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用C++编写的小游戏源代码

2023-05-19 23:34:14
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瑞瑞爱吃桃

五子棋的代码：

#include<iostream>

#include<stdio.h>

#include<stdlib.h>

#include <time.h>

using namespace std;

const int N=15;                 //15*15的棋盘

const char ChessBoardflag = " ";          //棋盘标志

const char flag1="o";              //玩家1或电脑的棋子标志

const char flag2="X";              //玩家2的棋子标志

typedef struct Coordinate          //坐标类

{   

int x;                         //代表行

int y;                         //代表列

}Coordinate;

class GoBang                    //五子棋类

{ 

public:

GoBang()                //初始化

{

InitChessBoard();

}

void Play()               //下棋

{

Coordinate Pos1;      // 玩家1或电脑

Coordinate Pos2;      //玩家2

int n = 0;

while (1)

{

int mode = ChoiceMode();

while (1)

{

if (mode == 1)       //电脑vs玩家

{

ComputerChess(Pos1,flag1);     // 电脑下棋

if (GetVictory(Pos1, 0, flag1) == 1)     //0表示电脑,真表示获胜

break;

PlayChess(Pos2, 2, flag2);     //玩家2下棋

if (GetVictory(Pos2, 2, flag2))     //2表示玩家2

break;

}

else            //玩家1vs玩家2

{

PlayChess(Pos1, 1, flag1);     // 玩家1下棋

if (GetVictory(Pos1, 1, flag1))      //1表示玩家1

break;

PlayChess(Pos2, 2, flag2);     //玩家2下棋

if (GetVictory(Pos2, 2, flag2))  //2表示玩家2

break;

}

}

cout << "***再来一局***" << endl;

cout << "y or n :";

char c = "y";

cin >> c;

if (c == "n")

break;

}       

}

protected:

int ChoiceMode()           //选择模式

{

int i = 0;

system("cls");        //系统调用，清屏

InitChessBoard();       //重新初始化棋盘

cout << "***0、退出  1、电脑vs玩家  2、玩家vs玩家***" << endl;

while (1)

{

cout << "请选择：";

cin >> i;

if (i == 0)         //选择0退出

exit(1);

if (i == 1 || i == 2)

return i;

cout << "输入不合法" << endl;

}

}

void InitChessBoard()      //初始化棋盘

{

for (int i = 0; i < N + 1; ++i)      

{

for (int j = 0; j < N + 1; ++j)

{

_ChessBoard[i][j] = ChessBoardflag;

}

}

}

void PrintChessBoard()    //打印棋盘，这个函数可以自己调整

{

system("cls");                //系统调用，清空屏幕

for (int i = 0; i < N+1; ++i)

{

for (int j = 0; j < N+1; ++j)

{

if (i == 0)                               //打印列数字

{

if (j!=0)

printf("%d  ", j);

else

printf("   ");

}

else if (j == 0)                //打印行数字

printf("%2d ", i);

else

{

if (i < N+1)

{

printf("%c |",_ChessBoard[i][j]);

}

}

}

cout << endl;

cout << "   ";

for (int m = 0; m < N; m++)

{

printf("--|");

}

cout << endl;

}

}

void PlayChess(Coordinate& pos, int player, int flag)       //玩家下棋

{

PrintChessBoard();         //打印棋盘

while (1)

{

printf("玩家%d输入坐标：", player);

cin >> pos.x >> pos.y;

if (JudgeValue(pos) == 1)          //坐标合法

break;

cout << "坐标不合法，重新输入" << endl;

}

_ChessBoard[pos.x][pos.y] = flag;

}

void ComputerChess(Coordinate& pos, char flag)       //电脑下棋

{

PrintChessBoard();         //打印棋盘

int x = 0;

int y = 0;

while (1)

{

x = (rand() % N) + 1;      //产生1~N的随机数

srand((unsigned int) time(NULL));

y = (rand() % N) + 1;     //产生1~N的随机数

srand((unsigned int) time(NULL));

if (_ChessBoard[x][y] == ChessBoardflag)      //如果这个位置是空的，也就是没有棋子

break;

}

pos.x = x;

pos.y = y;

_ChessBoard[pos.x][pos.y] = flag;

}

int JudgeValue(const Coordinate& pos)       //判断输入坐标是不是合法

{

if (pos.x > 0 && pos.x <= N&&pos.y > 0 && pos.y <= N)

{

if (_ChessBoard[pos.x][pos.y] == ChessBoardflag)

{

return 1;    //合法

}

}

return 0;        //非法

}

int JudgeVictory(Coordinate pos, char flag)           //判断有没有人胜负(底层判断)

{

int begin = 0;

int end = 0;

int begin1 = 0;

int end1 = 0;

//判断行是否满足条件

(pos.y - 4) > 0 ? begin = (pos.y - 4) : begin = 1;

(pos.y + 4) >N ? end = N : end = (pos.y + 4);

for (int i = pos.x, j = begin; j + 4 <= end; j++)

{

if (_ChessBoard[i][j] == flag&&_ChessBoard[i][j + 1] == flag&&

_ChessBoard[i][j + 2] == flag&&_ChessBoard[i][j + 3] == flag&&

_ChessBoard[i][j + 4] == flag)

return 1;

}

//判断列是否满足条件

(pos.x - 4) > 0 ? begin = (pos.x - 4) : begin = 1;

(pos.x + 4) > N ? end = N : end = (pos.x + 4);

for (int j = pos.y, i = begin; i + 4 <= end; i++)

{

if (_ChessBoard[i][j] == flag&&_ChessBoard[i + 1][j] == flag&&

_ChessBoard[i + 2][j] == flag&&_ChessBoard[i + 3][j] == flag&&

_ChessBoard[i + 4][j] == flag)

return 1;

}

int len = 0;

//判断主对角线是否满足条件

pos.x > pos.y ? len = pos.y - 1 : len = pos.x - 1;

if (len > 4)

len = 4;

begin = pos.x - len;       //横坐标的起始位置

begin1 = pos.y - len;      //纵坐标的起始位置

pos.x > pos.y ? len = (N - pos.x) : len = (N - pos.y);

if (len>4)

len = 4;

end = pos.x + len;       //横坐标的结束位置

end1 = pos.y + len;      //纵坐标的结束位置

for (int i = begin, j = begin1; (i + 4 <= end) && (j + 4 <= end1); ++i, ++j)

{

if (_ChessBoard[i][j] == flag&&_ChessBoard[i + 1][j + 1] == flag&&

_ChessBoard[i + 2][j + 2] == flag&&_ChessBoard[i + 3][j + 3] == flag&&

_ChessBoard[i + 4][j + 4] == flag)

return 1;

}

//判断副对角线是否满足条件

(pos.x - 1) >(N - pos.y) ? len = (N - pos.y) : len = pos.x - 1;

if (len > 4)

len = 4;

begin = pos.x - len;       //横坐标的起始位置

begin1 = pos.y + len;      //纵坐标的起始位置

(N - pos.x) > (pos.y - 1) ? len = (pos.y - 1) : len = (N - pos.x);

if (len>4)

len = 4;

end = pos.x + len;       //横坐标的结束位置

end1 = pos.y - len;      //纵坐标的结束位置

for (int i = begin, j = begin1; (i + 4 <= end) && (j - 4 >= end1); ++i, --j)

{

if (_ChessBoard[i][j] == flag&&_ChessBoard[i + 1][j - 1] == flag&&

_ChessBoard[i + 2][j - 2] == flag&&_ChessBoard[i + 3][j - 3] == flag&&

_ChessBoard[i + 4][j - 4] == flag)

return 1;

}

for (int i = 1; i < N + 1; ++i)           //棋盘有没有下满

{

for (int j =1; j < N + 1; ++j)

{

if (_ChessBoard[i][j] == ChessBoardflag)

return 0;                      //0表示棋盘没满

} 

}

return -1;      //和棋

}

bool GetVictory(Coordinate& pos, int player, int flag)   //对JudgeVictory的一层封装，得到具体那个玩家获胜

{

int n = JudgeVictory(pos, flag);   //判断有没有人获胜

if (n != 0)                    //有人获胜，0表示没有人获胜

{

PrintChessBoard();

if (n == 1)                //有玩家赢棋

{

if (player == 0)     //0表示电脑获胜，1表示玩家1,2表示玩家2

printf("***电脑获胜*** ");

else

printf("***恭喜玩家%d获胜*** ", player);

}

else

printf("***双方和棋*** ");

return true;      //已经有人获胜

}

return false;   //没有人获胜

}

private:

char _ChessBoard[N+1][N+1];      

