment": [], // 环境变量
+      "internalConsoleOptions": "neverOpen", // 如果不设为neverOpen，调试时会跳到“调试控制台”选项卡，你应该不需要对gdb手动输命令吧？
+      "MIMode": "gdb", // 指定连接的调试器，可以为gdb或lldb。但我没试过lldb
+      "miDebuggerPath": "gdb.exe", // 调试器路径，Windows下后缀不能省略，Linux下则不要
+      "setupCommands": [
+        { // 模板自带，好像可以更好地显示STL容器的内容，具体作用自行Google
+          "description": "Enable pretty-printing for gdb",
+          "text": "-enable-pretty-printing",
+          "ignoreFailures": false
+        }
+      ],
+      "preLaunchTask": "Compile" // 调试会话开始前执行的任务，一般为编译程序。与tasks.json的label相对应
+    }
+  ]
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/.vscode/settings.json
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/.vscode/settings.json
@@ -0,0 +1,112 @@
+{
+    "files.defaultLanguage": "c", // ctrl+N新建文件后默认的语言
+    "editor.formatOnType": true,  // 输入分号(C/C++的语句结束标识)后自动格式化当前这一行的代码
+    "editor.suggest.snippetsPreventQuickSuggestions": false, // clangd的snippets有很多的跳转点，不用这个就必须手动触发Intellisense了
+    "editor.acceptSuggestionOnEnter": "off", // 我个人的习惯，按回车时一定是真正的换行，只有tab才会接受Intellisense
+
+    "code-runner.runInTerminal": true, // 设置成false会在“输出”中输出，无法输入    
+    "code-runner.saveFileBeforeRun": true, // run code前保存
+    "code-runner.preserveFocus": true,     // 若为false，run code后光标会聚焦到终端上。如果需要频繁输入数据可设为false
+    "code-runner.clearPreviousOutput": false, // 每次run code前清空属于code runner的终端消息，默认false
+    "code-runner.ignoreSelection": true,   // 默认为false，效果是鼠标选中一块代码后可以单独执行，但C是编译型语言，不适合这样用
+
+    "C_Cpp.clang_format_sortIncludes": true,
+    "files.associations": {
+      "array": "cpp",
+      "atomic": "cpp",
+      "*.tcc": "cpp",
+      "cctype": "cpp",
+      "clocale": "cpp",
+      "cmath": "cpp",
+      "cstdarg": "cpp",
+      "cstddef": "cpp",
+      "cstdint": "cpp",
+      "cstdio": "cpp",
+      "cstdlib": "cpp",
+      "cwchar": "cpp",
+      "cwctype": "cpp",
+      "deque": "cpp",
+      "unordered_map": "cpp",
+      "vector": "cpp",
+      "exception": "cpp",
+      "algorithm": "cpp",
+      "memory": "cpp",
+      "memory_resource": "cpp",
+      "optional": "cpp",
+      "string": "cpp",
+      "string_view": "cpp",
+      "system_error": "cpp",
+      "tuple": "cpp",
+      "type_traits": "cpp",
+      "utility": "cpp",
+      "fstream": "cpp",
+      "initializer_list": "cpp",
+      "iosfwd": "cpp",
+      "iostream": "cpp",
+      "istream": "cpp",
+      "limits": "cpp",
+      "new": "cpp",
+      "ostream": "cpp",
+      "sstream": "cpp",
+      "stdexcept": "cpp",
+      "streambuf": "cpp",
+      "typeinfo": "cpp",
+      "valarray": "cpp",
+      "cassert": "cpp",
+      "cerrno": "cpp",
+      "cfloat": "cpp",
+      "chrono": "cpp",
+      "ciso646": "cpp",
+      "climits": "cpp",
+      "complex": "cpp",
+      "csetjmp": "cpp",
+      "csignal": "cpp",
+      "cstring": "cpp",
+      "ctime": "cpp",
+      "cuchar": "cpp",
+      "numeric": "cpp",
+      "ratio": "cpp",
+      "version": "cpp",
+      "bitset": "cpp",
+      "regex": "cpp",
+      "random": "cpp",
+      "unordered_set": "cpp",
+      "list": "cpp",
+      "any": "cpp",
+      "functional": "cpp",
+      "mutex": "cpp",
+      "shared_mutex": "cpp",
+      "thread": "cpp",
+      "variant": "cpp",
+      "map": "cpp",
+      "scoped_allocator": "cpp",
+      "future": "cpp",
+      "iomanip": "cpp",
+      "forward_list": "cpp",
+      "bit": "cpp",
+      "ccomplex": "cpp",
+      "cfenv": "cpp",
+      "charconv": "cpp",
+      "cinttypes": "cpp",
+      "codecvt": "cpp",
+      "condition_variable": "cpp",
+      "cstdalign": "cpp",
+      "cstdbool": "cpp",
+      "ctgmath": "cpp",
+      "set": "cpp",
+      "iterator": "cpp",
