'commit'

TangDou/KaoShi/.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

TangDou/KaoShi/.vscode/c_cpp_properties.conflict.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
+}

TangDou/KaoShi/.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
+}

TangDou/KaoShi/.vscode/extensions.json

@@ -0,0 +1,5 @@
+{
+  "recommendations": [
+    "MS-CEINTL.vscode-language-pack-zh-hans"
+  ]
+}

TangDou/KaoShi/.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
+    }
+  ]
+}

TangDou/KaoShi/.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）
+}

TangDou/KaoShi/.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
+      },
+    }
+  ]
+}

TangDou/KaoShi/01背包.cpp

@@ -0,0 +1,24 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 1010;
+
+int n, m;
+int v[N], w[N];
+int f[N];
+
+int main() {
+    //优化输入
+    ios::sync_with_stdio(false);
+    //物品个数n
+    cin >> n >> m;
+    //读入体积和重量
+    for (int i = 1; i <= n; i++) cin >> v[i] >> w[i];
+    // 01背包模板
+    for (int i = 1; i <= n; i++)
+        for (int j = m; j >= v[i]; j--)
+            f[j] = max(f[j], f[j - v[i]] + w[i]);
+
+    cout << f[m] << endl;
+    return 0;
+}

TangDou/KaoShi/24点.cpp

@@ -0,0 +1,69 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+/*
+10 2 4 8
+答案：2
+*/
+// 24点
+int main() {
+    int a, b, c, d;
+    cin >> a >> b >> c >> d;
+    int res = 0, x;
+    x = a + b + c + d;
+    if (x == 24) res++;
+    x = a + b + c - d;
+    if (x == 24) res++;
+    x = a + b + c * d;
+    if (x == 24) res++;
+    x = a + b - c + d;
+    if (x == 24) res++;
+    x = a + b - c - d;
+    if (x == 24) res++;
+    x = a + b - c * d;
+    if (x == 24) res++;
+    x = a + b * c + d;
+    if (x == 24) res++;
+    x = a + b * c - d;
+    if (x == 24) res++;
+    x = a + b * c * d;
+    if (x == 24) res++;
+    x = a - b + c + d;
+    if (x == 24) res++;
+    x = a - b + c - d;
+    if (x == 24) res++;
+    x = a - b + c * d;
+    if (x == 24) res++;
+    x = a - b - c + d;
+    if (x == 24) res++;
+    x = a - b - c - d;
+    if (x == 24) res++;
+    x = a - b - c * d;
+    if (x == 24) res++;
+    x = a - b * c + d;
+    if (x == 24) res++;
+    x = a - b * c - d;
+    if (x == 24) res++;
+    x = a - b * c * d;
+    if (x == 24) res++;
+    x = a * b + c + d;
+    if (x == 24) res++;
+    x = a * b + c - d;
+    if (x == 24) res++;
+    x = a * b + c * d;
+    if (x == 24) res++;
+    x = a * b - c + d;
+    if (x == 24) res++;
+    x = a * b - c - d;
+    if (x == 24) res++;
+    x = a * b - c * d;
+    if (x == 24) res++;
+    x = a * b * c + d;
+    if (x == 24) res++;
+    x = a * b * c - d;
+    if (x == 24) res++;
+    x = a * b * c * d;
+    if (x == 24) res++;
+    cout << res;
+    return 0;
+}

TangDou/KaoShi/3056.in

@@ -0,0 +1,4 @@
+2 2
+1001 3357 
+D 0
+Z 4388 852

TangDou/KaoShi/BABA_2.cpp

@@ -0,0 +1,11 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    int cnt = 0;
+    for (int i = 1; i < 100; i += 5)
+        cnt += i;
+
+    cout << cnt << endl;
+    return 0;
+}

TangDou/KaoShi/BABA_3.cpp

@@ -0,0 +1,47 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int INF=0x3f3f3f3f;
+const int N=1510;
+int n;
+int h[N],e[N],ne[N],idx;
+int w[N];
+int f[N][3];
+int in[N];
+void add(int a,int b){
+    e[idx]=b,ne[idx]=h[a],h[a]=idx++;
+}
+void dfs(int u){
+    f[u][0]=0;
+    f[u][1]=INF ;
+    f[u][2]=w[u];
+
+    for(int i=h[u];~i;i=ne[i]){
+        int v=e[i];
+        dfs(v);
+        f[u][0]+=min(f[v][1],f[v][2]);
+        f[u][2]+=min({f[v][0],f[v][1],f[v][2]});
+    }
+    for(int i=h[u];~i;i=ne[i]){
+        int v=e[i];
+        f[u][1]=min(f[u][1],f[v][2]+f[u][0]-min(f[v][1],f[v][2]));
+    }
+} 
+
+int main() {
+    memset(h,-1,sizeof h);
+    cin>>n;
+    for (int i = 1; i <= n; i++){
+        int x,m;
+        cin>>x>>w[x]>>m;
+        while(m--){
+            int y;
+            cin>>y;
+        }
+    }
+    int root=1;
+    while(in[root])root++;
+    dfs(root);
+    printf("%d\n",min(f[root][1],f[root][2]));
+    return 0;
+}

TangDou/KaoShi/BABA_4.cpp

@@ -0,0 +1,25 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N=1010;
+int a[N][N];
+
+int main() {
+int n,m,q;
+cin>>n>>m;
+for (int i = 1; i <= n; i++){
+    for (int j = 1; j <= m; j++){
+        int x;
+        cin>>x;
+        a[i][j]=a[i-1][j]+a[i][j-1]-a[i-1][j-1]+x;
+        
+    }
+} 
+cin>>q;
+while(q--){
+    int x1,x2,y1,y2;
+    cin>>x1>>x2>>y1>>y2;
+    printf("%d\n",a[x2][y2]-a[x1-1][y2]-a[x2][y1-1]+a[x1-1][y1-1]);
+}   
+    return 0;
+}

TangDou/KaoShi/BABA_5.cpp

@@ -0,0 +1,32 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    int n;
+    cin >> n;
+
+    vector<int> count(81, 0);
+    for (int i = 0; i < n; ++i) {
+        int num;
+        cin >> num;
+        count[num]++;
+    }
+
+    for (int i = 1; i <= 80; ++i) {
+        if (count[i] >= 2) {
+            count[i + 1] += count[i] / 2;
+            count[i] %= 2;
+        }
+    }
+
+    int max_num = 0;
+    for (int i = 1; i <= 80; ++i) {
+        if (count[i] > 0) {
+            max_num = i;
+        }
+    }
+
+    cout << max_num << endl;
+
+    return 0;
+}