};

chessboard

扩展资料：

设计思路

1、进行问题分析与设计，计划实现的功能为，开局选择人机或双人对战，确定之后比赛开始。

2、比赛结束后初始化棋盘，询问是否继续比赛或退出，后续可加入复盘、悔棋等功能。

3、整个过程中，涉及到了棋子和棋盘两种对象，同时要加上人机对弈时的AI对象，即涉及到三个对象。

S笔记

以下是贪吃蛇源代码：

#include<iostream.h>

#include<windows.h>

#include<time.h>

#include<stdlib.h>

#include<conio.h>

#define N 21

void gotoxy(int x,int y)//位置函数

{

COORD pos;

pos.X=2*x;

pos.Y=y;

SetConsoleCursorPosition(GetStdHandle(STD_OUTPUT_HANDLE),pos);

}

void color(int a)//颜色函数

{

SetConsoleTextAttribute(GetStdHandle(STD_OUTPUT_HANDLE),a);

}

void init(int apple[2])//初始化函数（初始化围墙、显示信息、苹果）

{

int i,j;//初始化围墙

int wall[N+2][N+2]={{0}};

for(i=1;i<=N;i++)

{

for(j=1;j<=N;j++)

wall[i][j]=1;

}

color(11);

for(i=0;i<N+2;i++)

{

for(j=0;j<N+2;j++)

{

if(wall[i][j])

cout<<"■";

else cout<<"□" ;

}

cout<<endl;

}

gotoxy(N+3,1);//显示信息

color(20);

cout<<"按 W S A D 移动方向"<<endl;

gotoxy(N+3,2);

color(20);

cout<<"按任意键暂停"<<endl;

gotoxy(N+3,3);

color(20);

cout<<"得分："<<endl;

apple[0]=rand()%N+1;//苹果

apple[1]=rand()%N+1;

gotoxy(apple[0],apple[1]);

color(12);

cout<<"●"<<endl;

}

int main()

{

int i,j;

int** snake=NULL;

int apple[2];

int score=0;

int tail[2];

int len=3;

char ch="p";

srand((unsigned)time(NULL));

init(apple);

snake=(int**)realloc(snake,sizeof(int*)*len);

for(i=0;i<len;i++)

snake[i]=(int*)malloc(sizeof(int)*2);

for(i=0;i<len;i++)

{

snake[i][0]=N/2;

snake[i][1]=N/2+i;

gotoxy(snake[i][0],snake[i][1]);

color(14);

cout<<"★"<<endl;

}

while(1)//进入消息循环

{

tail[0]=snake[len-1][0];

tail[1]=snake[len-1][1];

gotoxy(tail[0],tail[1]);

color(11);

cout<<"■"<<endl;

for(i=len-1;i>0;i--)

{

snake[i][0]=snake[i-1][0];

snake[i][1]=snake[i-1][1];

gotoxy(snake[i][0],snake[i][1]);

color(14);

cout<<"★"<<endl;

}

if(kbhit())

{

gotoxy(0,N+2);

ch=getche();

}

switch(ch)

{

case "w":snake[0][1]--;break;

case "s":snake[0][1]++;break;

case "a":snake[0][0]--;break;

case "d":snake[0][0]++;break;

default: break;

}

gotoxy(snake[0][0],snake[0][1]);

color(14);

cout<<"★"<<endl;

Sleep(abs(200-0.5*score));

if(snake[0][0]==apple[0]&&snake[0][1]==apple[1])//吃掉苹果后蛇分数加1，蛇长加1

{

score++;

len++;

snake=(int**)realloc(snake,sizeof(int*)*len);

snake[len-1]=(int*)malloc(sizeof(int)*2);

apple[0]=rand()%N+1;

apple[1]=rand()%N+1;

gotoxy(apple[0],apple[1]);

color(12);

cout<<"●"<<endl;

gotoxy(N+5,3);

color(20);

cout<<score<<endl;

}

if(snake[0][1]==0||snake[0][1]==N||snake[0][0]==0||snake[0][0]==N)//撞到围墙后失败

{

gotoxy(N/2,N/2);

color(30);

cout<<"失败！！！"<<endl;

for(i=0;i<len;i++)

free(snake[i]);

Sleep(INFINITE);

exit(0);

}

}

return 0;