+      "filesystem": "cpp",
+      "ios": "cpp",
+      "locale": "cpp",
+      "queue": "cpp",
+      "stack": "cpp",
+      "typeindex": "cpp",
+      "compare": "cpp",
+      "concepts": "cpp",
+      "coroutine": "cpp",
+      "numbers": "cpp",
+      "ranges": "cpp",
+      "span": "cpp",
+      "stop_token": "cpp"
+    }, // 格式化时调整include的顺序（按字母排序）
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/.vscode/tasks.json
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/.vscode/tasks.json
@@ -0,0 +1,40 @@
+// https://code.visualstudio.com/docs/editor/tasks
+{
+  "version": "2.0.0",
+  "tasks": [
+    {
+      "label": "Compile", // 任务名称，与launch.json的preLaunchTask相对应
+      "command": "g++", // 要使用的编译器，C++用g++
+      "args": [
+        "${file}",
+        "-o", // 指定输出文件名，不加该参数则默认输出a.exe，Linux下默认a.out
+        "${fileDirname}\\${fileBasenameNoExtension}.exe",
+        "-g", // 生成和调试有关的信息
+        "-Wall", // 开启额外警告
+        "-static-libgcc", // 静态链接libgcc，一般都会加上
+        "-fexec-charset=GBK", // 生成的程序使用GBK编码，不加这一条会导致Win下输出中文乱码
+        "-std=c++17", // C++最新标准为c++17，或根据自己的需要进行修改
+      ], // 编译的命令，其实相当于VSC帮你在终端中输了这些东西
+      "type": "process", // process是vsc把预定义变量和转义解析后直接全部传给command；shell相当于先打开shell再输入命令，所以args还会经过shell再解析一遍
+      "group": {
+        "kind": "build",
+        "isDefault": true // 不为true时ctrl shift B就要手动选择了
+      },
+      "presentation": {
+        "echo": true,
+        "reveal": "always", // 执行任务时是否跳转到终端面板，可以为always，silent，never。具体参见VSC的文档
+        "focus": false, // 设为true后可以使执行task时焦点聚集在终端，但对编译C/C++来说，设为true没有意义
+        "panel": "shared" // 不同的文件的编译信息共享一个终端面板
+      },
+    },
+    {
+      "label": "commit gitee",
+      "type": "shell",
+      "command": "D:\\python\\一键推拉.cmd",
+      "group": {
+        "kind": "build",
+        "isDefault": true
+      },
+    }
+  ]
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/1.cpp
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/1.cpp
@@ -0,0 +1,12 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+int n, cnt;
+int main() {
+    cin >> n;
+    for (int i = 3; i <= n; i++)
+        if (i % 10 == 3) cnt++;
+
+    cout << cnt << endl;
+    return 0;
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/2.cpp
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/2.cpp
@@ -0,0 +1,14 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 110;
+char a[N];
+int main() {
+    int n;
+    cin >> n;
+    for (int i = 1; i <= n; i++) cin >> a[i];
+    sort(a + 1, a + 1 + n);
+
+    for (int i = 1; i <= n; i++) cout << a[i];
+    return 0;
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/3.md
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/3.md
@@ -0,0 +1,54 @@
+### 一、 理论知识
+裴蜀定理：任意两个数的组合必定是它们最大公约数的倍数。
+
+* 推广：如果$n$个数的最大公约数是$d$，那么它们的组合是$d$的倍数，如果$d≠1$，则必然有无限个数无法被组合出来
+
+* 如果每笼包子数目的最大公约数不是$1$，应该输出$INF$
+  
+* 如果给出的这些包子数互质，即最大公约数为$1$，那么由定理可知，是有有限个数不能被凑出来的，应该输出个数。
+  
+* 赛瓦维斯特定理 
+  
+  如果这些包子数互质，那么最大不能被表示出来的数字就是 (最小$a$和次小$b$) $a*b-a-b$
+
+
+### 二、形象理解
+  * 不互质栗子:
+  比如$3$ 和 $6$两种规格，能组装的就是 $3,6,9,12,15,...$,这个变化就是它们的最大公约数3。
+    
+  此时 ，类似于 $1,2,?,4,5,?,7,8,?,10,11,?,13,14,?,16,17,?,19,20, ,...$之类的数字均是无法凑出的，有无穷多个。
+
+  * 互质栗子
+  
+    比如 $3$ 和 $4$ 
+    $11$->$1*3+2*4$  
+    $12-> 3*4$  
+    $13->1*4+3*3$      
+    $14->4*2+3*2$      
+    $15->3*5$      
+    $16->4*4$      
+    $17->3*3+4*2$  
+     ...  
+
+    再来 $4$ 和 $7$      
+    $29->3*7+4*2$  
+    $30->2*7+4*4$  
+    $31->1*7+4*6$  
+    $32->4*8$  
+    $33->3*7+4*3$  
+    $34->2*7+4*5$
+    
+     ...  
+
+    再来 $5$ 和 $7$  
+    $31->3*7+5*2$  
+    $32->1*7+5*5$    
+    $33->4*7+1*5$   
+    $34->2*7+4*5$  
+    $35->5*7$  
+    $36->3*7+3*5$  
+     ...
