五子棋：白棋要怎样取胜征答：题目如图.只需答出第14手即可!采纳第一个正答者为最佳答案.附：如果不仅给出了正确的14手,

这道题不算难!告诉你答案,好好理解吧!
说明：C(M,m)表示在M个中取m个的组合
在棋盘上能形成N子的方式如下
横竖线每条线上有16-N种,全部有(16-N)*15*2
对角线有1+2+3+...+(15-N)+(16-N)+(15-N)+...+1=(16-N)^2,全部有2*(16-N)^2
共有2*(16-N)(31-N)种.
(注意：当N=1时对角线与横竖线是重复,只要计算其中一种15*15=225!)
由于全随机落子,最后形成N子与次序无关!
黑棋先走的人有113个子,白棋只有112个子,全部共有C(225,113)种
可以看成先取N个放在能成N个子的位置上,再把剩余的棋子随机放置!
黑：2*(16-N)(31-N)*C(225-N,113-N)
P=2*(16-N)(31-N)*C(225-N,113-N) /C(225,113)
白：2*(16-N)(31-N)*C(225-N,112-N)
P=2*(16-N)(31-N)*C(225-N,112-N) /C(225,112)
有什么不明白,自己想半小时再问我!

必然先手有优势.先说第一个问题,问在数学板块里并不好,你还需要多学习呀.数学家再解释任何一个数学概念时都喜欢用集合论来解释.现在把象棋归为数学问题还不现实.
对于很多博弈类游戏一般都是先手方优势.象棋也是这样的,怎样达到公平是一个值得讨论的话题.对于围棋来说并不是仅仅贴几目就可以完成公平的.比如武宫正树的宇宙流在白棋时基本没法用.
以前比赛都是下两盘,每次交换先后手,平了就加赛快棋.但是这样后来大量的和棋导致了象棋生命力的下降.胡荣华提出了红贴分黑贴时的办法,就是红方用时加长黑用时减短,但是和棋的话黑得分要多.网友推荐的答案没说贴时.而且中炮对屏风马一直都是屏风马方在防御,你说均势绝对是红方的失败.就像棋谱中所说黑方取得可以抗衡的局面时往往红不满.对于国际象棋来说黑方的定式往往叫什么什么防御,如卡罗康防御,法兰西防御,西西里防御,古印度防御.而白方的定式都不以防御冠名.这足以说明先手的威力.值得一说的是在国际象棋残局时有一种奇怪的现象就是有时会出现谁先走谁吃亏的情况叫“触网”.

In this paper, computer Gobang game system as a subject, in accordance with artificial intelligence and computer game design of the general principles of a Gobang game system's basic model, and based on the analysis of the plane tree Game Game. Gobang game and the algorithm a number of effective research, analysis of the game tree Minimax search technology, and on this basis, the Alpha-Beta pruning and pruning process optimization problem. It also has Alpha-Beta pruning to improve the algorithm NegaScout algorithm, the algorithm first to adopt a limited window of the Alpha-Beta, to determine the scope of the actual valuation, and then in this small area in search of the actual estimated value, because In the smaller scope of the search, efficiency can be improved.
The system can be a variety of algorithm Game features, including Minimax algorithm, Alpha-Beta search, based on historical inspiration optimize the Alpha-Beta search, based on the replacement table and history inspired to explore enhanced window (NegaScout) 4种Algorithm. Procedures at the same time can be adjusted Game tree depth of the search, the algorithm is time-consuming and nodes to Traversal statistics, intuitive reaction different game difference between the algorithm.

1
（1）两枚正面都朝上的可能性是：1÷4=——
4
1
（2）两枚反面都朝上的可能性是：1÷4=——
4
2 1
（3）一枚正面朝上、一枚反面朝上的可能性是：2÷4= —— =——
4 2

（3n-2)和（2n+1）都是正确答案,因为这两个式子其实是一样的

数学解法我不会,但我会用五子棋的思路给你解.
1楼思路是最好的：用防守效率最高的八卦阵来解决这个问题再恰当不过了.
如果没有差错的话,在一张标准棋盘上布满八卦阵,需要45个棋子（这个确实有以偏盖全的可能,因为随着棋形的偏移不排除增多或减少棋子的可能,例如44个,或46个,毕竟我不是用数学的方法来思考的）.
八卦阵的特点是：如果棋盘上此阵完全布满,那么棋盘中再无成五可能.所以说,楼主问题的答案应该是225-45=180.
我对五子棋非常熟,但说实话数学不算强,因此并不能给出数学算法.关于八卦阵这条思路,楼主可以拿个棋盘自己实验一下.