}参考资料：从C++吧看来的

小菜G

像素射击：int anim abc_fade_in 0x7f010001

int anim abc_fade_out 0x7f010002

int anim abc_grow_fade_in_from_bottom 0x7f010003

int anim abc_popup_enter 0x7f010004

int anim abc_popup_exit 0x7f010005

int anim abc_shrink_fade_out_from_bottom 0x7f010006

int anim abc_slide_in_bottom 0x7f010007

int anim abc_slide_in_top 0x7f010008

int anim abc_slide_out_bottom 0x7f010009

int anim abc_slide_out_top 0x7f01000a

int anim abc_tooltip_enter 0x7f01000b

int anim abc_tooltip_exit 0x7f01000c

int attr actionBarDivider 0x7f040001

int attr actionBarItemBackground 0x7f040002

int attr actionBarPopupTheme 0x7f040003

int attr actionBarSize 0x7f040004

int attr actionBarSplitStyle 0x7f040005

int attr actionBarStyle 0x7f040006

int attr actionBarTabBarStyle 0x7f040007

int attr actionBarTabStyle 0x7f040008

int attr actionBarTabTextStyle 0x7f040009

int attr actionBarTheme 0x7f04000a

int attr actionBarWidgetTheme 0x7f04000b

int attr actionButtonStyle 0x7f04000c

int attr actionDropDownStyle 0x7f04000d

int attr actionLayout 0x7f04000e

int attr actionMenuTextAppearance 0x7f04000f

int attr actionMenuTextColor 0x7f040010

int attr actionModeBackground 0x7f040011

int attr actionModeCloseButtonStyle 0x7f040012

int attr actionModeCloseDrawable 0x7f040013

int attr actionModeCopyDrawable 0x7f040014

int attr actionModeCutDrawable 0x7f040015

int attr actionModeFindDrawable 0x7f040016

int attr actionModePasteDrawable 0x7f040017

int attr actionModePopupWindowStyle 0x7f040018

int attr actionModeSelectAllDrawable 0x7f040019

int attr actionModeShareDrawable 0x7f04001a

int attr actionModeSplitBackground 0x7f04001b

int attr actionModeStyle 0x7f04001c

int attr actionModeWebSearchDrawable 0x7f04001d

int attr actionOverflowButtonStyle 0x7f04001e

int attr actionOverflowMenuStyle 0x7f04001f

int attr actionProviderClass 0x7f040020

int attr actionViewClass 0x7f040021

int attr activityChooserViewStyle 0x7f040022

int attr alertDialogButtonGroupStyle 0x7f040023

int attr alertDialogCenterButtons 0x7f040024

int attr alertDialogStyle 0x7f040025

int attr alertDialogTheme 0x7f040026

int attr allowStacking 0x7f040027

int attr alpha 0x7f040028

int attr alphabeticModifiers 0x7f040029

int attr arrowHeadLength 0x7f04002a

int attr arrowShaftLength 0x7f04002b

int attr autoCompleteTextViewStyle 0x7f04002c

int attr autoSizeMaxTextSize 0x7f04002d

int attr autoSizeMinTextSize 0x7f04002e

int attr autoSizePresetSizes 0x7f04002f

int attr autoSizeStepGranularity 0x7f040030

int attr autoSizeTextType 0x7f040031

int attr background 0x7f040032

int attr backgroundSplit 0x7f040033

int attr backgroundStacked 0x7f040034

int attr backgroundTint 0x7f040035

int attr backgroundTintMode 0x7f040036

int attr barLength 0x7f040037

int attr borderlessButtonStyle 0x7f040038

int attr buttonBarButtonStyle 0x7f040039

int attr buttonBarNegativeButtonStyle 0x7f04003a

int attr buttonBarNeutralButtonStyle 0x7f04003b

int attr buttonBarPositiveButtonStyle 0x7f04003c

int attr buttonBarStyle 0x7f04003d

int attr buttonGravity 0x7f04003e

int attr buttonIconDimen 0x7f04003f

int attr buttonPanelSideLayout 0x7f040040

int attr buttonStyle 0x7f040041

int attr buttonStyleSmall 0x7f040042

int attr buttonTint 0x7f040043

int attr buttonTintMode 0x7f040044

int attr checkboxStyle 0x7f040045

int attr checkedTextViewStyle 0x7f040046

int attr closeIcon 0x7f040047

int attr closeItemLayout 0x7f040048

int attr collapseContentDescription 0x7f040049

int attr collapseIcon 0x7f04004a

int attr color 0x7f04004b

int attr colorAccent 0x7f04004c

int attr colorBackgroundFloating 0x7f04004d

int attr colorButtonNormal 0x7f04004e

int attr colorControlActivated 0x7f04004f

int attr colorControlHighlight 0x7f040050

int attr colorControlNormal 0x7f040051

int attr colorError 0x7f040052

int attr colorPrimary 0x7f040053

int attr colorPrimaryDark 0x7f040054

int attr colorSwitchThumbNormal 0x7f040055

int attr commitIcon 0x7f040056

int attr contentDescription 0x7f040057

int attr contentInsetEnd 0x7f040058

int attr contentInsetEndWithActions 0x7f040059

int attr contentInsetLeft 0x7f04005a

int attr contentInsetRight 0x7f04005b

int attr contentInsetStart 0x7f04005c

int attr contentInsetStartWithNavigation 0x7f04005d

int attr controlBackground 0x7f04005e

int attr coordinatorLayoutStyle 0x7f04005f

int attr customNavigationLayout 0x7f040060

int attr defaultQueryHint 0x7f040061

int attr dialogCornerRadius 0x7f040062

int attr dialogPreferredPadding 0x7f040063

int attr dialogTheme 0x7f040064

int attr displayOptions 0x7f040065

int attr divider 0x7f040066

int attr dividerHorizontal 0x7f040067

int attr dividerPadding 0x7f040068

int attr dividerVertical 0x7f040069

int attr drawableSize 0x7f04006a

int attr drawerArrowStyle 0x7f04006b

int attr dropDownListViewStyle 0x7f04006c

int attr dropdownListPreferredItemHeight 0x7f04006d

int attr editTextBackground 0x7f04006e

int attr editTextColor 0x7f04006f

int attr editTextStyle 0x7f040070

int attr elevation 0x7f040071

int attr expandActivityOverflowButtonDrawable 0x7f040072

int attr firstBaselineToTopHeight 0x7f040073

int attr font 0x7f040074

int attr fontFamily 0x7f040075

int attr fontProviderAuthority 0x7f040076

int attr fontProviderCerts 0x7f040077

int attr fontProviderFetchStrategy 0x7f040078

int attr fontProviderFetchTimeout 0x7f040079

int attr fontProviderPackage 0x7f04007a

int attr fontProviderQuery 0x7f04007b

int attr fontStyle 0x7f04007c

int attr fontVariationSettings 0x7f04007d

int attr fontWeight 0x7f04007e

int attr gapBetweenBars 0x7f04007f

int attr goIcon 0x7f040080

int attr height 0x7f040081

int attr hideOnContentScroll 0x7f040082

int attr homeAsUpIndicator 0x7f040083

int attr homeLayout 0x7f040084

int attr icon 0x7f040085

int attr iconTint 0x7f040086

int attr iconTintMode 0x7f040087

int attr iconifiedByDefault 0x7f040088

int attr imageButtonStyle 0x7f040089

int attr indeterminateProgressStyle 0x7f04008a

int attr initialActivityCount 0x7f04008b

int attr isLightTheme 0x7f04008c

int attr itemPadding 0x7f04008d

int attr keylines 0x7f04008e

int attr lastBaselineToBottomHeight 0x7f04008f

int attr layout 0x7f040090

int attr layout_anchor 0x7f040091

int attr layout_anchorGravity 0x7f040092

int attr layout_behavior 0x7f040093

int attr layout_dodgeInsetEdges 0x7f040094

int attr layout_insetEdge 0x7f040095

int attr layout_keyline 0x7f040096

int attr lineHeight 0x7f040097

int attr listChoiceBackgroundIndicator 0x7f040098

int attr listDividerAlertDialog 0x7f040099

int attr listItemLayout 0x7f04009a

int attr listLayout 0x7f04009b

int attr listMenuViewStyle 0x7f04009c

int attr listPopupWindowStyle 0x7f04009d

int attr listPreferredItemHeight 0x7f04009e

int attr listPreferredItemHeightLarge 0x7f04009f

int attr listPreferredItemHeightSmall 0x7f0400a0

int attr listPreferredItemPaddingLeft 0x7f0400a1

int attr listPreferredItemPaddingRight 0x7f0400a2

int attr logo 0x7f0400a3

int attr logoDescription 0x7f0400a4

int attr maxButtonHeight 0x7f0400a5

int attr measureWithLargestChild 0x7f0400a6

int attr multiChoiceItemLayout 0x7f0400a7

int attr navigationContentDescription 0x7f0400a8

int attr navigationIcon 0x7f0400a9

int attr navigationMode 0x7f0400aa

int attr numericModifiers 0x7f0400ab

int attr overlapAnchor 0x7f0400ac

int attr paddingBottomNoButtons 0x7f0400ad

int attr paddingEnd 0x7f0400ae

int attr paddingStart 0x7f0400af

int attr paddingTopNoTitle 0x7f0400b0

int attr panelBackground 0x7f0400b1

int attr panelMenuListTheme 0x7f0400b2

int attr panelMenuListWidth 0x7f0400b3

int attr popupMenuStyle 0x7f0400b4

int attr popupTheme 0x7f0400b5

int attr popupWindowStyle 0x7f0400b6

int attr preserveIconSpacing 0x7f0400b7

int attr progressBarPadding 0x7f0400b8

int attr progressBarStyle 0x7f0400b9

int attr queryBackground 0x7f0400ba

int attr queryHint 0x7f0400bb

int attr radioButtonStyle 0x7f0400bc

int attr ratingBarStyle 0x7f0400bd

int attr ratingBarStyleIndicator 0x7f0400be

int attr ratingBarStyleSmall 0x7f0400bf

int attr searchHintIcon 0x7f0400c0

int attr searchIcon 0x7f0400c1

int attr searchViewStyle 0x7f0400c2

int attr seekBarStyle 0x7f0400c3

int attr selectableItemBackground 0x7f0400c4

int attr selectableItemBackgroundBorderless 0x7f0400c5

int attr showAsAction 0x7f0400c6

int attr showDividers 0x7f0400c7

int attr showText 0x7f0400c8

int attr showTitle 0x7f0400c9

int attr singleChoiceItemLayout 0x7f0400ca

int attr spinBars 0x7f0400cb

int attr spinnerDropDownItemStyle 0x7f0400cc

int attr spinnerStyle 0x7f0400cd

int attr splitTrack 0x7f0400ce

int attr srcCompat 0x7f0400cf

int attr state_above_anchor 0x7f0400d0

int attr statusBarBackground 0x7f0400d1

int attr subMenuArrow 0x7f0400d2

int attr submitBackground 0x7f0400d3

int attr subtitle 0x7f0400d4

int attr subtitleTextAppearance 0x7f0400d5

int attr subtitleTextColor 0x7f0400d6

int attr subtitleTextStyle 0x7f0400d7

int attr suggestionRowLayout 0x7f0400d8

int attr switchMinWidth 0x7f0400d9

int attr switchPadding 0x7f0400da

int attr switchStyle 0x7f0400db

int attr switchTextAppearance 0x7f0400dc

int attr textAllCaps 0x7f0400dd

int attr textAppearanceLargePopupMenu 0x7f0400de

int attr textAppearanceListItem 0x7f0400df

int attr textAppearanceListItemSecondary 0x7f0400e0

int attr textAppearanceListItemSmall 0x7f0400e1

int attr textAppearancePopupMenuHeader 0x7f0400e2

int attr textAppearanceSearchResultSubtitle 0x7f0400e3

int attr textAppearanceSearchResultTitle 0x7f0400e4

int attr textAppearanceSmallPopupMenu 0x7f0400e5

int attr textColorAlertDialogListItem 0x7f0400e6

int attr textColorSearchUrl 0x7f0400e7

int attr theme 0x7f0400e8

int attr thickness 0x7f0400e9

int attr thumbTextPadding 0x7f0400ea

int attr thumbTint 0x7f0400eb

int attr thumbTintMode 0x7f0400ec

int attr tickMark 0x7f0400ed

int attr tickMarkTint 0x7f0400ee

int attr tickMarkTintMode 0x7f0400ef

int attr tint 0x7f0400f0

int attr tintMode 0x7f0400f1

int attr title 0x7f0400f2

int attr titleMargin 0x7f0400f3

int attr titleMarginBottom 0x7f0400f4

int attr titleMarginEnd 0x7f0400f5

int attr titleMarginStart 0x7f0400f6

int attr titleMarginTop 0x7f0400f7

int attr titleMargins 0x7f0400f8

int attr titleTextAppearance 0x7f0400f9

int attr titleTextColor 0x7f

tt白

想用这种c语言编写一个小游戏可以实现，但是非语言编写出的游戏一般的话都是比较大的，占内存比较短。

苏萦
* 回复内容中包含的链接未经审核，可能存在风险，暂不予完整展示！

使用语言：C++使用工具：vs2019

再也不做稀饭了

用c++编写小程序的源代码，你可以找一个程序员用c++编写一个小程序，其实是很容易的。

max笔记

我有个2899行的，发不上去

北有云溪

#include<iostream>

#include<windows.h>

#include<time.h>

#include<stdlib.h>

#include<conio.h>

#define N 21

void gotoxy(int x,int y)//位置函数

{

COORD pos;

pos.X=2*x;

pos.Y=y;