+
+
+* ### 二、完全背包求方案数
+[$AcWing$ $1371$. 货币系统](https://www.cnblogs.com/littlehb/p/16199799.html)
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/3.mmap
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/3.mmap
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/3_1.cpp
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/3_1.cpp
@@ -0,0 +1,55 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+const int N = 30;
+const int M = 1e4 + 10;
+/*
+2
+3 5
+
+答案：
+4
+
+最大不能表示的应该是：3*5-3-5=7
+四个不能表示的数是：1 2 4 7
+*/
+int f[N][M];
+int w[N];
+int res;
+
+int gcd(int a, int b) {
+    if (b == 0) return a;
+    return gcd(b, a % b);
+}
+
+int main() {
+    int n;
+    cin >> n;
+    for (int i = 1; i <= n; i++) cin >> w[i];
+
+    sort(w + 1, w + 1 + n);
+    int m1 = w[1], m2 = w[2];
+    int M = m1 * m2 - m1 - m2; // a*b-a-b =7
+
+    int g = w[1];
+    for (int i = 2; i <= n; i++) g = gcd(g, w[i]);
+
+    if (g > 1) {
+        cout << -1 << endl;
+        return 0;
+    }
+
+    // 完全背包二维写法
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= M; j++)
+            if (j >= w[i])
+                f[i][j] = max(f[i - 1][j], f[i][j - w[i]] + w[i]);
+            else
+                f[i][j] = f[i - 1][j];
+
+    for (int i = 1; i <= M; i++)
+        if (f[n][i] < i) res++;
+
+    printf("%d\n", res);
+    return 0;
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/3_2.cpp
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/3_2.cpp
@@ -0,0 +1,51 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+const int N = 30;
+const int M = 1e4 + 10;
+/*
+2
+3 5
+
+答案：
+4
+
+最大不能表示的应该是：3*5-3-5=7
+四个不能表示的数是：1 2 4 7
+*/
+int f[M];
+int w[N];
+int res;
+
+int gcd(int a, int b) {
+    if (b == 0) return a;
+    return gcd(b, a % b);
+}
+
+int main() {
+    int n;
+    cin >> n;
+    for (int i = 1; i <= n; i++) cin >> w[i];
+
+    sort(w + 1, w + 1 + n);
+    int m1 = w[1], m2 = w[2];
+    int M = m1 * m2 - m1 - m2;
+
+    int g = w[1];
+    for (int i = 2; i <= n; i++) g = gcd(g, w[i]);
+
+    if (g > 1) {
+        cout << "-1" << endl;
+        return 0;
+    }
+    //完全背包一维写法
+    for (int i = 1; i <= n; i++)
+        for (int j = w[i]; j <= M; j++)
+            f[j] = max(f[j], f[j - w[i]] + w[i]);
+
+    for (int i = 1; i <= M; i++)
+        if (f[i] < i) res++;
+
+    printf("%d\n", res);
+    return 0;
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/4.cpp
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/4.cpp
@@ -0,0 +1,36 @@
+#include <bits/stdc++.h>
+/**
+ 5 2
+ 200 145 240 50 45
+ 150 300
+ 答案
+ 4
+ */
+using namespace std;
+int m, n;
+const int N = 310;
+int a[N], b[N];
+int res;
+void dfs(int u, int s) {
+    if (u == n + 1) {
+        res = max(res, s);
+        return;
+    }
+
+    dfs(u + 1, s);
+
+    for (int i = 1; i <= m; i++)
+        if (a[u] <= b[i]) {
+            b[i] -= a[u]; //回溯
+            dfs(u + 1, s + 1);
+            b[i] += a[u];
+        }
+}
+int main() {
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++) cin >> a[i];
+    for (int i = 1; i <= m; i++) cin >> b[i];
+    dfs(1, 0);
+    printf("%d\n", res);
+    return 0;
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/5_bfs_1.cpp
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/5_bfs_1.cpp
@@ -0,0 +1,48 @@
+#include <bits/stdc++.h>
+/**
+ 5 4
+ 1 2
+ 1 3
+ 2 4
+ 2 5
+
+ 答案
+ 2 1 3 2
+*/
+using namespace std;
+const int INF = 0x3f3f3f3f;
+const int N = 110;
+int n, m;
+int g[N][N];
+int f[N];
+bool st[N];
+
+void bfs() {
+    queue<int> q;
+    q.push(n);
+    st[n] = true;
+    while (q.size()) {
+        auto u = q.front();
+        q.pop();
+        for (int i = 1; i < n; i++)
+            if (!st[i] && g[u][i] != INF) {
+                f[i] = f[u] + 1;
+                q.push(i);
+                st[i] = true;
+            }
+    }
+}
+
+int main() {
+    memset(g, 0x3f, sizeof g); // 地图初始化
+
+    cin >> n >> m;
+    while (m--) {
+        int a, b;
+        cin >> a >> b;
+        g[a][b] = g[b][a] = 1;
+    }
+    bfs();
+    for (int i = 1; i < n; i++) cout << f[i] << " ";
+    return 0;
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/5_bfs_2.cpp
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/5_bfs_2.cpp