TangDou/KaoShi/TangDou_1.cpp

@@ -0,0 +1,69 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+/*
+2015 1 4
+*/
+int months_1[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
+int months_2[13] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
+
+bool LeapYear(int year) {
+    return ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0);
+}
+/*
+2023 1 1 --> 7
+2023 7 1 --> 6
+
+2023 7 7
+2023 5 7 --->1?
+*/
+int main() {
+    // 1、是不是需要判断闰年？
+    int year, month;
+    int ydxq, month_1_xq;
+    // 2、year年1月1日 是星期xq
+    cin >> year >> month >> ydxq;
+    cout << "S   M   T   W   T   F   S" << endl;
+    int days = 0;
+    if (LeapYear(year)) {
+        // 比如month=5,要的是5月1号的，应该是1~4月份相加的和
+        for (int i = 1; i < month; i++) days += months_2[i];
+    } else {
+        for (int i = 1; i < month; i++) days += months_1[i];
+    }
+
+    // days含义：days就是当前年份的第几天-1
+    month_1_xq = (ydxq + days) % 7;
+
+    // 是指定月份的第1天是星期几0~6
+    // cout << month_1_xq << endl;
+
+    // 打空格
+    for (int i = 1; i <= month_1_xq; i++) cout << "    ";
+    // 打第一行的剩余日期
+    // 1~ 6
+    // 2~ 5
+    // 3~ 4
+    // 4~ 3
+    // 5~ 2
+    // 6~ 1
+    // 0~ 7
+    for (int i = 1; i <= 7 - month_1_xq; i++) printf("%-4d", i);
+    cout << endl;
+
+    int this_month_days = 0;
+    if (LeapYear(year)) {
+        this_month_days = months_2[month];
+    } else {
+        this_month_days = months_1[month];
+    }
+
+    int cnt = 0;
+    for (int i = 7 - month_1_xq + 1; i <= this_month_days; i++) {
+        cnt++;
+        printf("%-4d", i);
+        if (cnt % 7 == 0) cout << endl;
+    }
+
+    return 0;
+}

TangDou/KaoShi/TangDou_2.cpp

@@ -0,0 +1,25 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 110;
+int x[N], y[N];
+int n;
+int maxv, maxp;
+int main() {
+    cin >> n;
+    for (int i = 1; i <= n; i++) cin >> x[i] >> y[i];
+    for (int i = 1; i <= n; i++){
+        int cnt = 0;
+        for (int j = 1; j <= n; j++)
+            if (x[i] > x[j] && y[i] > y[j]) cnt++;
+        if(cnt>=maxv){
+            maxv = cnt;
+            maxp = i;
+        }
+        cout << cnt;
+    
+    }
+    cout << endl;
+    cout << maxp << endl;
+    return 0;
+}

TangDou/KaoShi/TangDou_3.cpp

@@ -0,0 +1,22 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+int n;
+
+int main() {
+    cin >> n;
+    while (1) {
+        if (n == 1) break;
+        if (n % 2 == 0) {
+            n /= 2;
+            printf("%d\n", n);
+        }
+
+        else if (n % 2 == 1) {
+            n = n * 3 + 1;
+            printf("%d\n", n);
+        }
+    }
+
+    return 0;
+}

TangDou/KaoShi/TangDou_5.cpp

@@ -0,0 +1,35 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+// 给出一个数字，比如(11)，将其转为2进制，计算出它的十进制是多少,答案应该是3
+//1234  
+int convert(string s, int k) {
+    int res = 0;
+    for (int i = 0; i < s.size(); i++) {
+        int x = s[i] - '0';
+        if (x >= k) return -1;
+        res += x * pow(k, s.size() - 1 - i); // 105
+    }
+    return res;
+}
+
+int main() {
+    string a, b, c;
+    cin >> a >> b >> c;
+
+    bool flag = false;
+    for (int k = 2; k <= 16; k++) {
+        int x = convert(a, k);
+        int y = convert(b, k);
+        int z = convert(c, k);
+        if (x * y == z) {
+            cout << k << endl;
+            flag = true;
+            break;
+        }
+    }
+    if (!flag) cout << 0 << endl;
+
+    return 0;
+}

TangDou/KaoShi/Test.cpp

@@ -0,0 +1,23 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+int n = 9;
+const int INF = 0x3f3f3f3f;
+int res = INF, a1, b1, c1;
+int main() {
+    int a, b, c;
+    for (int a = 1; a * a * a <= n; a++) {
+        for (int b = a; a * b * b <= n; b++) {
+            int c = n / a / b;
+            if(a*b*c==n){
+                if(2*c-2*a<res){
+                    res = 2 * c - 2 * a;
+                    a1 = a, b1 = b, c1 = c;
+                }
+            }
+        }
+    }
+    int ans = (a1 * b1 + a1 * c1 + b1 * c1) * 2;
+    cout << ans << endl;
+    return 0;
+}

TangDou/KaoShi/买瓜子.cpp

@@ -0,0 +1,27 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+const int N = 1010;
+int n, m;
+int v[N], w[N];
+int f[N];
+/**
+ 测试用例:
+ 80 2
+ 18 10
+ 30 20
+ */
+
+//完全背包问题
+int main() {
+    cin >> m >> n;
+    for (int i = 1; i <= n; i++) cin >> v[i] >> w[i];
+
+    for (int i = 1; i <= n; i++)
+        for (int j = v[i]; j <= m; j++)
+            f[j] = max(f[j], f[j - v[i]] + w[i]);
+
+    cout << f[m] << endl;
+    return 0;
+}

TangDou/KaoShi/倒序输出字符串.cpp

@@ -0,0 +1,20 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+int main() {
+    string a;
+
+    //问题1：错误的使用了getline
+    //getline(cin,a);
+    //完全没有必要使用getline,因为题目没有可能包含空格，使用cin即可
+
+    //问题2：错误的使用for循环重复读取a,因为a是string,是一次性cin读取。
+
+    //问题3：a的界限是 a.size()-1 ,0 而不是a.size(),1
+
+    cin >> a;
+    for (int i = a.size() - 1; i >= 0; i--) printf("%c", a[i]);
+
+    return 0;
+}