SetConsoleCursorPosition(GetStdHandle(STD_OUTPUT_HANDLE),pos);

}

void color(int a)//颜色函数

{

SetConsoleTextAttribute(GetStdHandle(STD_OUTPUT_HANDLE),a);

}

void init(int apple[2])//初始化函数（初始化围墙、显示信息、苹果）

{

int i,j;//初始化围墙

int wall[N+2][N+2]={{0}};

for(i=1;i<=N;i++)

{

for(j=1;j<=N;j++)

wall[i][j]=1;

}

color(11);

for(i=0;i<N+2;i++)

{

for(j=0;j<N+2;j++)

{

if(wall[i][j])

std::cout<< "■";

else std::cout<<"□" ;

}

std::cout<< std::endl;

}

gotoxy(N+3,1);//显示信息

color(20);

std::cout<<"按 W S A D 移动方向"<< std::endl;

gotoxy(N+3,2);

color(20);

std::cout<<"按任意键暂停"<< std::endl;

gotoxy(N+3,3);

color(20);

std::cout<<"得分："<< std::endl;

apple[0]=rand()%N+1;//苹果

apple[1]=rand()%N+1;

gotoxy(apple[0],apple[1]);

color(12);

std::cout<<"●"<< std::endl;

}

int main()

{

int i,j;

int** snake=NULL;

int apple[2];

int score=0;

int tail[2];

int len=3;

char ch="p";

srand((unsigned)time(NULL));

init(apple);

snake=(int**)realloc(snake,sizeof(int*)*len);

for(i=0;i<len;i++)

snake[i]=(int*)malloc(sizeof(int)*2);

for(i=0;i<len;i++)

{

snake[i][0]=N/2;

snake[i][1]=N/2+i;

gotoxy(snake[i][0],snake[i][1]);

color(14);

std::cout<<"★"<< std::endl;

}

while(1)//进入消息循环

{

tail[0]=snake[len-1][0];

tail[1]=snake[len-1][1];

gotoxy(tail[0],tail[1]);

color(11);

std::cout<<"■"<< std::endl;

for(i=len-1;i>0;i--)

{

snake[i][0]=snake[i-1][0];

snake[i][1]=snake[i-1][1];

gotoxy(snake[i][0],snake[i][1]);

color(14);

std::cout<<"★"<< std::endl;

}

if(kbhit())

{

gotoxy(0,N+2);

ch=getche();

}

switch(ch)

{

case "w":snake[0][1]--;break;

case "s":snake[0][1]++;break;

case "a":snake[0][0]--;break;

case "d":snake[0][0]++;break;

default: break;

}

gotoxy(snake[0][0],snake[0][1]);

color(14);

std::cout<<"★"<< std::endl;

Sleep(abs(200-0.5*score));

if(snake[0][0]==apple[0]&&snake[0][1]==apple[1])//吃掉苹果后蛇分数加1，蛇长加1

{

score++;

len++;

snake=(int**)realloc(snake,sizeof(int*)*len);

snake[len-1]=(int*)malloc(sizeof(int)*2);

apple[0]=rand()%N+1;

apple[1]=rand()%N+1;

gotoxy(apple[0],apple[1]);

color(12);

std::cout<<"●"<< std::endl;

gotoxy(N+5,3);

color(20);

std::cout<<score<< std::endl;

}

if(snake[0][1]==0||snake[0][1]==N||snake[0][0]==0||snake[0][0]==N)//撞到围墙后失败

{

gotoxy(N/2,N/2);

color(30);

std::cout<<"失败！！！"<< std::endl;

for(i=0;i<len;i++)

free(snake[i]);

Sleep(INFINITE);

exit(0);

}

}

return 0;

}

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2023-01-12 07:53:571

骑士游历问题8*8的棋盘，骑士走日子，要把棋盘全部走一遍，不重复。不知道哪里错了，急！！！！

你最好写上注释,语法还错了不少;这个是我之前做的算法，但然我也是参考了网上提供的思想来做的。这个思想很广泛。把你输入的一个点为第一步，接着搜他可走8个方向中每一个可走的点，并记录这样可走的点的位置。再把这些记录的点再搜他可走的8个方向，但这些只要记录8个方向中可走方向的数目。接着就走方向最少的一条路。这样难走的路在前面已经走了，后面的路就好走很多了，一般都能找到方向。当然还有递归这种算法，但高维度的递归是没有意议的，电脑是算不出来的。下面是我的算法#include<iostream>using namespace std; //定义棋盘的大小#define N 8#define LOSE 0#define SUCCEED 1void printknightgo(); //输出结果int knightgo(int x,int y); //运算走法int chessboard[8][8] = { 0 }; //建立棋盘void main(){ int x,y; cout<<"请输入坐标"<<endl; cin>>x>>y; //输入坐标   if(knightgo(x,y)){ //开始运算走法，如果成功则返回成功，否则返回失败 cout<<"成功完成"<<endl; } else{ cout<<"失败"<<endl; } printknightgo(); //输出结果}//输出结果的算法void printknightgo(){ for(int i = 0;i < N; i++){ for(int j = 0;j < N;j++){ cout<<chessboard[i][j]<<" "; } cout<<endl; }}int knightgo(int x,int y){ chessboard[x][y] = 1;//将第一步标为1 int step;//走的步数 int nextx[8] = {-1,1,-2,2,-2,2,-1,1};//走的X的步法 int nexty[8] = {2,2,1,1,-1,-1,-2,-2};//走的Y的步法 int recordNextx[8]={0};//记住下步可走X的位置，并用count来记数 int recordNexty[8]={0};//记住下步可走Y的位置，并用count来记数 int tempx,tempy;//用于临时记住X和Y int i,j;//临时计数变量 int count = 0;//记住可循环的个数 int exist[8]={0};//第2次找8个方向每个可走的记录 for(int step = 2;step <=N*N;step++){//把输进来或循环的位置，找下一个能走的位置,并记住这些位置和可走的个数 for(i = 0;i < 8;i++){ //把上次记录重置0; recordNextx[i] = 0; recordNexty[i] = 0; exist[i] = 0; } count = 0; for(i = 0;i < 8;i++){//第一次循环，找可走的个位置，并记录可走方向的个数 tempx = x + nextx[i];//tempx为临时记录X tempy = y + nexty[i];//tempy为临时记录Y if(chessboard[tempx][tempy] == 0 && tempx >= 0 && tempx < N && tempy >= 0 && tempy < N){//当这个位置没走过和不走出盘外时，才能记录 recordNextx[count] = tempx; recordNexty[count] = tempy;//记录可走方向的x,y坐标 count++; //记录可以走的方向的个数 } } //把记住的位置,找他们的下个能走位置的个数,就是重复上面循环，只需记录可走方向的个数 if(count == 0){//如果没有出路则返回失败 return LOSE; } else {//如果只有一条路，则走这一条路 if(count == 1){ x = recordNextx[0];//因为可走方向只有一个，所记录中就有recordNext(x,y)[0]; y = recordNexty[0]; chessboard[x][y] = step; //把第几步写入这个棋盘的位置 } else{//有多条路可以走的话 for(j = 0;j < count;j++){//第一个点记录的可走的方向，并找出们下一个方向可以走的的方向的个数 for(i = 0;i < 8;i++){//找第2个点8个方向可走的方向 tempx = recordNextx[j] + nextx[i]; tempy = recordNexty[j] + nexty[i]; if(chessboard[tempx][tempy] == 0 && tempx >= 0 && tempx < N && tempy >= 0 && tempy < N){//当这个位置没走过和不走出盘外时，才能记录 exist[j]++;//记录第2个点可走方向的个数 } } } //找最方向个数最小的一条路 int min = exist[0]; int last = 0;//用于记住，recordNext(x,y)中可走方向中，能走方向数目最小的一个方向 for(i = 1;i < count;i++){//找出方向数目最小的数和方向 if(exist[i]<min){ min = exist[i]; last = i; } } x = recordNextx[last];//将这个方向给x,y; y = recordNexty[last]; chessboard[x][y] = step; //将这个步数写出这个棋盘 } } } return SUCCEED;}
2023-01-12 07:54:021