@@ -0,0 +1,54 @@
+#include <bits/stdc++.h>
+/**
+ 5 4
+ 1 2
+ 1 3
+ 2 4
+ 2 5
+
+ 答案
+ 2 1 3 2
+
+ bfs求的是无权图或边权相等图的最短路径算法
+*/
+using namespace std;
+const int INF = 0x3f3f3f3f;
+typedef pair<int, int> PII;
+const int N = 110;
+int n, m;
+int g[N][N]; // 邻接矩阵装图
+int f[N];    // 离n点的距离
+bool st[N];  // 是不是走过
+
+void bfs() {
+    queue<PII> q;   // PII: first sencond
+    q.push({n, 0}); // first:n号点是起点,second:离出发点的距离
+    st[n] = true;
+    while (q.size()) {
+        auto u = q.front();
+        q.pop();
+        for (int i = 1; i < n; i++)
+            // 1、没走过
+            // 2、有路可走
+            if (!st[i] && g[u.first][i] != INF) {
+                f[i] = u.second + 1;
+                q.push({i, f[i]});
+                st[i] = true;
+            }
+    }
+}
+
+int main() {
+    cin >> n >> m;
+    memset(g, 0x3f, sizeof g); // 初始化每两个点之间的距离无穷大
+    memset(f, -1, sizeof f);
+    while (m--) {
+        int a, b;
+        cin >> a >> b;
+        g[a][b] = g[b][a] = 1;
+    }
+    bfs();
+    for (int i = 1; i < n; i++)
+        cout << f[i] << " ";
+    return 0;
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/5_floyd.cpp
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/5_floyd.cpp
@@ -0,0 +1,38 @@
+#include <bits/stdc++.h>
+/**
+ 5 4
+ 1 2
+ 1 3
+ 2 4
+ 2 5
+ 答案
+ 2 1 3 2
+ *
+ */
+using namespace std;
+const int INF = 0x3f3f3f3f;
+const int N = 110;
+int d[N][N];
+int n, m;
+void floyd() {
+    for (int k = 0; k < n; k++)
+        for (int i = 1; i <= n; i++)
+            for (int j = 1; j <= n; j++)
+                d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
+}
+int main() {
+    cin >> n >> m;
+    memset(d, 0x3f, sizeof d);
+    for (int i = 0; i < m; i++) {
+        int a, b;
+        cin >> a >> b;
+        d[a][b] = d[b][a] = 1;
+    }
+    floyd();
+    for (int i = 1; i < n; i++)
+        if (d[n][i] == INF)
+            cout << -1 << " ";
+        else
+            cout << d[n][i] << " ";
+    return 0;
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/6_0.cpp
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/6_0.cpp
@@ -0,0 +1,52 @@
+#include <bits/stdc++.h>
+/*
+测试用例I：
+5 6
+1 0 1 0 0 0
+0 1 0 1 0 0
+1 0 0 0 1 0
+0 1 0 0 0 1
+1 0 1 0 0 0
+
+答案
+4
+
+测试用例II：
+5 6
+1 0 1 0 0 0
+0 1 0 1 1 0
+1 0 0 0 1 0
+0 1 0 0 0 1
+1 0 1 0 0 0
+
+答案
+3
+*/
+using namespace std;
+const int INF = 0x3f3f3f3f;
+const int N = 110;
+int a[N][N];
+int res;
+
+int main() {
+    int n, m;
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++)
+            cin >> a[i][j];
+
+    // 1e8 =100000000
+    //     1500000000
+    for (int i = 1; i <= n; i++)                   // 1000
+        for (int j = 1; j <= m; j++) {             // 1000
+            for (int k = 0;; k++) {                // 1000 *SQRT(2)=1500
+                if (i + k > n || j + k > m) break; //越界
+                if (a[i + k][j + k] == 0) break;
+                //从[i,j]出发，边长长度为k+1的矩形，内部数字为1的有多少个
+
+                res = max(res, k + 1);
+            }
+        }
+    cout << res << endl;
+    return 0;
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/6_1.cpp
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/6_1.cpp
@@ -0,0 +1,55 @@
+#include <bits/stdc++.h>
+/*
+测试用例I：
+5 6
+1 0 1 0 0 0
+0 1 0 1 0 0
+1 0 0 0 1 0
+0 1 0 0 0 1
+1 0 1 0 0 0
+
+答案
+4
+
+测试用例II：
+5 6
+1 0 1 0 0 0
+0 1 0 1 1 0
+1 0 0 0 1 0
+0 1 0 0 0 1
+1 0 1 0 0 0
+
+答案
+2
+*/
+using namespace std;
+const int INF = 0x3f3f3f3f;
+const int N = 110;
+int a[N][N];
+int s[N][N];
+int res;
+
+int main() {
+    int n, m;
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++) {
+            cin >> a[i][j];
+            s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + a[i][j];
+        }
+
+    for (int i = 1; i <= n; i++)       // 100
+        for (int j = 1; j <= m; j++) { // 100
+            for (int r = 0;; r++) {
+                if (i + r > n || j + r > m) break; // 越界
+                if (a[i + r][j + r] == 0) break;
+                
+                int x1 = i, y1 = j, x2 = i + r, y2 = j + r;
+                int sum = s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1];
+                if (sum != r + 1) break;