TangDou/KaoShi/分口罩.cpp

@@ -0,0 +1,33 @@
+#include <iostream>
+
+using namespace std;
+const int N = 15;
+const int INF = 0x3f3f3f3f;
+int a[N];
+int n;
+int cha = INF;
+int res1, res2;
+/*
+3
+200 300 110
+*/
+void dfs(int u, int s1, int s2) {
+    if (u == n) {
+        if (s1 < s2) swap(s1, s2);
+        if (s1 - s2 < cha) {
+            res1 = s1, res2 = s2;
+            cha = s1 - s2;
+        }
+        return;
+    }
+    dfs(u + 1, s1 + a[u], s2);
+    dfs(u + 1, s1, s2 + a[u]);
+}
+
+int main() {
+    cin >> n;
+    for (int i = 0; i < n; i++) cin >> a[i];
+    dfs(0, 0, 0);
+    printf("%d %d\n", res1, res2);
+    return 0;
+}

TangDou/KaoShi/分组背包.cpp

@@ -0,0 +1,33 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 110;
+
+int n;       // n类物品
+int m;       //背包上限为m
+int v[N][N]; //价值
+int w[N][N]; //体积
+int s[N];    //个数
+int f[N];    // dp数组，最大值
+
+//分组背包
+int main() {
+    //优化输入
+    ios::sync_with_stdio(false);
+    cin >> n >> m;
+
+    for (int i = 1; i <= n; i++) {      //枚举每类物品,比如：水果，蔬菜，肉类...
+        cin >> s[i];                    //此类物品中物品的个数,比如2，可能是苹果、香蕉
+        for (int j = 1; j <= s[i]; j++) //读入苹果、香蕉的价值和体积
+            cin >> v[i][j] >> w[i][j];
+    }
+
+    for (int i = 1; i <= n; i++)            //遍历每类物品
+        for (int j = m; j >= 0; j--)        //遍历每个可能的体积
+            for (int k = 1; k <= s[i]; k++) //遍历第k个物品
+                if (v[i][k] <= j)
+                    f[j] = max(f[j], f[j - v[i][k]] + w[i][k]);
+
+    cout << f[m] << endl;
+    return 0;
+}

TangDou/KaoShi/分解质因数.cpp

@@ -0,0 +1,31 @@
+#include <iostream>
+#include <cmath>
+
+using namespace std;
+// 欧拉筛
+const int N = 1e5 + 10;
+int primes[N], cnt; // primes[]存储所有素数
+bool st[N];         // st[x]存储x是否被筛掉
+void get_primes(int n) {
+    for (int i = 2; i <= n; i++) {
+        if (!st[i]) primes[cnt++] = i;
+        for (int j = 0; primes[j] * i <= n; j++) {
+            st[primes[j] * i] = true;
+            if (i % primes[j] == 0) break;
+        }
+    }
+}
+int main() {
+    int n;
+    cin >> n;
+    get_primes(sqrt(n));
+    for (int i = 0; i < cnt; i++) {
+        while (n % primes[i] == 0) {
+            cout << primes[i] << endl;
+            n /= primes[i];
+        }
+    }
+    if (n > 1) cout << n << endl;
+
+    return 0;
+}

TangDou/KaoShi/判断年份重叠.cpp

@@ -0,0 +1,13 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 10;
+int main() {
+    string s;
+    cin >> s;
+    if (s[0] == s[3] && s[1] == s[2])
+        cout << "YES";
+    else
+        cout << "NO";
+    return 0;
+}

TangDou/KaoShi/剪绳子.cpp

@@ -0,0 +1,13 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 30;
+int a[N] = {2};
+
+int main() {
+    int n;
+    cin >> n;
+    for (int i = 1; i <= n; i++) a[i] = a[i - 1] + pow(2, i - 1);
+    printf("%d", a[n]);
+    return 0;
+}

TangDou/KaoShi/加密.cpp

@@ -0,0 +1,10 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+int main() {
+    string a;
+    cin >> a;
+    for (int i = 0; i < a.size(); i++) cout << char(a[i] + 3);
+    return 0;
+}

TangDou/KaoShi/区间因数.cpp

@@ -0,0 +1,39 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+const int N = 1e5 + 10;
+
+int primes[N], cnt; // primes[]存储所有素数
+bool st[N];         // st[x]存储x是否被筛掉
+
+int factor[N]; // 存储每一个数的质因子个数
+
+void get_primes(int n) {
+    memset(st, 0, sizeof st);
+    for (int i = 2; i <= n; i++) {
+        if (!st[i]) {
+            primes[cnt++] = i;
+            factor[i] = 1; // 质数的质因子个数是1
+        }
+        for (int j = 0; primes[j] * i <= n; j++) {
+            st[primes[j] * i] = true;              // 标识primes[j] * i是个合数
+            factor[primes[j] * i] = factor[i] + 1; // t=primes[j]*i,那么t的质因子个数=i的质因子个数+1
+            if (i % primes[j] == 0) break;         // 只被最小的质数因子筛掉
+        }
+    }
+}
+
+int main() {
+    int n, m;
+    cin >> n >> m;
+
+    get_primes(m);
+
+    int res = 0;
+    for (int i = n; i <= m; i++) {
+        res = max(res, factor[i]);
+        // printf("The count of prime factor(s) of %d is: %d\n", i, factor[i]);
+    }
+    cout << res << endl;
+    return 0;
+}

TangDou/KaoShi/合并麦子.cpp

@@ -0,0 +1,37 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+// OJ地址
+// https://www.acwing.com/problem/content/150/
+
+// 黄海题解
+// https://www.cnblogs.com/littlehb/p/15488791.html
+
+//小顶堆
+priority_queue<int, vector<int>, greater<int>> q;
+
+int res;
+
+int main() {
+    int n;
+    cin >> n;
+
+    while (n--) {
+        int x;
+        cin >> x;
+        q.push(x);
+    }
+
+    while (q.size() > 1) {
+        int a = q.top();
+        q.pop();
+        int b = q.top();
+        q.pop();
+        res += a + b;
+        q.push(a + b);
+    }
+
+    printf("%d\n", res);
+    return 0;
+}

TangDou/KaoShi/合数的和.cpp

@@ -0,0 +1,18 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+bool isPrime(int n) {
+    for (int i = 2; i <= n / i; i++)
+        if (n % i == 0) return false;
+    return true;
+}
+
+int main() {
+    int n, s = 0;
+    cin >> n;
+    for (int i = 4; i <= n; i++)
+        if (!isPrime(i))s += i;
+    cout << s << endl;
+    return 0;
+}