分治算法的应用实例

下面通过实例加以说明： 给你一个装有1 6个硬币的袋子。1 6个硬币中有一个是伪造的，并且那个伪造的硬币比真的硬币要轻一些。你的任务是找出这个伪造的硬币。为了帮助你完成这一任务，将提供一台可用来比较两组硬币重量的仪器，利用这台仪器，可以知道两组硬币的重量是否相同。比较硬币1与硬币2的重量。假如硬币1比硬币2轻，则硬币1是伪造的；假如硬币2比硬币1轻，则硬币2是伪造的。这样就完成了任务。假如两硬币重量相等，则比较硬币3和硬币4。同样，假如有一个硬币轻一些，则寻找伪币的任务完成。假如两硬币重量相等，则继续比较硬币5和硬币6。按照这种方式，可以最多通过8次比较来判断伪币的存在并找出这一伪币。另外一种方法就是利用分而治之方法。假如把1 6硬币的例子看成一个大的问题。第一步，把这一问题分成两个小问题。随机选择8个硬币作为第一组称为A组，剩下的8个硬币作为第二组称为B组。这样，就把1 6个硬币的问题分成两个8硬币的问题来解决。第二步，判断A和B组中是否有伪币。可以利用仪器来比较A组硬币和B组硬币的重量。假如两组硬币重量相等，则可以判断伪币不存在。假如两组硬币重量不相等，则存在伪币，并且可以判断它位于较轻的那一组硬币中。最后，在第三步中，用第二步的结果得出原先1 6个硬币问题的答案。若仅仅判断硬币是否存在，则第三步非常简单。无论A组还是B组中有伪币，都可以推断这1 6个硬币中存在伪币。因此，仅仅通过一次重量的比较，就可以判断伪币是否存在。假设需要识别出这一伪币。把两个或三个硬币的情况作为不可再分的小问题。注意如果只有一个硬币，那么不能判断出它是否就是伪币。在一个小问题中，通过将一个硬币分别与其他两个硬币比较，最多比较两次就可以找到伪币。这样，1 6硬币的问题就被分为两个8硬币（A组和B组）的问题。通过比较这两组硬币的重量，可以判断伪币是否存在。如果没有伪币，则算法终止。否则，继续划分这两组硬币来寻找伪币。假设B是轻的那一组，因此再把它分成两组，每组有4个硬币。称其中一组为B1，另一组为B2。比较这两组，肯定有一组轻一些。如果B1轻，则伪币在B1中，再将B1又分成两组，每组有两个硬币，称其中一组为B1a，另一组为B1b。比较这两组，可以得到一个较轻的组。由于这个组只有两个硬币，因此不必再细分。比较组中两个硬币的重量，可以立即知道哪一个硬币轻一些。较轻的硬币就是所要找的伪币。 在n个元素中找出最大元素和最小元素。我们可以把这n个元素放在一个数组中，用直接比较法求出。算法如下：void maxmin1(int A[],int n,int *max,int *min){ int i;*min=*max=A[0];for(i=0;i <= n;i++){ if(A[i]> *max) *max= A[i];if(A[i] < *min) *min= A[i];}}上面这个算法需比较2(n-1)次。能否找到更好的算法呢？我们用分治策略来讨论。把n个元素分成两组：A1={A[1],...,A[int(n/2)]}和A2={A[INT(N/2)+1],...,A[N]}分别求这两组的最大值和最小值，然后分别将这两组的最大值和最小值相比较，求出全部元素的最大值和最小值。如果A1和A2中的元素多于两个，则再用上述方法各分为两个子集。直至子集中元素至多两个元素为止。例如有下面一组元素：-13，13，9，-5，7，23，0，15。用分治策略比较的算法如下：void maxmin2(int A[],int i,int j,int *max,int *min)/*A存放输入的数据，i，j存放数据的范围，初值为0，n-1，*max,*min 存放最大和最小值*/{ int mid,max1,max2,min1,min2;if (j==i) {最大和最小值为同一个数;return;}if (j-1==i) {将两个数直接比较，求得最大会最小值；return；}mid=(i+j)/2;求i~mid之间的最大最小值分别为max1，min1;求mid+1~j之间的最大最小值分别为max2，min2;比较max1和max2，大的就是最大值;比较min1和min2，小的就是最小值;} 题目：在一个(2^k)*(2^k)个方格组成的棋盘上，有一个特殊方格与其他方格不同，称为特殊方格，称这样的棋盘为一个特殊棋盘。我们要求对棋盘的其余部分用L型方块填满（注：L型方块由3个单元格组成。即围棋中比较忌讳的愚形三角，方向随意），且任何两个L型方块不能重叠覆盖。L型方块的形态如下：题目的解法使用分治法，即子问题和整体问题具有相同的形式。我们对棋盘做一个分割，切割一次后的棋盘如图1所示，我们可以看到棋盘被切成4个一样大小的子棋盘，特殊方块必定位于四个子棋盘中的一个。假设如图1所示，特殊方格位于右上角，我们把一个L型方块（灰色填充）放到图中位置。这样对于每个子棋盘又各有一个“特殊方块”，我们对每个子棋盘继续这样分割，直到子棋盘的大小为1为止。用到的L型方块需要（4^k-1)/3 个，算法的时间是O（4^k)，是渐进最优解法。本题目的C语言的完整代码如下（TC2.0下调试），运行时，先输入k的大小，(1<=k<=6)，然后分别输入特殊方格所在的位置(x,y), 0<=x,y<=(2^k-1)。 #include<stdio.h>//#include<conio.h>//#include<math.h>inttitle=1;intboard[64][64];voidchessBoard(inttr,inttc,intdr,intdc,intsize){ints,t;if(size==1)return;t=title++;s=size/2;if(dr<tr+s&&dc<tc+s)chessBoard(tr,tc,dr,dc,s);else{board[tr+s-1][tc+s-1]=t;chessBoard(tr,tc,tr+s-1,tc+s-1,s);}if(dr<tr+s&&dc>=tc+s)chessBoard(tr,tc+s,dr,dc,s);else{board[tr+s-1][tc+s]=t;chessBoard(tr,tc+s,tr+s-1,tc+s,s);}if(dr>=tr+s&&dc<tc+s)chessBoard(tr+s,tc,dr,dc,s);else{board[tr+s][tc+s-1]=t;chessBoard(tr+s,tc,tr+s,tc+s-1,s);}if(dr>=tr+s&&dc>=tc+s)chessBoard(tr+s,tc+s,dr,dc,s);else{board[tr+s][tc+s]=t;chessBoard(tr+s,tc+s,tr+s,tc+s,s);}}voidmain(){intdr=0,dc=0,s=1,i=0,j=0;printf(printinthesizeofchess: );scanf(%d,&s);printf(printinspecalpointx,y: );scanf(%d%d,&dr,&dc);if(dr<s&&dc<s){chessBoard(0,0,dr,dc,s);for(i=0;i<s;i++){for(j=0;j<s;j++){printf(%4d,board[i][j]);}printf( );}}elseprintf(thewrongspecalpoint!! );getch();}
2023-01-12 07:54:081

c# 如何把画好的线放到容器上去?

form上放一个Button,在其单击事件里的代码如下private void button1_Click(object sender, EventArgs e){ ChessBoard c = new ChessBoard(); c.picctrl = pictureBox1; c.ShowChessBoard();}棋盘画的不错,呵呵.
2023-01-12 07:54:173

含有ch的单词

check children choose change child cheer chance chessboard cheese chessman Chinese
2023-01-12 07:54:296

l型骨牌的个数是

在一个2k×2k个方格组成的棋盘中，恰有一个方格与其他方格不同，称该方格为一特殊方格，且称该棋盘为一特殊棋盘。在棋盘覆盖问题中，要用图示的4种不同形态的L型骨牌覆盖给定的特殊棋盘上除特殊方格以外的所有方格，且任何2个L型骨牌不得重叠覆盖。易知，覆盖任意一个2k×2k的特殊棋盘，用到的骨牌数恰好为(4K-1)/3。
2023-01-12 07:54:512

unity中给自定义类中的属性赋值，报错“Object reference not set to an instance of an object”

是不是没有实例化
2023-01-12 07:54:592

c语言编的五子棋源代码

楼上的(),请问#include <bios.h> 中的bios.h是什么头文件啊?调试没通过,好象也没听人讲过有这个头文件呢.请指教.
2023-01-12 07:55:084