+                res = max(res, r + 1);
+            }
+        }
+    printf("%d\n", res);
+    return 0;
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/6_2.cpp
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/6_2.cpp
@@ -0,0 +1,63 @@
+#include <bits/stdc++.h>
+/*
+测试用例：
+5 6
+1 0 1 0 0 0
+0 1 0 1 0 0
+1 0 0 0 1 0
+0 1 0 0 0 1
+1 0 1 0 0 0
+
+
+测试用例II：
+5 6
+1 0 1 0 0 0         2 0 4 0 0 0
+0 1 0 1 1 0         0 1 0 3 1 0
+1 0 0 0 1 0         3 0 0 0 2 0
+0 1 0 0 0 1         0 2 0 0 0 1
+1 0 1 0 0 0         1 0 1 0 0 0
+
+5 6
+1 0 1 0 0 0
+0 1 0 1 1 0
+1 0 0 0 1 0
+0 1 0 0 0 1
+1 0 1 0 0 0
+
+
+测试用例III：
+5 6
+1 0 1 0 0 0
+0 1 0 1 1 0
+1 0 0 0 1 0
+0 1 0 1 0 1
+1 0 1 0 1 0
+
+*/
+using namespace std;
+const int N = 110;
+int a[N][N];
+int res;
+
+int main() {
+    int n, m;
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++)
+            cin >> a[i][j];
+
+    //直接在原数组上递推
+    for (int i = n; i >= 1; i--)            //从倒数第二行开始
+        for (int j = 1; j <= m; j++)        //这个需要考虑一下边界问题
+            if (a[i][j])                    //只有当前位置大于0才有考虑的价值
+                a[i][j] += a[i + 1][j + 1]; //递推式
+
+    cout << "====================" << endl;
+    //打印出这个新的递推数组看看是什么情况
+    for (int i = 1; i <= n; i++) {
+        for (int j = 1; j <= m; j++)
+            cout << a[i][j] << " ";
+        cout << endl;
+    }
+    return 0;
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/6_3.cpp
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/6_3.cpp
@@ -0,0 +1,57 @@
+#include <bits/stdc++.h>
+/*
+测试用例：
+5 6
+1 0 1 0 0 0
+0 1 0 1 0 0
+1 0 0 0 1 0
+0 1 0 0 0 1
+1 0 1 0 0 0
+
+答案
+4
+
+
+测试用例II：
+5 6
+1 0 1 0 0 0
+0 1 0 1 1 0
+1 0 0 0 1 0
+0 1 0 0 0 1
+1 0 1 0 0 0
+
+答案
+2
+*/
+using namespace std;
+const int N = 110;
+int a[N][N];
+int s[N][N];
+int res;
+
+int main() {
+    int n, m;
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++) {
+            cin >> a[i][j];
+            //维护二维前缀和
+            s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + a[i][j];
+        }
+
+    //直接在原数组上递推
+    for (int i = n - 1; i >= 1; i--)        //从倒数第二行开始
+        for (int j = 1; j <= m; j++)        //这个需要考虑一下边界问题
+            if (a[i][j]) {                  //只有当前位置大于0才有考虑的价值
+                a[i][j] += a[i + 1][j + 1]; //递推式,当前位置可以向下扩展的最长对角线长度
+
+                //利用二维前缀和计算区域的总和
+                int x1 = i, y1 = j, x2 = i + a[i][j] - 1, y2 = j + a[i][j] - 1;
+                int cnt = s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1];
+                if (cnt == a[i][j])
+                    res = max(res, a[i][j]); //在递推过程中不断求最大值
+            }
+    //输出
+    cout << res << endl;
+    return 0;
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/6_递推示意图.png
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/6_递推示意图.png
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/7.cpp
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/7.cpp
@@ -0,0 +1,36 @@
+/*
+小马需要将N件物品从河的一岸搬运到河的另一岸，每次搬运的物品为1到3件。
+请问小马将N件物品全部搬运过去有多少种方案。
+例如：N=3，将3件物品全部搬运过去有4种方案：
+方案一：第一次搬运1件，第二次搬运1件，第三次搬运1件；
+方案二：第一次搬运1件，第二次搬运2件；
+方案三：第一次搬运2件，第二次搬运1件；
+方案四：一次搬运3件。
+*/
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 110;
+int a[N];
+int res;
+int n;
+
+/**
+ * @brief
+ *
+ * @param step 第几次
+ * @param r  剩余件数
+ */
+void dfs(int u, int r) {
+    if (r == 0) res++; //剩余为0，就得到一组答案
+
+    //本轮运几件
+    for (int i = 1; i <= 3; i++)
+        if (r - i >= 0) dfs(u + 1, r - i);
+}
+int main() {
+    cin >> n;
+    dfs(0, n);
+    printf("%d", res);
+    return 0;
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/8.cpp
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/8.cpp
@@ -0,0 +1,63 @@
+/*
+有一片海域划分为N*M个方格，其中有些海域已被污染（用0表示），
+有些海域没被污染（用1表示）。请问这片N*M海域中有几块是没被污染的独立海域
+（没被污染的独立海域是指该块海域上下左右被已污染的海域包围，
+且N*M以外的海域都为已被污染的海域）
+例如：N=4，M=5，4*5的海域中，已被污染海域和没被污染的海域如下图：
+这块4*5的海域，有3块海域（绿色）没被污染，因为每一块的上下左右都被污染的海域包围。
+
+4 5
+1 1 0 0 0
+1 0 1 0 0
+1 0 0 0 0
+1 1 0 1 1
+
+样例输出
+3
+*/
+#include <bits/stdc++.h>
+using namespace std;
+#define x first
+#define y second
+typedef pair<int, int> PII;
+const int N = 110;
+int a[N][N];
+int n, m;
+
+int dx[] = {-1, 0, 1, 0}; //上右下左
+int dy[] = {0, 1, 0, -1}; //上右下左