TangDou/KaoShi/合数的因子个数.cpp

@@ -0,0 +1,21 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+int n;
+bool check(int x) {
+    int cnt = 0;
+    for (int i = 1; i <= x; i++)
+        if (x % i == 0) cnt++;
+    return cnt == n;
+}
+
+int main() {
+    cin >> n;
+    for (int i = 3; i <= 100; i++) {
+        if (check(i)) {
+            printf("%d", i);
+            return 0;
+        }
+    }
+    return 0;
+}

TangDou/KaoShi/埃拉筛.cpp

@@ -0,0 +1,25 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int INF = 0x3f3f3f3f;
+const int N = 1e5 + 10;
+int primes[N], cnt; // primes[]存储所有素数
+bool st[N];         // st[x]存储x是否被筛掉
+//埃拉筛
+void get_primes(int n) {
+    for (int i = 2; i <= n; i++)
+        if (!st[i]) {
+            primes[cnt++] = i;                  //记录素数
+            for (int j = 2 * i; j <= n; j += i) //成倍数的标识
+                st[j] = true;
+        }
+}
+/*
+ get_primes(200);
+    for (int i = 0; i < cnt; i++)
+        if (primes[i] > 100)
+cout << primes[i] << " ";
+*/
+int main() {
+    return 0;
+}

TangDou/KaoShi/多重背包.cpp

@@ -0,0 +1,28 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+//多重背包问题，每件物品的数量是有限制的，不是无穷，也不是只有1个。
+// 状态表示  集合，属性
+// 状态计算
+
+const int N = 110;
+
+int n, m;
+int v[N], w[N], s[N];
+int f[N];
+
+int main() {
+    //优化输入
+    ios::sync_with_stdio(false);
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++) cin >> v[i] >> w[i] >> s[i];
+
+    for (int i = 1; i <= n; i++)
+        for (int j = m; j >= 0; j--)
+            for (int k = 0; k <= s[i] && k * v[i] <= j; k++) //穷举所有可能性，尝试拿到最大的合理值
+                f[j] = max(f[j], f[j - v[i] * k] + w[i] * k);
+
+    cout << f[m] << endl;
+    return 0;
+}

TangDou/KaoShi/多重背包_二进制优化.cpp

@@ -0,0 +1,44 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 12010, M = 2010;
+
+int n, m;
+int v[N], w[N];
+int f[M];
+
+//多重背包的二进制优化
+int main() {
+    //优化输入
+    ios::sync_with_stdio(false);
+    cin >> n >> m;
+
+    int cnt = 0;
+    for (int i = 1; i <= n; i++) {
+        int a, b, s;
+        cin >> a >> b >> s;
+        //二进制优化,能打包则打包之，1,2,4,8,16,...
+        int k = 1;
+        while (k <= s) {
+            cnt++;
+            v[cnt] = a * k;
+            w[cnt] = b * k;
+            s -= k;
+            k *= 2;
+        }
+        //剩下的
+        if (s > 0) {
+            cnt++;
+            v[cnt] = a * s;
+            w[cnt] = b * s;
+        }
+    }
+    n = cnt; //数量减少啦
+    // 01背包
+    for (int i = 1; i <= n; i++)
+        for (int j = m; j >= v[i]; j--)
+            f[j] = max(f[j], f[j - v[i]] + w[i]);
+
+    cout << f[m] << endl;
+    return 0;
+}

TangDou/KaoShi/大小写转换.cpp

@@ -0,0 +1,14 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+int main() {
+    string s;
+    cin >> s;
+    for (int i = s.size() - 1; i >= 0; i--)
+        if (s[i] >= 'A' && s[i] <= 'Z')
+            cout << s[i] + 32;
+        else
+            cout << s[i];
+    return 0;
+}

TangDou/KaoShi/字符串对比.cpp

@@ -0,0 +1,26 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+string a, b;
+
+bool check(string a, string b) {
+    for (int i = 0; i < a.size(); i++) {
+        if (a[i] >= 'A' && a[i] <= 'Z') a[i] = a[i] + 32;
+        if (b[i] >= 'A' && b[i] <= 'Z') b[i] = b[i] + 32;
+        if (a[i] != b[i]) return false;
+    }
+    return true;
+}
+
+int main() {
+    cin >> a >> b;
+    if (a.size() != b.size())
+        cout << 1 << endl;
+    else if (a == b)
+        cout << 2 << endl;
+    else if (a.size() == b.size() && check(a, b))
+        cout << 3 << endl;
+    else
+        cout << 4 << endl;
+    return 0;
+}

TangDou/KaoShi/完全数.cpp

@@ -0,0 +1,16 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+//完数
+int isWMS(int n) {
+    int s = 0;
+    for (int i = 1; i <= n / 2; i++)
+        if (n % i == 0) s += i;
+    return s == n ? 1 : 0;
+}
+int main() {
+    for (int i = 1; i <= 1000; i++)
+        if (isWMS(i)) cout << i << " ";
+    return 0;
+}

TangDou/KaoShi/完全背包.cpp

@@ -0,0 +1,25 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+const int N = 1010;
+
+int n, m;
+int v[N];
+int w[N];
+int f[N];
+
+// 完全背包问题
+int main() {
+    //优化输入
+    ios::sync_with_stdio(false);
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++) cin >> v[i] >> w[i];
+
+    for (int i = 1; i <= n; i++)
+        for (int j = v[i]; j <= m; j++) //一个一个加上来，求一个最大值
+            f[j] = max(f[j], f[j - v[i]] + w[i]);
+
+    cout << f[m] << endl;
+    return 0;
+}

TangDou/KaoShi/小蓝突发奇想1.cpp

@@ -0,0 +1,34 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 110;
+int a[N];
+int n, m;
+int cnt;
+
+/**
+  测试数据：
+  5 1
+  答案：
+  3
+ */
+void dfs(int u, int s1, int s2) {
+    //收集答案
+    if (u == n + 1) {
+        if (s1 - s2 == m) cnt++;
+        return;
+    }
+    //选择了当前的数字
+    dfs(u + 1, s1 + a[u], s2);
+
+    //放弃了当前的数字
+    dfs(u + 1, s1, s2 + a[u]);
+}
+
+int main() {
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++) a[i] = i;
+    dfs(1, 0, 0);
+    cout << cnt << endl;
+    return 0;
+}