约客 古诗用了什么手法

　　百科名片　　《约客》是赵师秀的著名作品。该诗写的是诗人在一个风雨交加的夏夜独自期客的情景。诗歌采用写景寄情的写法，表达了诗人内心复杂的思想感情。是一首情景交融、清新隽永、耐人寻味的精妙小诗。　　诗词原文　　《约客》 南宋 赵师秀 黄梅时节家家雨， 青草池塘处处蛙。 有约不来过夜半， 闲敲棋子落灯花。　　编辑本段诗歌注释　　①约客：约请客人来相会。 ②黄梅时节：农历四、五月间，江南梅子黄了，熟了，大都是阴雨连连的时候，成为“梅雨季节”，所以称江南雨季为“黄梅时节”。意思就是夏初江南梅子黄熟的时节。 ③家家雨：家家户户都赶上下雨。形容处处都在下雨。 ④处处蛙：到处是蛙跳蛙鸣。 ⑤有约：即邀约友人。 ⑥落灯花：旧时以油灯照明，灯心烧残，落下来时好像一朵闪亮的小花。落：使……掉落。灯花：灯芯燃尽结成的花状物。 约客选自《清苑斋集》（《南宋群贤小集》本）。　　编辑本段译文　　一个梅雨绵绵的夜晚，乡村池塘中传来阵阵蛙鸣。这时已经是半夜了，诗人等候朋友如约来下棋。他无聊地敲着棋子，灯灰震落在棋盘上，心中有些失落。　　编辑本段英语翻译　　Amid a rainy night, Village pond, spreading waves of frogs. If waiting for friends to play chess at about middle of the night has passed, Bored tapping piece, light gray fall down on the chessboard.　　编辑本段诗文赏析　　前二句交待了当时的环境和时令。“黄梅”、“雨”、“池塘”、“蛙声”，写出了江南梅雨季节的夏夜之景：雨声不断，蛙声一片。读来使人如身临其境，仿佛细雨就在身边飘，蛙声就在身边叫。这看似表现得很“热闹”的环境，实际上诗人要反衬出它的“寂静”。 后二句点出了人物和事情。主人耐心地而又有几分焦急地等着，没事可干，“闲敲”棋子，静静地看着闪闪的灯花。第三句“有约不来过夜半”，用“有约”点出了诗人曾“约客”来访，“过夜半”说明了等待时间之久，本来期待的是约客的叩门声，但听到的却只是一阵阵的雨声和蛙声，比照之下更显示出作者焦躁的心情。第四句“闲敲棋子”是一个细节描写，诗人约客久候不到，灯芯很长，诗人百无聊赖之际，下意识地将黑白棋子在棋盘上轻轻敲打，而笃笃的敲棋声又将灯花都震落了。这种姿态貌似闲逸，其实反映出诗人内心的焦躁。 全诗通过对撩人思绪的环境及“闲敲棋子”这一细节动作的渲染，既写了诗人雨夜候客来访的情景，也写出约客未至的一种怅惘的心情，可谓形神兼备。全诗生活气息较浓，又摆脱了雕琢之习，清丽可诵。　　另一种：　　首句“黄梅时节家家雨”，交待了当时的环境。黄梅时节乃是立夏后数日梅子由青转黄之时，江南多雨，俗称黄梅天。其时细雨绵绵，正所谓“自在飞花轻似梦，无边丝雨细如愁”。对于视觉，是一种低沉的安慰。至于雨敲在鳞鳞千瓣的瓦上，由远而近，轻轻重重轻轻，夹着一股股的细流沿瓦槽与屋檐潺潺泄下，各种敲击音与滑音密织成网”，仿佛“谁的千指百指在按摩耳轮”，心情异常恬静安祥。 “青草池塘处处蛙”这句，诗人的注意力从霏霏淫雨，自然而然地转到了远远近近，此起彼伏的片片蛙声，正是这处处蛙声，烘托出了当时周遭的清静，试想，如非心如止水，神游物外，而是焦灼烦躁，何以知微渺“虫声”今夜“新透绿窗纱”？ 再看第三句“有约不来过夜半”。我猜想，书中之所以得出“焦灼”结论，多半便依了这句。朋友过了夜半还不来，倘是你我，当然不免焦灼。但这是赵师秀，是“永嘉四灵”之一，人称“鬼才”的赵师秀。赵师秀，字紫芝，又字灵秀，光宗绍熙元年进士，曾任上元县主薄，筠州推官。他虽寄身仕宦，但失意消沉，常与僧道同游山水之间，向往恬静淡泊的生活，甚至还想与陶渊明一样“归寻故园”（《九客一羽衣泛舟，分韵得尊字，就送朱几仲》）。他死后，江湖派巨子戴复古作《哭赵紫芝》，说他是“东晋时人物”。当不致于“有约不来过夜半”便焦灼不安吧？ 最后一句“闲敲棋子落灯花”。我不知道前人是怎么理解“闲”字的。我是这样想，“闲敲”之“闲”，应当仿佛我们偶凭小几，百无聊赖，适见案头笔墨，于是顺手拿过，随随便便，漫不经心，信笔涂去，一如陆游“矮纸斜行闲作草”之意趣。赵师秀也便这样坐于灯前，遥等客人不至，百无聊赖，适见局中棋子，于是顺手拈起，随随便便，漫不经心，信手敲去，何来焦灼之感？　　《约客》这首诗究竟营造了一个什么样的意境?且看——江南的夏夜，梅雨纷飞，蛙声齐鸣，诗人约了友人来下棋，然而，时过夜半，约客未至，诗人闲敲棋子，静静等候……此时，诗人的心情如何呢？我看主要不是或根本就没有什么焦躁和烦闷的情绪，而更可能是一种闲逸、散淡和恬然自适的心境。也许曾有那么一会儿焦躁过（这种焦躁情绪怎么会持续到“过夜半”呢？），但现在，诗人被眼前江南夏夜之情之景感染了：多情的梅雨，欢快的哇鸣，闪烁的灯火，清脆的棋子敲击声……这是一幅既热闹又冷清、既凝重又飘逸的画面。也许诗人已经忘了他是在等友人，而完全沉浸到内心的激荡和静谧中。应该感谢友人的失约，让诗人享受到了这样一个独处的美妙的不眠之夜。　　作者简介　　赵师秀(1170～1219) 南宋诗人。字紫芝、灵芝，号灵秀，又号天乐。永嘉（今浙江温州）人。光宗绍熙元年(1190)进士,与徐照（字灵晖）、徐玑（字灵渊）、翁卷（字灵舒）并称“永嘉四灵”，开创了“江湖派”一代诗风。宁宗庆元元年(1195)任上元主簿，后为筠州推官。晚年宦游，逝于临安。有《赵师秀集》2卷，别本《天乐堂集》1卷，已佚。其《清苑斋集》1卷，有《南宋群贤小集》本,《永嘉诗人祠堂丛刻》本。 赵师秀是“永嘉四灵”中较出色的作家。诗学姚合、贾岛，尊姚、贾为“二妙”。所编《二妙集》选姚诗121首、贾诗 81首。绝大部分是五言诗。又编《众妙集》，从沈□期起，共76家，排除杜甫，而选刘长卿诗却多达23首，编选宗旨与《二妙集》相似，大约是对《二妙集》的补充。作诗尚白描，反对江西派的“资书以为诗”。然而他自己在写诗实践中，对姚、贾诗及与姚、贾风格近似的诗，往往或袭其命意、或套用其句法。如姚合《送宗慎言》“驿路多连水，州城半在云。”赵诗《薛氏瓜庐》则有“野水多于地,春山半是云。”即是一例(《诗人玉屑》卷十九引黄□语)。他比较擅长五律，中间两联描写景物，偶有警句，如《桐柏观》中的“瀑近春风湿，松多晓日青”之类。然而通体完整者不多。自称“一篇幸止有四十字，更增一字，吾末如之何矣！”（刘克庄《野谷集序》引）可见枯窘之状。七绝《约客》写道：“黄梅时节家家雨，青草池塘处处蛙，有约不来过夜半，闲敲棋子落灯花。”明净自然，被不少选本选录。 代表作 【约客】 黄梅时节家家雨，青草池塘处处蛙。 有约不来过夜半，闲敲棋子落灯花。 【数日】 数日秋风欺病夫，尽吹黄叶下庭芜。 林疏放得遥山出，又被云遮一半无。 【庵西】 数里庵西路，东风去更吹。稻寒生叶细，松老作花迟。 小憩嫌无侣，曾游忆有诗。新萤光尚小，未暗出空池。 【岩居僧】 开扉在石层，尽日少人登。一鸟过寒木，数花摇翠藤。 茗煎冰下水，香炷佛前灯。吾亦逃名者，何因似此僧。　　创作背景　　全诗通过对撩人思绪的环境及“闲敲棋子”这一细节动作的渲染，与人约会而久候不至，既写了诗人雨夜候客来访的情景，也写出约客未至的一种怅惘,稍有些失落的心情，可谓形神兼备。全诗生活气息较浓，又摆脱了雕琢之习，语言清丽可诵。
2023-01-12 07:55:223

关于java写程序显示不出窗体

从异常情况来看，问题可能处在这一行代码：image=ImageIO.read(getClass().getResource("0.pgn"));我推测有以下两种可能：1、0.pgn 文件位置错误，应该放到 src 下，编译后在 class 文件夹下。2、0.pgn 文件名错误，应该是 0.png
2023-01-12 07:55:331

棋盘checkerboard
2023-01-12 07:55:436

makefile生成的文件为什么不是可执行文件

当然不会生成。因为你没有生成ttt的规则啊。ttt : main.o chessboard.o fileio.o neuralnetwork.o base.o engine.o train.ogccflags = -g -wall -i/usr/local/include -i$(curdir)/includelibs = -l/usr/local/lib 这段应该改为gccflags = -g -wall -i/usr/local/include -i$(curdir)/includelibs = -l/usr/local/libttt : main.o chessboard.o fileio.o neuralnetwork.o base.o engine.o train.o g++ -o $@ $^ $(gccflags) $(libs)
2023-01-12 07:56:042

棋盘覆盖问题(高分回报)

看不懂
2023-01-12 07:56:132

里,最激励人的是哪句台词呢?