+void bfs(int x, int y) {
+    a[x][y] = 0;
+    queue<PII> q;
+    q.push({x, y});
+    while (q.size()) {
+        auto u = q.front();
+        q.pop();
+        for (int i = 0; i <= 3; i++) {
+            int tx = u.x + dx[i], ty = u.y + dy[i];
+            if (tx < 1 || tx > n || ty < 1 || ty > m) continue;
+            if (a[tx][ty]) {
+                a[tx][ty] = 0;
+                q.push({tx, ty});
+            }
+        }
+    }
+}
+int res;
+int main() {
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++)
+            cin >> a[i][j];
+
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++) {
+            if (a[i][j]) {
+                bfs(i, j);
+                res++;
+            }
+        }
+
+    printf("%d", res);
+    return 0;
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/BaBa.cpp
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/BaBa.cpp
@@ -0,0 +1,22 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+int cnt;
+const int N = 1010;
+int f[N];
+
+int main() {
+    int n = 1000, m = 1000;
+    for (int x = 1; x <= n; x++)
+        for (int y = 1; y <= m; y++)
+            for (int len = 1;; len++) {
+                int tx = x + len - 1, ty = y + len - 1;
+                if (tx > n || ty > m) break;
+                f[len]++;
+            }
+
+    for (int i = 1; i <= min(n, m); i++)
+        cout << f[i] << " ";
+
+    return 0;
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/MaiTangYuan_BEIBAO.cpp
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/MaiTangYuan_BEIBAO.cpp
@@ -0,0 +1,72 @@
+#include <bits/stdc++.h>
+using namespace std;
+/**
+ <20><><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>
+ 2
+ 3 5
+
+ <20><><EFBFBD><EFBFBD>
+ 4
+ */
+const int N = 30;
+
+int f[N];  //<2F><>ʾ<EFBFBD>ճ<EFBFBD>j<EFBFBD><6A><EFBFBD><EFBFBD><EFBFBD>ӵķ<D3B5><C4B7><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>
+int a[N]; //ÿ<>ֹ<EFBFBD><D6B9><EFBFBD><EFBFBD>а<EFBFBD><D0B0>ӵ<EFBFBD><D3B5><EFBFBD><EFBFBD><EFBFBD>
+int ans;  //<2F><>װ<EFBFBD><D7B0><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>İ<EFBFBD><C4B0>Ӹ<EFBFBD><D3B8><EFBFBD>
+
+//<2F><><EFBFBD><EFBFBD>
+bool isHz(int a, int b) {
+	for (int i = 2; i <= min(a, b); i++) //a,b<><62><EFBFBD>ܴ<EFBFBD><DCB4>ڵĹ<DAB5><C4B9><EFBFBD><EFBFBD><EFBFBD>
+		if (a % i == 0 && b % i == 0)
+			return false;
+	return true;
+}
+
+int main() {
+	int n;
+	cin >> n;
+	for (int i = 1; i <= n; i++)
+		cin >> a[i];
+
+	//<2F><><EFBFBD><EFBFBD>
+	sort(a + 1, a + 1 + n);
+
+	//a<><61><EFBFBD><EFBFBD><EFBFBD>У<EFBFBD><D0A3>ǲ<EFBFBD><C7B2>Ǵ<EFBFBD><C7B4><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>ʵ<EFBFBD><CAB5><EFBFBD>
+	bool flag = false;
+	int mx;//<2F><><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>
+	for (int i = 1; i <= n; i++) {
+		for (int j = i + 1; j <= n; j++)
+			if (isHz(a[i], a[j])) {
+				flag = true;
+				mx = a[i] * a[j] - a[i] - a[j];
+				break;
+			}
+		if (flag)
+			break;
+	}
+
+
+	if (!flag) {
+		cout << -1 << endl;
+		return 0;
+	}
+
+	/**
+	  <20><>ȫ<EFBFBD><C8AB><EFBFBD><EFBFBD>
+	  f[j]<5D><>ʾ<EFBFBD>ճ<EFBFBD>j<EFBFBD><6A><EFBFBD><EFBFBD><EFBFBD>ӵķ<D3B5><C4B7><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>f[j]>0<>ʹյó<D5B5><C3B3><EFBFBD>f[j]=0<>ʹղ<CDB4><D5B2><EFBFBD><EFBFBD><EFBFBD>
+	*/
+	f[0] = 1; //  0<><30><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>״̬<D7B4>ǺϷ<C7BA><CFB7>ģ<EFBFBD><C4A3>ǿ<EFBFBD><C7BF>Ե<EFBFBD><D4B5><EFBFBD><EFBFBD><EFBFBD>״̬<D7B4><CCAC><EFBFBD><EFBFBD><EFBFBD><EFBFBD>ֻ<EFBFBD><D6BB><EFBFBD><EFBFBD>һ<EFBFBD>ַ<EFBFBD><D6B7><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>ǲ<EFBFBD><C7B2><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>ֹ<EFBFBD><D6B9><EFBFBD><EFBFBD><EFBFBD>һ<EFBFBD>Ų<EFBFBD>Ҫ
+	for (int i = 1; i <= n; i++)
+		for (int j = 1; j <= mx; j++)
+			if (j >= a[i] && f[j - a[i]])//<2F><><EFBFBD><EFBFBD><EFBFBD><EFBFBD>װ<EFBFBD>£<EFBFBD><C2A3><EFBFBD><EFBFBD><EFBFBD>ǰ<EFBFBD><C7B0><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>ɴ<C9B4><EFA3AC><EFBFBD><EFBFBD>ǰ<EFBFBD><C7B0><EFBFBD>Լ<EFBFBD><D4BC><EFBFBD><EFBFBD>ƽ<EFBFBD>
+				f[j]++;
+