TangDou/KaoShi/小蓝突发奇想2.cpp

@@ -0,0 +1,38 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+int n, m;
+int cnt;
+vector<int> p1, p2;
+/**
+  测试数据：
+  5 1
+  答案：
+  3
+ */
+void dfs(int u) {
+    //收集答案
+    if (u == n + 1) {
+        int s1 = 0, s2 = 0;
+        for (int i = 0; i < p1.size(); i++)
+            s1 += p1[i];
+        for (int i = 0; i < p2.size(); i++)
+            s2 += p2[i];
+        if (s1 - s2 == m)
+            cnt++;
+        return;
+    }
+    p1.push_back(u);
+    dfs(u + 1);
+    p1.pop_back();
+
+    p2.push_back(u);
+    dfs(u + 1);
+    p2.pop_back();
+}
+int main() {
+    cin >> n >> m;
+    dfs(1);
+    cout << cnt << endl;
+    return 0;
+}

TangDou/KaoShi/开关门.cpp

@@ -0,0 +1,18 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 1010;
+int a[N];
+int n, cnt;
+
+int main() {
+    cin >> n;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= 2 * n; j++)
+            if (j % i == 0)a[j] = !a[j];
+
+    for (int i = 1; i <= 2 * n; ++i)
+        if (a[i]) cnt++;
+    printf("%d", cnt);
+    return 0;
+}

TangDou/KaoShi/打印Y.cpp

@@ -0,0 +1,21 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+int main() {
+    int n;
+    cin >> n;
+    for (int i = 1; i < n; i++) {
+        for (int j = 1; j < 2 * n; j++) {
+            if (i == j || i + j == 2 * n)
+                cout << "*";
+            else
+                cout << " ";
+        }
+        cout << endl;
+    }
+    for (int i = 1; i <= n; i++) {
+        for (int j = 1; j < n; j++) cout << " ";
+        cout << "*" << endl;
+    }
+    return 0;
+}

TangDou/KaoShi/找位置.cpp

@@ -0,0 +1,26 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 2000;
+/*
+4
+-1 3 5 3
+3
+*/
+int a[N];
+int k, n, x;
+int main() {
+    cin >> n;
+    for (int i = 1; i <= n; i++) cin >> a[i];
+    cin >> x;
+    for (int i = 1; i <= n; i++) {
+        if (a[i] == x) {
+            k = i;
+            printf("%d", k);
+            break;
+        }
+    }
+    if (k == 0)
+        printf("-1");
+    return 0;
+}

TangDou/KaoShi/找字符串.cpp

@@ -0,0 +1,35 @@
+#include<iostream>
+
+using namespace std;
+string a, b;
+int cnt, p = -1;
+
+/**
+ * to tomato不能匹配！
+ * @return
+ */
+int main() {
+    //读入带空格的字符串
+    getline(cin, a);
+    getline(cin, b);
+    //a字符串转小写
+    a = ' ' + a + ' ';
+    for (int i = 0; i < a.size(); i++)
+        if (a[i] >= 'A' && a[i] <= 'Z') a[i] = a[i] + 32;
+    //b字符串转小写
+    b = ' ' + b+' ';
+    for (int i = 0; i < b.size(); i++)
+        if (b[i] >= 'A' && b[i] <= 'Z') b[i] = b[i] + 32;
+
+    //在b字符串查找
+    for (int i = 0; i < b.size(); i++)
+        if (b.substr(i, a.size()) == a) {
+            cnt++;
+            //记录第一个位置
+            if (p == -1)p = i;
+        }
+    //输出
+    if (cnt > 0) cout << cnt << " " << p << endl;
+    else cout << -1 << endl;
+    return 0;
+}

TangDou/KaoShi/投篮游戏I.cpp

@@ -0,0 +1,50 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 1010;
+int a[N];
+int n;
+//最大值
+int Max;
+
+/**
+ 测试用例：
+ 3
+ 130 200 55
+
+ 最多投进球的个数
+ 答案：2
+ */
+
+/**
+ * 功能：深度优先搜索
+ * @param step      选择了哪个篮筐
+ * @param last      前面最后一个数字是多少
+ * @param cnt       已经投中了多少个球
+ */
+void dfs(int u, int last, int cnt) {
+    //如果越界
+    if (u == n + 1) {
+        //更新最大长度
+        Max = max(Max, cnt);
+        return;
+    }
+
+    //如果可以选择当前篮筐，并且选择了当前篮筐
+    if (a[u] < last)
+        dfs(u + 1, a[u], cnt + 1);
+
+    //放弃当前篮筐
+    dfs(u + 1, last, cnt);
+}
+
+int main() {
+    //输入
+    cin >> n;
+    for (int i = 1; i <= n; i++) cin >> a[i];
+    //深度优先
+    dfs(1, INT_MAX, 0);
+    //输出最大值
+    cout << Max << endl;
+    return 0;
+}

TangDou/KaoShi/投篮游戏II.cpp

@@ -0,0 +1,52 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 1010;
+int a[N];
+int n;
+//最大值
+int cnt;
+//路径
+vector<int> path;
+/**
+ 测试用例：
+ 3
+ 130 200 55
+
+ 最多投进球的个数
+ 答案：2
+ */
+
+/**
+ * 功能：深度优先搜索
+ * @param step    选择了哪个篮筐
+ */
+void dfs(int step) {
+    //如果越界
+    if (step == n + 1) {
+        //更新最大长度
+        cnt = max(cnt, (int) path.size());
+        return;
+    }
+
+    //如果选择了当前篮筐
+    //是不是可以选择这个位置
+    if (path.empty() || a[step] < a[path.back()]) {
+        path.push_back(step);//选择了第几个篮筐
+        dfs(step + 1);
+        path.pop_back();//回溯
+    }
+    //不选择当前篮筐
+    dfs(step + 1);
+}
+
+int main() {
+    //输入
+    cin >> n;
+    for (int i = 1; i <= n; i++) cin >> a[i];
+    //深度优先
+    dfs(1);
+    //输出最大值
+    cout << cnt << endl;
+    return 0;
+}