Hope is a good thing, maybe the best of things, and no good thing ever dies.
2023-01-12 07:56:216

第十三届全国青少年信息学奥林匹克联赛初赛试题及答案及答案

2023-01-12 07:56:421

雅思口语part2卡片能看着说吗

雅思口语对于雅思备考的同学来讲都是一个不小的难点。雅思口语PART3是很难的，小编建议大家在应对这块的时候，要学会随机应变。一般来说雅思口语考一共有三个部分，其中PART3算是特别难的，这方面也综合考察烤鸭的多个方面，所以考生在这一部分，必须要做好应变的准备。PART3那么难，该如何准备呢?下面小编给大家分享关于该如何备考雅思口语part3的相关内容，希望可以帮到正在雅思备考的同学。雅思口语part3那么难该如何备考雅思口语三部分的问题难度是有所增加的，有的很抽象，有的令人困惑，雅思考官的脑洞是什么问题都问得出来的，因此你要学会随机应变。在应对雅思口语考的时候，对于比较难的三部分的问题，考生们要先听懂问题，若实在没有确定好其含义或没有明白，可以提问以求确定，然后尽量避免在没有想清楚或是没有反应出一个关键词的含义时，草率作答。多看例题的分析，需要大明确的是，抓住问题中的关键词是非常重要的，此外，环球教育雅思周末班老师建议大尽量以例子来使自己的答案更清楚、更“接地气”，然后借用雅思写作中的一些答题思路，使用简单的语言，将自己的见解表达清楚，这样才能在三部分提高你的分数，达到提升的效果。通过题目和分析，我们不难看出，雅思口语三部分的问题难度是在增加的，尤其是近期所出现的话题，有抽象的比如happiness,有令人困惑的比如音乐噪音等等，这些都是我们平时在准备时需要多考虑，多练习的。很多同学由于在平时的备考中，过于偏重二部分的准备，导致一、二、三部分不均衡，三部分完全暴露了自己的英语临场发挥能力缺陷，因此直接影响了二部分的成绩。其实，大只要平时多思考问题，结合雅思写作中关于社会话题的分析，多积累素材，是可以应对这些题目的。
2023-01-12 07:56:494

请大神帮忙设计一个完整的c语言程序，实现”在4乘4的棋盘上放置8个棋,要求每一行每一列上只能放置2个”

楼主你好。见丑了。。我用了一个穷举的方法。我写的代码的好处在于易于扩展，N代表了棋盘的大小，你可以任意修改，不止是4。【经过几次优化后变为以下代码，采用回溯法】#include <stdio.h>#include <stdlib.h>#define N 6#define NOPRINTING //标志是否打印数组//如果只是计算可能的情况的话，根本不需要N×N的棋盘数组。void printMatrix(int* matrix, int m, int n){int i,j;for(i=0;i<m;i++){for(j=0;j<n;j++){printf("%d ",matrix[n*i+j]);}printf(" ");}}void recusion(int* chessBoard, int n, int iter, int* count, int* col){int i,j;#ifndef NOPRINTINGint* currentRow = &chessBoard[n*(n-iter)];#endifif(iter == 0){(*count)++;#ifndef NOPRINTINGprintMatrix(chessBoard, n, n);printf("=============================== ");#endifreturn;}for(i=0;i<n-1;i++){//布置每行第一个棋子for(j=i+1;j<n;j++){//布置每行第二个棋子#ifndef NOPRINTINGcurrentRow[i]=currentRow[j]=1;//重写当前行#endifif(col[i]+1 <=2 && col[j]+1 <=2){//有效col[i]++;col[j]++;recusion(chessBoard, n, iter-1, count, col);//若当前仍满足要求，对下一行进行递归。col[i]--;col[j]--;}#ifndef NOPRINTINGcurrentRow[i]=currentRow[j]=0;//清空当前行#endif}}}void putChess(int* chessBoard, int n, int* count){int* col=(int*)calloc(1,sizeof(int)*n);//标记每一列放置的棋子数recusion(chessBoard, n, n, count, col);free(col);}void main(){int count=0;int chessBoard[N][N]={0};printf("==========START TO PUT CHESS!=========== ");putChess((int*)chessBoard, N, &count);printf("All %d possible deploy! ", count);}输出：【N为3且选择打印】==========START TO PUT CHESS!===========1 1 01 0 10 1 1===============================1 1 00 1 11 0 1===============================1 0 11 1 00 1 1===============================1 0 10 1 11 1 0===============================0 1 11 1 01 0 1===============================0 1 11 0 11 1 0===============================All 6 possible deploy!Press any key to continue【N为8的时候，有近两亿种情况，花了将近一分钟的时间算完。】==========START TO PUT CHESS!===========All 187530840 possible deploy!Press any key to continue
2023-01-12 07:57:036

求、博尔赫斯的《 棋》（部分）西班牙文 译 英文

先此说明--以下内容转载它处（网址在参考资料栏中）In their solemn corner, the playersgovern the lingering pieces. The chessboarddelays them until daybreak in its severesphere in which colors are hateful.Inside they radiate magical severitythe forms: Homeric tower, lighthorse, armed queen, last king,oblique bishop and attacking pawns.When the players will have gone,when time will have consumed them,certainly the ritual will have not ceased.In the Orient this war was litwhich amphitheater is today all the earth.As the other, this game is infinite.2Fainting king, slanting bishop, fiercequeen, straightforward tower and cunning pawnon the black and white pathsearching and fighting their armed battle.They ignore the player"s pointing handgoverns his destiny,they ignore that a tamed severityholds his will and day.The player is himself a prisoner(the sentence is Omar"s) of another boardof dark nights and light days.God moves the player, and he, the chess piece.Which God behind God begins the conspiracyof dust and time and dream and agony?
2023-01-12 07:57:251

chessboard造句

chessboard造句：He hit my nose with the chess board.他用棋盘打我的鼻子。32 dominoes can cover the whole chess board.32张骨牌能盖住整个棋盘。These are the two metals I used for the squares on the chess board.这是我为两个金属棋盘上使用的平方。Enter another novelty: the battlefield looks similar to a chess board.
2023-01-12 07:57:351

棋盘的英语翻译 棋盘用英语怎么说

chessboard 棋盘
2023-01-12 07:57:522

国际象棋英语介绍

Chess is a recreational and competitive game played between two players. Sometimes called Western chess or international chess to distinguish it from its predecessors and other chess variants, the current form of the game emerged in Southern Europe during the second half of the 15th century after evolving from similar, much older games of Indian and Persian origin. Today, chess is one of the world"s most popular games, played by millions of people worldwide in clubs, online, by correspondence, in tournaments and informally.The game is played on a square chequered chessboard with 64 squares arranged in an eight-by-eight square. At the start, each player (one controlling the white pieces, the other controlling the black pieces) controls sixteen pieces: one king, one queen, two rooks, two knights, two bishops, and eight pawns. The object of the game is to checkmate the opponent"s king, whereby the king is under immediate attack (in "check") and there is no way to remove it from attack on the next move.The tradition of organized competitive chess started in the sixteenth century and has developed extensively. Chess today is a recognized sport of the International Olympic Committee. The first official World Chess Champion, Wilhelm Steinitz, claimed his title in 1886; Viswanathan Anand is the current World Champion. Theoreticians have developed extensive chess strategies and tactics since the game"s inception. Aspects of art are found in chess composition.One of the goals of early computer scientists was to create a chess-playing machine, and today"s chess is deeply influenced by the abilities of current chess programs and by the possibility to play online. In 1997, a match between Garry Kasparov, then World Champion, and a computer proved for the first time that machines are able to beat even the strongest human players.
2023-01-12 07:58:001