+	//ö<><C3B6>ÿ<EFBFBD><C3BF><EFBFBD><EFBFBD><EFBFBD>ܵ<EFBFBD><DCB5><EFBFBD><EFBFBD>֣<EFBFBD><D6A3><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>ִ<EFBFBD><D6B4><EFBFBD>1<EFBFBD><31><EFBFBD><EFBFBD><EFBFBD>ϵķ<CFB5><C4B7><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>ǿ<EFBFBD><C7BF>Դճ<D4B4><D5B3><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>޷<EFBFBD><DEB7>ճ<EFBFBD>
+	for (int i = 1; i <= mx; i++)
+		if (!f[i])
+			ans++;
+
+	//<2F><><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>
+	cout << ans << endl;
+	return 0;
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/MaiTangYuan_DFS.cpp
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/MaiTangYuan_DFS.cpp
@@ -0,0 +1,73 @@
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 30;
+int a[N];
+int n;
+//<2F><><EFBFBD>մ<EFBFBD><D5B4><EFBFBD>
+int res;
+
+//<2F><><EFBFBD><EFBFBD>
+bool isHz(int a, int b) {
+	for (int i = 2; i <= min(a, b); i++) //a,b<><62><EFBFBD>ܴ<EFBFBD><DCB4>ڵĹ<DAB5><C4B9><EFBFBD><EFBFBD><EFBFBD>
+		if (a % i == 0 && b % i == 0)
+			return false;
+	return true;
+}
+
+bool found = false; //<2F><><EFBFBD><EFBFBD><EFBFBD>ǲ<EFBFBD><C7B2><EFBFBD><EFBFBD>ܱ<EFBFBD>ʾ<EFBFBD><CABE><EFBFBD><EFBFBD> found=true<75>ǿ<EFBFBD><C7BF>Ա<EFBFBD><D4B1>found=false<73>ǲ<EFBFBD><C7B2>ܱ<EFBFBD><DCB1><EFBFBD>
+
+//<2F>ܱ<EFBFBD>ʾ
+void dfs(int sum, int u) { //<2F><>a[]<5D><><EFBFBD><EFBFBD><EFBFBD><EFBFBD>Ԫ<EFBFBD><D4AA><EFBFBD><EFBFBD><EFBFBD>ϣ<EFBFBD><CFA3>ܲ<EFBFBD><DCB2>ܱ<EFBFBD>ʾ<EFBFBD><CABE>x<EFBFBD><78><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>֣<EFBFBD><D6A3><EFBFBD><EFBFBD><EFBFBD><EFBFBD>ܷ<EFBFBD><DCB7><EFBFBD>true,<2C><><EFBFBD>򷵻<EFBFBD>false
+	if (u == n + 1) {
+		if (sum == 0)
+			found = true;
+		return;
+	}
+
+	//<2F><><EFBFBD>Ե<EFBFBD>ǰa[u]ѡ<><D1A1>i<EFBFBD><69>
+	for (int i = 0; sum >= i * a[u]; i++) //i<><69>ö<EFBFBD>ٵ<EFBFBD>a[u]<5D><><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>ĸ<EFBFBD><C4B8><EFBFBD>
+		dfs(sum - i * a[u], u + 1);
+
+}
+
+
+int main() {
+
+	cin >> n;
+	for (int i = 1; i <= n; i++)
+		cin >> a[i];
+
+	//<2F><>һ<EFBFBD><D2BB><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>С<EFBFBD><D0A1><EFBFBD><EFBFBD>
+	sort(a + 1, a + 1 + n);
+
+	//a<><61><EFBFBD><EFBFBD><EFBFBD>У<EFBFBD><D0A3>ǲ<EFBFBD><C7B2>Ǵ<EFBFBD><C7B4><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>ʵ<EFBFBD><CAB5><EFBFBD>
+	bool flag = false;
+
+
+	int mx;//<2F><><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD><EFBFBD>
+	for (int i = 1; i <= n; i++) {
+		for (int j = i + 1; j <= n; j++)
+			if (isHz(a[i], a[j])) {
+				flag = true;
+				mx = a[i] * a[j] - a[i] - a[j];
+				break;
+			}
+		if (flag)
+			break;
+	}
+
+
+	if (flag) {
+		for (int i = 1; i <= mx; i++) {
+			//һ<><D2BB>Ҫ<EFBFBD>ǵ<EFBFBD><C7B5><EFBFBD><EFBFBD><EFBFBD>
+			found = false;
+			dfs(i, 1);
+			if (!found)
+				res++;
+		}
+		cout << res << endl;
+	} else
+		cout << -1 << endl;
+
+	return 0;
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/TD.cpp
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/TD.cpp
@@ -0,0 +1,19 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+int cnt;
+const int N = 30000;
+int f[N];
+int main() {
+    int n = 3000, m = 3000;
+    for (int x = 1; x <= n; x++)
+        for (int y = 1; y <= m; y++)
+            for (int len = 1;; len++) {
+                int tx = x + len - 1, ty = y + len - 1;
+                if (tx > n || ty > m) break;
+                f[len]++;
+            }
+            for (int i = 1; i <= min(n,m); i++)
+        printf("%d ", f[i]);
+    return 0;
+}
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/今日课程内容2023-02-22.md
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/今日课程内容2023-02-22.md
@@ -0,0 +1,76 @@
+## 1、激光炸弹
+
+#### 录入训练
+https://www.acwing.com/problem/content/101/
+
+#### 温故知新
+- 点坐标如何修改为数组方格坐标
+- 二维前缀和
+- 不能开两个数组，因为每个数组是$5000x5000$这么大，就是$5000x5000x4bytes，5000x5000x4/1024/1024=95MB$，这题的空间是$168MB$，所以两个数组会$MLE$。
+一个数组就行了，在自己身上求前缀和。
+- $xy$坐标是从$0$开始的，二维前缀和需要坐标从$(1,1)$开始，需要$x++,y++$
+- 炸弹的范围会超过$r>M$，会导致计算二维前缀和的两层循环失效，造成结果错误。正确的作法是时刻小心观察，循环什么情况下不会执行，然后加上特判。
+- 只枚举右下角即可完成所有边长为$r$的正方形枚举
+
+### 蓝桥杯第13届省赛中级组第6题
+
+![](http://dsideal.obs.cn-north-1.myhuaweicloud.com/HuangHai/BlogImages/2023/02/b15ecc654c62371bfd8433381b1a3110.png)