TangDou/KaoShi/排位置.cpp

@@ -0,0 +1,25 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 15;
+double a[N], c[N], d[N];
+int b[N], cl, dl;
+int main() {
+    int n = 10;
+    for (int i = 1; i <= n; i++) cin >> a[i];
+    for (int i = 1; i <= n; i++) cin >> b[i];
+    for (int i = 1; i <= n; i++)
+        if (b[i] == 1)
+            c[++cl] = a[i];
+        else
+            d[++dl] = a[i];
+    sort(c + 1, c + 1 + 5);
+    sort(d + 1, d + 1 + 5, greater<double>());
+
+    for (int i = 1; i <= 5; i++)
+        printf("%d %.2f\n", 1, c[i]);
+
+        for (int i = 1; i <= 5; i++)
+        printf("%d %.2f\n", 1, c[i]);
+    return 0;
+}

TangDou/KaoShi/摘苹果[洪水覆盖].cpp

@@ -0,0 +1,53 @@
+/*
+
+果园里有两种水果，苹果（1）和梨子（2），起点（6）也是苹果。
+从起点开始，不能碰到梨子，求最多能摘到多少个苹果。
+输入：
+3 4
+2 1 2 1
+1 6 1 2
+1 1 1 2
+输出：
+7
+*/
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 110;
+typedef pair<int, int> PII;
+int a[N][N];
+int n, m;
+int x, y;
+int dx[] = {-1, 0, 1, 0}; //上右下左
+int dy[] = {0, 1, 0, -1}; //上右下左
+int bfs(int x, int y) {
+    queue<PII> q;
+    int cnt = 1;
+    q.push({x, y});
+    a[x][y] = 2;
+
+    while (q.size()) {
+        auto t = q.front();
+        q.pop();
+        for (int i = 0; i < 4; i++) {
+            int tx = t.first + dx[i], ty = t.second + dy[i];
+            if (tx == 0 || tx > n || ty == 0 || ty > m) continue;
+            if (a[tx][ty] == 1) {
+                q.push({tx, ty});
+                a[tx][ty] = 2;
+                cnt++;
+            }
+        }
+    }
+    return cnt;
+}
+int main() {
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++) {
+            cin >> a[i][j];
+            if (a[i][j] == 6) x = i, y = j;
+        }
+    cout << bfs(x, y) << endl;
+    return 0;
+}

TangDou/KaoShi/数位递增数.cpp

@@ -0,0 +1,28 @@
+#include <bits/stdc++.h>
+/*
+一个正整数如果任何一个数位不大于右边相邻的数位，则称为一个数位递增的数，
+例如1135是一个数位递增的数，而1024不是一个数位递增的数。
+给定正整数 n，请问在整数 11 至 n 中有多少个数位递增的数？
+*/
+using namespace std;
+int n, cnt;
+
+bool iSDiZeng(int t) {
+    //哨兵
+    int x = 10; //这里定义x为10，是为了便于将个位数的大小与t作比较，
+    //个位数永远小于10
+    while (t) {
+        if (t % 10 > x) return false;
+        x = t % 10; // x一直在更新
+        t /= 10;    // n分解出一个数便舍去一个数
+    }
+    return true;
+}
+int main() {
+    cin >> n;
+    for (int i = 11; i <= n; i++)
+        if (iSDiZeng(i)) cnt++;
+
+    cout << cnt << endl;
+    return 0;
+}

TangDou/KaoShi/数字分拆.cpp

@@ -0,0 +1,25 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+/*
+输入：8
+答案：3
+*/
+int n;
+int res;
+// r:待分拆的数字
+// last:记录上一个分拆的数字是什么
+void dfs(int u, int r, int last) {
+    if (u == 4) {
+        if (r == 0) res++; //如果完成分拆的数字恰好等于n,说明是成功的分拆，否则就是没用完有剩余
+        return;
+    }
+    for (int i = last + 1; i <= r; i++)             //必须大于上一个，才能保证唯一不重复
+        if (i != 3 && i != 7) dfs(u + 1, r - i, i); //当前位置不能分配3和7
+}
+int main() {
+    cin >> n;
+    dfs(1, n, 0); //站在第1个箱子面前，待分拆的数是n,上一个分拆完的数字是0，
+    printf("%d\n", res);
+    return 0;
+}

TangDou/KaoShi/欧拉筛.cpp

@@ -0,0 +1,25 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int INF = 0x3f3f3f3f;
+//欧拉筛
+const int N = 1e5 + 10;
+int primes[N], cnt; // primes[]存储所有素数
+bool st[N];         // st[x]存储x是否被筛掉
+void get_primes(int n) {
+    for (int i = 2; i <= n; i++) {
+        if (!st[i]) primes[cnt++] = i;
+        for (int j = 0; primes[j] * i <= n; j++) {
+            st[primes[j] * i] = true;
+            if (i % primes[j] == 0) break;
+        }
+    }
+}
+
+int main() {
+    get_primes(200);
+    for (int i = 0; i < cnt; i++)
+        if (primes[i] > 100)
+            cout << primes[i] << " ";
+    return 0;
+}

TangDou/KaoShi/比身高.cpp

@@ -0,0 +1,45 @@
+#include <iostream>
+using namespace std;
+/**
+3 0
+165 175 163
+155 165 157
+答案：4
+*/
+
+int n, k;
+
+const int N = 110;
+int a[N], b[N];
+int cnt;
+int st[N];
+
+void dfs(int u, int s) {
+    if (u == n + 1) {
+        cnt++;
+        return;
+    }
+    for (int i = 1; i <= n; i++) {
+        if (!st[i]) {
+            if (a[u] >= b[i]) {
+                st[i] = 1;
+                dfs(u + 1, s);
+                st[i] = 0;
+            }
+
+            if (a[u] < b[i] && s < k) {
+                st[i] = 1;
+                dfs(u + 1, s + 1);
+                st[i] = 0;
+            }
+        }
+    }
+}
+int main() {
+    cin >> n >> k;
+    for (int i = 1; i <= n; i++) cin >> a[i];
+    for (int i = 1; i <= n; i++) cin >> b[i];
+    dfs(1, 0);
+    printf("%d\n", cnt);
+    return 0;
+}

TangDou/KaoShi/水仙花数.cpp

@@ -0,0 +1,30 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+//是不是水仙花数
+bool isSXH(int n) {
+    int s = 0, m;
+    m = n;
+    while (m) {
+        int a = m % 10;
+        s += a * a * a;
+        m /= 10;
+    }
+    return s == n;
+}
+/**
+ * 测试用例：1000
+ 输出:
+ 153
+ 370
+ 371
+ 407
+ */
+int main() {
+    int n;
+    cin >> n;
+    for (int i = 2; i <= n; i++)
+        if (isSXH(i)) cout << i << endl;
+    return 0;
+}