国际象棋中的车用英语表示？

rook或者castle
2023-01-12 07:58:063

左边一个木字右边一个秋读什么

楸qiū部首笔画部首:木 部外笔画:9 总笔画:13五笔86:STOY 五笔98:STOY 仓颉:DHDF笔顺编号:1234312344334 四角号码:49980 Unicode:CJK 统一汉字 U+6978基本字义1. 落叶乔木，干高叶大，木材质地致密，耐湿，可造船，亦可做器具：～局。～枰（棋盘。古代多用楸木做成）。详细字义〈名〉1. 落叶乔木,干高叶大,夏天开黄绿色细花,木材质地致密,可做器具 [Chinese catalpa]揪,梓也。——《说文》望长楸而太息兮。——《楚辞·哀郢》2. 棋盘 [chessboard]。如:楸局(楸木制的围棋盘);楸玉局(苍青色玉石制的围棋盘);楸枰(棋盘)
2023-01-12 07:58:171

c语言，怎样讲整形的数字赋值给一个字符串数组？

可以用sprintf示例：char a[100];int b =312348;sprintf(a,"%d",b);
2023-01-12 07:58:292

python解决四皇后问题

import randomimport pprintchessboard = [ [" "," "," "," "," "," "," "," ",], [" "," "," "," "," "," "," "," ",], [" "," "," "," "," "," "," "," ",], [" "," "," "," "," "," "," "," ",], [" "," "," "," "," "," "," "," ",], [" "," "," "," "," "," "," "," ",], [" "," "," "," "," "," "," "," ",], [" "," "," "," "," "," "," "," ",], ]def putqueen(pos): global chessboard i,j = pos delta = i-j if chessboard[i][j] != " ": return False for m in range(8): if chessboard[m][j] == " ": chessboard[m][j] = "x" if chessboard[i][m] == " ": chessboard[i][m] = "x" if 0 <= m+i-j < 8 and chessboard[m+i-j][m] == " ": chessboard[m+i-j][m] = "x" if 0 <= i+j-m < 8 and chessboard[m][i+j-m] == " ": chessboard[m][i+j-m] = "x" chessboard[i][j] = "O" print pos return Truedef getRandPos(): return random.randint(0,7), random.randint(0,7)def show(): global chessboard print " ".join([" ".join(r) for r in chessboard])# 1stputqueen(getRandPos())# 2ndwhile True: if putqueen(getRandPos()): break# 3rdwhile True: if putqueen(getRandPos()): break# 4thwhile True: if putqueen(getRandPos()): breakshow()
2023-01-12 07:58:462

makefile 可执行文件没生成

当然不会生成。因为你没有生成TTT的规则啊。TTT : main.o ChessBoard.o FileIO.o NeuralNetwork.o Base.o Engine.o Train.oGCCFLAGS = -g -Wall -I/usr/local/include -I$(CURDIR)/includeLIBS = -L/usr/local/lib 这段应该改为GCCFLAGS = -g -Wall -I/usr/local/include -I$(CURDIR)/includeLIBS = -L/usr/local/libTTT : main.o ChessBoard.o FileIO.o NeuralNetwork.o Base.o Engine.o Train.o    g++ -o $@ $^  $(GCCFLAGS) $(LIBS)
2023-01-12 07:58:571

国际象棋中象英语怎么拼? 最好有车马象3个的英语说法

国际象棋 chess 棋盘 chessboard 棋子 piece 王 king 后 queen 车 chariot 象 bishop 马 knight; horse 兵 pawn
2023-01-12 07:59:031

国际象棋英语介绍

Chess is a recreational and competitive game played between two players. Sometimes called Western chess or international chess to distinguish it from its predecessors and other chess variants, the current form of the game emerged in Southern Europe during the second half of the 15th century after evolving from similar, much older games of Indian and Persian origin. Today, chess is one of the world"s most popular games, played by millions of people worldwide in clubs, online, by correspondence, in tournaments and informally.The game is played on a square chequered chessboard with 64 squares arranged in an eight-by-eight square. At the start, each player (one controlling the white pieces, the other controlling the black pieces) controls sixteen pieces: one king, one queen, two rooks, two knights, two bishops, and eight pawns. The object of the game is to checkmate the opponent"s king, whereby the king is under immediate attack (in "check") and there is no way to remove it from attack on the next move.The tradition of organized competitive chess started in the sixteenth century and has developed extensively. Chess today is a recognized sport of the International Olympic Committee. The first official World Chess Champion, Wilhelm Steinitz, claimed his title in 1886; Viswanathan Anand is the current World Champion. Theoreticians have developed extensive chess strategies and tactics since the game"s inception. Aspects of art are found in chess composition.One of the goals of early computer scientists was to create a chess-playing machine, and today"s chess is deeply influenced by the abilities of current chess programs and by the possibility to play online. In 1997, a match between Garry Kasparov, then World Champion, and a computer proved for the first time that machines are able to beat even the strongest human players.
2023-01-12 07:59:081

求宇多田光 colors 歌词

宇多田光 - Colors作词者名 宇多田ヒカル作曲者名 宇多田ヒカルミラーが映し出す幻を気にしながらいつの间にか速度上げてるのさどこへ行ってもいいと言われると半端な愿望には标识も全部灰色だ炎の揺らめき 今宵も梦を描くあなたの笔先 渇いていませんか青い空が见えぬなら青い伞広げていいじゃないか キャンパスは君のもの白い旗はあきらめた时にだけかざすの今は真っ赤に 诱う闘牛士のようにカラーも色褪せる蛍光灯の下白黒のチェスボードの上で君に出会った仆らは一时 迷いながら寄り添ってあれから一月 忆えていますかオレンジ色の夕日を隣で见てるだけでよかったのにな 口は灾いの元黒い服は死者に祈る时にだけ着るのわざと真っ赤に残したルージュの痕もう自分には梦の无い絵しか描けないと言うなら涂り溃してよ キャンパスを何度でも白い旗はあきらめた时にだけかざすの今の私はあなたの知らない色终わるMIRAA ga utsushidasu maboroshi woki ni shinagara itsunomanika sokudo ageteru no sadoko e ittemo iito iwarerutohanpa na ganbou niwa hyoushiki mo zenbu haiiro dahonou no yurameki koyoi mo yume wo egakuanata no fudesaki kawaite imasenka ?aoi sora ga mienu nara aoi kasa hirogeteiija naika KYANBASU wa kimi no monoshiroi hata wa akirameta toki ni dake kazasu noima wa makka ni sasou tougyushi no you niKARAA mo iroaseru keikoutou no motoshirokuro no chessboard no ue de kimi ni deattabokura wa hitotoki mayoinagara yorisottearekara hitotsuki oboete imasuka ?ORENJI iro no yuuhi wo tonari de miteru dake deyokatta noni ahh kuchi wa wazawai no motokuroi fuku wa shisya ni inoru toki ni dake kiru nowazato makka ni nokoshita RUJU no atomou jibun ni wa yume no nai e shika kakenai to iu naranuritsubushite yo KYANBASU wo nando demoshiroi hata wa akirameta toki ni dake kazasu noima no watashi wa anata no shiranai iro注意著镜中映照的幻影不知不觉间 我已在加速被告知去那儿也很好即使不认真的热望中充满的 全是灰色的交通讯号火在摇晃 今夜你也在绘梦你的笔尖 是否已经乾涸?如果看不到蔚蓝的天空 开一把蓝色的伞没问题的 帆布是你的东西白色的旗 是否你在放弃时手握的东西?像一个现在以深红诱惑的斗牛士一样色彩皆消失无踪的萤光灯下我在黑白色的棋盘上跟你会面我们曾在一时迷茫中拥抱一个月过去了 记忆还在吗?只是看在你身旁的橙色日落已很好 口里出来的是邪恶穿著黑色的衣服时你是否只为死者默哀?残留的胭脂是为了些什麼而深红的如果说你只描绘没有梦的画再在帆布上重覆涂上直至你满意白色的旗 是否你在放弃时手握的东西?现在的我 是你所不知道的颜色
2023-01-12 07:59:171

......"chess piece" 是什么意思?

chess 棋 piece 块
2023-01-12 07:59:223

我的程序的参数是正确的，却出现编译错误：“AfxBeginThread”: 2 个重载中没有一个可以转换所有参数类型

threadFun的声明不符合AfxBeginThread的需求要按这个需求：UINT MyControllingFunction( LPVOID pParam );修改为这个看看：UINT __cdecl threadFun( LPVOID pParam ){ ChessBoard(0,0,1,1,16);}
2023-01-12 07:59:361

用java写一个国际象棋的棋盘，输出结果要是一张 国际象棋的棋盘

0
2023-01-12 07:59:512

java数组调用的问题

首先应该调用初始化方法initBoard()Chessboard chessboard=new Chessboard();chessboard.initBoard();String[][] board = chessboard.getBoard();或者把数组的初始化放到构造函数中去
2023-01-12 08:00:072

怎么可以让java在五子棋盘上面随机产生棋子的位置？ 下面是棋盘的代码

看下棋盘的row和column各多少行，然后建立一个2d数组boolean[][] pos = new boolean[row][column];初始化2d数组为boolean并且全部赋值为false你要做的就是随机产生row和column，用Random里的nextInt(row 或者column)随机产生点，并将pos[row][column] = true;
2023-01-12 08:00:151

正式国际象棋比赛中用的棋盘尺寸是多大？有多少个棋子？

48*4832个棋子
2023-01-12 08:00:214

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