+
+![](http://dsideal.obs.cn-north-1.myhuaweicloud.com/HuangHai/BlogImages/2023/02/7d8e8d0f52943660628d42fa572d70ea.png)
+
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+/*
+测试用例I：
+5 6
+1 0 1 0 0 0
+0 1 0 1 0 0
+1 0 0 0 1 0
+0 1 0 0 0 1
+1 0 1 0 0 0
+
+答案
+4
+
+测试用例II：
+5 6
+1 0 1 0 0 0
+0 1 0 1 1 0
+1 0 0 0 1 0
+0 1 0 0 0 1
+1 0 1 0 0 0
+
+答案
+2
+*/
+using namespace std;
+const int INF = 0x3f3f3f3f;
+const int N = 110;
+int a[N][N];
+int s[N][N];
+int res;
+
+int main() {
+    int n, m;
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++) {
+            cin >> a[i][j];
+            s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + a[i][j];
+        }
+
+    for (int i = 1; i <= n; i++)       // 100
+        for (int j = 1; j <= m; j++) { // 100
+            for (int k = 0;; k++) {
+                if (i + k > n || j + k > m) break; // 越界
+                if (a[i + k][j + k] == 0) break;
+                int x1 = i, y1 = j, x2 = i + k, y2 = j + k;
+                int cnt = s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1];
+                if (cnt != k + 1) break;
+                res = max(res, cnt);
+            }
+        }
+    printf("%d\n", res);
+    return 0;
+}
+```
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/1.1.png
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/1.1.png
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/1.2.png
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/1.2.png
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/2.1.png
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/2.1.png
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/2.2.png
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/2.2.png
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/3.1.png
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/3.1.png
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/3.2.png
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/3.2.png
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/4.1.png
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/4.1.png
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/4.2.png
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/4.2.png
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/4.3.png
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/4.3.png
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/5.1.png
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/5.1.png
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/5.2.png
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/5.2.png
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/5.3.png
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/5.3.png
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/6.1.png
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/6.1.png
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/6.2.png
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/编程题/6.2.png
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/裴蜀(贝祖)定理及其证明.md
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/裴蜀(贝祖)定理及其证明.md
@@ -0,0 +1,25 @@
+### 裴蜀定理(贝祖定理)
+
+**定理**
+$ax+by=c,x∈Z^+,y∈Z^+$成立的<font color='red'>充要条件</font>是$gcd(a,b)|c$。
+ 
+**证明**
+
+* 充分性
+设$s=gcd(a,b)$，显然$s|a$，并且$s|b$
+因为$x,y∈Z^+$
+所以$s|ax,s|by$
+显然要使得之前的式子成立，则必须满足$c$是$a$和$b$的公约数的倍数。
+    
+* 必要性
+又因为$x$和$y$是正整数
+所以$c$必然是$a,b$最大公约数的倍数。
+
+因此，证得该定理成立
+
+ 
+
+说明
+该定理完全可以推广到若干数的线性组合
+
+例题的话，请看这里：[$Luogu$ $4549$ 裴蜀定理/Min](https://www.luogu.com.cn/problem/P4549)
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/选择题/1.jpg
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/选择题/1.jpg
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/选择题/2.jpg
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/选择题/2.jpg
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/选择题/3.jpg
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/选择题/3.jpg
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/选择题/4.jpg
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/选择题/4.jpg
--- a/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/选择题/5.jpg
+++ b/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/选择题/5.jpg