TangDou/KaoShi/油漆问题.cpp

@@ -0,0 +1,79 @@
+#include <bits/stdc++.h>
+/*油漆面积
+X星球的一批考古机器人正在一片废墟上考古。该区域的地面坚硬如石、平整如镜。
+管理人员为方便，建立了标准的直角坐标系。每个机器人都各有特长、身怀绝技。它们感兴趣的内容也不相同。
+经过各种测量，每个机器人都会报告一个或多个矩形区域，作为优先考古的区域。
+矩形的表示格式为(x1,y1,x2,y2)，代表矩形的两个对角点坐标。
+为了醒目，总部要求对所有机器人选中的矩形区域涂黄色油漆。
+小明并不需要当油漆工，只是他需要计算一下，一共要耗费多少油漆。
+其实这也不难，只要算出所有矩形覆盖的区域一共有多大面积就可以了。
+注意，各个矩形间可能重叠。
+本题的输入为若干矩形，要求输出其覆盖的总面积。
+输入格式：
+第一行，一个整数n，表示有多少个矩形(1<=n<10000)
+接下来的n行，每行有4个整数x1 y1 x2 y2，空格分开，表示矩形的两个对角顶点坐标。
+(0<= x1,y1,x2,y2 <=10000)
+输出格式：
+一行一个整数，表示矩形覆盖的总面积面积。
+例如，
+测试用例：
+2
+2 2 9  5
+6 1 12 9
+
+答案：
+60
+
+输入：
+3
+1 5 10 10
+3 1 20 20
+2 7 15 17
+程序应该输出：
+340
+
+再例如，
+输入：
+3
+5 2 10 6
+2 7 12 10
+8 1 15 15
+程序应该输出：
+128
+
+I 暴力解法
+https://www.bilibili.com/video/BV1VV411z7S2?p=11
+
+II 高端解法
+https://www.bilibili.com/video/BV1VV411z7S2?p=12
+线段树+扫描线 模板题 (百度一下线段树+扫描线，可以看到还有很多求周长的东西~)
+
+*/
+using namespace std;
+const int N = 1e4 + 10; //矩形的数量
+int n, sum;
+// 65536kb /1024 =64mb
+// int p[N][N]; //10000*10000*4 byte =4 0000 0000  byte /1024/1024=381MB
+// 0 1
+bool p[N][N]; //这里需要估算一下空间，因为int是8个bit,bool是1个bit。
+
+//模拟，对于(x1,y1)->(x2,y2)内的所有单元格，都标识为已被刷过油漆
+//注意一下这里坐标与单元格的映射关系，就把每个单元格的左下角坐标认为是单元格的编号
+void paint(int x1, int y1, int x2, int y2) {
+    for (int i = x1; i < x2; i++)
+        for (int j = y1; j < y2; j++)
+            p[i][j] = 1;
+}
+int main() {
+    cin >> n;
+    for (int i = 0; i < n; i++) {
+        int x1, y1, x2, y2;
+        cin >> x1 >> y1 >> x2 >> y2;
+        paint(x1, y1, x2, y2);
+    }
+    for (int i = 0; i < N; i++)
+        for (int j = 0; j < N; j++)
+            if (p[i][j]) sum++;
+    cout << sum << endl;
+    return 0;
+}

TangDou/KaoShi/玻璃球乘积.cpp

@@ -0,0 +1,54 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+int n;
+const int N = 110;
+const int INF = 0x3f3f3f3f;
+int st[N];
+int a[N];
+int res = -INF;
+/*
+测试用例：
+5　
+
+答案：
+6　
+*/
+void dfs(int u, int r, int last) {
+    //递归出口
+    if (r == 0) {
+        if (u >= 2) {
+            //　输出每个拆分办法
+            // for (int i = 0; i < u - 1; i++) cout << a[i] << " * ";
+            // cout << a[u - 1];
+
+            //计算累乘积
+            int t = 1;
+            for (int i = 0; i < u; i++) t *= a[i];
+            // cout << " = " << t << endl;
+
+            //猴子选大王
+            res = max(res, t);
+        }
+        return;
+    }
+
+    for (int i = last + 1; i <= r; i++) {
+        if (!st[i]) {
+            st[i] = true;
+            a[u] = i;
+
+            dfs(u + 1, r - i, i);
+
+            a[u] = 0;
+            st[i] = false;
+        }
+    }
+}
+
+int main() {
+    cin >> n;
+    dfs(0, n, 0);
+    printf("%d\n", res);
+    return 0;
+}

TangDou/KaoShi/由大到小排序带逗号.cpp

@@ -0,0 +1,16 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 10;
+int a[N];
+/*
+3 2 5 5 4
+*/
+int main() {
+    int n = 5;
+    for (int i = 1; i <= n; i++) cin >> a[i];
+    sort(a + 1, a + 1 + n, greater<int>());
+    for (int i = 1; i < n; i++) printf("%d,", a[i]);
+    printf("%d", a[n]);
+    return 0;
+}

TangDou/KaoShi/病毒繁殖.cpp

@@ -0,0 +1,25 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+/*
+测试用例：
+10
+
+数量：
+10
+
+示例：1 1 1 1 2 3 4 5 7 10
+*/
+const int N = 110;
+int a[N] = {0, 1, 1, 1, 1};
+int main() {
+    int n;
+    cin >> n;
+    for (int i = 5; i <= n; i++)
+        a[i] = a[i - 1] + a[i - 4];
+
+    // for (int i = 1; i <= n; i++)        printf("%d ", a[i]);
+    // puts("");
+    printf("%d\n", a[n]);
+    return 0;
+}

TangDou/KaoShi/秒数转时分秒.cpp

@@ -0,0 +1,14 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+// 7259
+int main() {
+    int n;
+    cin >> n;
+    int h = n / 3600;
+    int m = (n - h * 3600) / 60;
+    int s = n - h * 3600 - m * 60;
+    printf("%02d:%02d:%02d\n", h, m, s);
+
+    return 0;
+}

TangDou/KaoShi/空瓶换饮料.cpp

@@ -0,0 +1,23 @@
+#include <iostream>
+using namespace std;
+
+int n;   // 换回来的饮料数量
+int sum; // 喝到总数
+int t;   // 空瓶数量
+
+int main() {
+    cin >> n;
+
+    while (n) {
+        //① 统计上轮结果
+        sum += n;  // 上轮换回来的可以喝到
+        t += n;    // 当前手中酒瓶 = 原空瓶数量+上轮换回
+
+        //② 本轮换取开始
+        n = t / 3; // 换回来的饮料数量
+        t %= 3;    // 剩余空瓶
+    }
+    //输出
+    printf("%d", sum);
+    return 0;
+}

TangDou/KaoShi/穿越方格.cpp

@@ -0,0 +1,43 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+struct Node {
+    int num;
+    int step;
+};
+
+const int N = 1010;
+int n, m;
+queue<Node> q;
+int st[N];
+
+void bfs() {
+    q.push({n, 0});
+    while (q.size()) {
+        auto x = q.front();
+        q.pop();
+        if (x.num == m) {
+            printf("%d\n", x.step);
+            break;
+        }
+        if (!st[x.num + 1]) {
+            q.push({x.num + 1, x.step + 1});
+            st[x.num + 1] = 1;
+        }
+        if (!st[x.num - 1]) {
+            q.push({x.num - 1, x.step + 1});
+            st[x.num - 1] = 1;
+        }
+        if (!st[x.num * 2]) {
+            q.push({x.num * 2, x.step + 1});
+            st[x.num * 2] = 1;
+        }
+    }
+}
+
+int main() {
+    cin >> n >> m;
+    bfs();
+    return 0;
+}

TangDou/KaoShi/算字符.cpp

@@ -0,0 +1,12 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+int main() {
+    char c;
+    int n;
+    scanf("%c %d", &c, &n);
+    if (c + n > 'z') printf("*");
+    else printf("%c", c + n);
+    return 0;
+}

TangDou/KaoShi/约瑟夫环经典版.cpp

@@ -0,0 +1,25 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+int n;
+queue<int> q;
+int cnt; //当前轮次第几个
+
+int main() {
+    cin >> n;
+    //为啥这里把所有人员都入队列呢？当面要解释清楚
+    for (int i = 1; i <= n; i++) q.push(i);
+    //如果还剩余大于1个
+    while (q.size() > 1) {
+        //取出一个
+        int t = q.front();
+        q.pop();
+        cnt++;//点了一个数
+
+        if (cnt == 3) cnt = 0;
+        else  q.push(t);
+    }
+    //输出最后一个
+    cout << q.front();
+    return 0;
+}

TangDou/KaoShi/约瑟夫环蓝桥国赛版.cpp

@@ -0,0 +1,44 @@
+#include <bits/stdc++.h>
+/*
+测试用例I:
+5 3 3
+
+答案：
+2 4 5 7 8
+
+测试用例II：
+5 4 3
+
+5 7 8 1 2
+这个测试用例可以看出由小到大排序的意义。
+*/
+using namespace std;
+const int N = 1010;
+int a[N], al;
+
+int main() {
+    int x, y, k;
+    cin >> x >> y >> k;
+    queue<int> q;
+    int cnt = 0;
+    for (int i = 1; i <= x + y; i++) q.push(i);
+    while (q.size() > x) {
+        auto t = q.front();
+        q.pop();
+        cnt++;
+        if (cnt % k > 0)
+            q.push(t);
+        else
+            cnt = 0;
+    }
+
+    while (q.size()) {
+        auto t = q.front();
+        q.pop();
+        a[al++] = t;
+    }
+    sort(a, a + al);
+    for (int i = 0; i < al; i++) cout << a[i] << " ";
+
+    return 0;
+}

TangDou/KaoShi/结构体排序.cpp

@@ -0,0 +1,33 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+/*
+3
+3 84
+1 87
+2 84
+
+输出样例
+1 87
+2 84
+3 84
+
+*/
+const int N = 1010;
+struct Node {
+    int id;
+    int score;
+    const bool operator<(const Node &b) const {
+        if (score == b.score) return id < b.id;
+        return score > b.score;
+    }
+} a[N];
+int main() {
+    int n;
+    cin >> n;
+    for (int i = 1; i <= n; i++) cin >> a[i].id >> a[i].score;
+    sort(a + 1, a + 1 + n);
+
+    for (int i = 1; i <= n; i++) printf("%d %d\n", a[i].id, a[i].score);
+    return 0;
+}

TangDou/KaoShi/蓝桥杯个数.cpp

@@ -0,0 +1,16 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+string s;
+int main() {
+    cin >> s;
+    for (int i = 0; i < s.size(); i++)
+        if (s[i] >= 'A' && s[i] <= 'Z') s[i] += 32;
+          
+    int cnt = 0;
+    for (int i = 0; i < s.size(); i++)
+        if (s.substr(i, 4) == "lqbs") cnt++;
+
+    cout << cnt << endl;
+    return 0;
+}

TangDou/KaoShi/质数.cpp

@@ -0,0 +1,19 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int INF = 0x3f3f3f3f;
+//判断一个数是不是质数
+bool isPrime(int n) {
+    if (n < 2) return false;
+    for (int i = 2; i <= n / i; i++)
+        if (n % i == 0) return false;
+    return true;
+}
+/*
+ for (int i = 100; i <= 200; i++)
+        if (isPrime(i)) cout << i << " ";
+*/
+
+int main() {
+    return 0;
+}

TangDou/KaoShi/近似值.cpp

@@ -0,0 +1,23 @@
+#include <iostream>
+
+using namespace std;
+const int N = 110;
+int a[N];
+/*
+9 4 3 8 8 9 11 9 8 5
+10
+*/
+int n;
+int res = -1;
+int main() {
+    for (int i = 1; i <= 10; i++) cin >> a[i];
+    cin >> n;
+
+    for (int i = 1; i <= 10; i++)
+        for (int j = i + 1; j <= 10; j++)
+            if (a[i] + a[j] <= n)
+                res = max(res, a[i] + a[j]);
+
+    printf("%d", res);
+    return 0;
+}

TangDou/KaoShi/选择题判定.cpp

@@ -0,0 +1,15 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+int main() {
+    string ans = "DCBAD";
+    int s = 0;
+    string c;
+    cin >> c;
+    for (int i = 0; i < 5; i++)
+        if (c[i] == ans[i]) s += 30;
+
+    cout << s << endl;
+    return 0;
+}