2盘
A和B、C、D、E分别1盘
B和A、C、E分别1盘
C和A、B分别1盘
D和A
E和A、B

#include #include #include #include #include #define CROSSRU 0xbf /*右上角点*/ #define CROSSLU 0xda /*左上角点*/ #define CROSSLD 0xc0 /*左下角点*/ #define CROSSRD 0xd9 /*右下角点*/ #define CROSSL 0xc3 /*左边*/ #define CROSSR 0xb4 /*右边*/ #define CROSSU 0xc2 /*上边*/ #define CROSSD 0xc1 /*下边*/ #define CROSS 0xc5 /*十字交叉点*/ /*定义棋盘左上角点在屏幕上的位置*/ #define MAPXOFT 5 #define MAPYOFT 2 /*定义1号玩家的操作键键码*/ #define PLAY1UP 0x1157/*上移--'W'*/ #define PLAY1DOWN 0x1f53/*下移--'S'*/ #define PLAY1LEFT 0x1e41/*左移--'A'*/ #define PLAY1RIGHT 0x2044/*右移--'D'*/ #define PLAY1DO 0x3920/*落子--空格键*/ /*定义2号玩家的操作键键码*/ #define PLAY2UP 0x4800/*上移--方向键up*/ #define PLAY2DOWN 0x5000/*下移--方向键down*/ #define PLAY2LEFT 0x4b00/*左移--方向键left*/ #define PLAY2RIGHT 0x4d00/*右移--方向键right*/ #define PLAY2DO 0x1c0d/*落子--回车键Enter*/ /*若想在游戏中途退出, 可按 Esc 键*/ #define ESCAPE 0x011b /*定义棋盘上交叉点的状态, 即该点有无棋子 */ /*若有棋子, 还应能指出是哪个玩家的棋子 */ #define CHESSNULL 0 /*没有棋子*/ #define CHESS1 'O'/*一号玩家的棋子*/ #define CHESS2 'X'/*二号玩家的棋子*/ /*定义按键类别*/ #define KEYEXIT 0/*退出键*/ #define KEYFALLCHESS 1/*落子键*/ #define KEYMOVECURSOR 2/*光标移动键*/ #define KEYINVALID 3/*无效键*/ /*定义符号常量: 真, 假 --- 真为1, 假为0 */ #define TRUE 1 #define FALSE 0 /**********************************************************/ /* 定义数据结构 */ /*棋盘交叉点坐标的数据结构*/ struct point { int x,y; }; /**********************************************************/ /*自定义函数原型说明 */ void Init(void); int GetKey(void); int CheckKey(int press); int ChangeOrder(void); int ChessGo(int Order,struct point Cursor); void DoError(void); void DoOK(void); void DoWin(int Order); void MoveCursor(int Order,int press); void DrawCross(int x,int y); void DrawMap(void); int JudgeWin(int Order,struct point Cursor); int JudgeWinLine(int Order,struct point Cursor,int direction); void ShowOrderMsg(int Order); void EndGame(void); /**********************************************************/ /**********************************************************/ /* 定义全局变量 */ int gPlayOrder; /*指示当前行棋方 */ struct point gCursor; /*光标在棋盘上的位置 */ char gChessBoard[19][19];/*用于记录棋盘上各点的状态*/ /**********************************************************/ /**********************************************************/ /*主函数*/ void main() { int press; int bOutWhile=FALSE;/*退出循环标志*/ printf("Welcome www.***.com"); Init();/*初始化图象，数据*/ while(1) { press=GetKey();/*获取用户的按键值*/ switch(CheckKey(press))/*判断按键类别*/ { /*是退出键*/ case KEYEXIT: clrscr();/*清屏*/ bOutWhile = TRUE; break; /*是落子键*/ case KEYFALLCHESS: if(ChessGo(gPlayOrder,gCursor)==FALSE)/*走棋*/ DoError();/*落子错误*/ else { DoOK();/*落子正确*/ /*如果当前行棋方赢棋*/ if(JudgeWin(gPlayOrder,gCursor)==TRUE) { DoWin(gPlayOrder); bOutWhile = TRUE;/*退出循环标志置为真*/ } /*否则*/ else /*交换行棋方*/ ChangeOrder(); ShowOrderMsg(gPlayOrder); } break; /*是光标移动键*/ case KEYMOVECURSOR: MoveCursor(gPlayOrder,press); break; /*是无效键*/ case KEYINVALID: break; } if(bOutWhile==TRUE) break; } /*游戏结束*/ EndGame(); } /**********************************************************/ /*界面初始化，数据初始化*/ void Init(void) { int i,j; char *Msg[]= { "Player1 key:", " UP----w", " DOWN--s", " LEFT--a", " RIGHT-d", " DO----space", "", "Player2 key:", " UP----up", " DOWN--down", " LEFT--left", " RIGHT-right", " DO----ENTER", "", "exit game:", " ESC", NULL, }; /* 先手方为1号玩家 */ gPlayOrder = CHESS1; /* 棋盘数据清零, 即棋盘上各点开始的时候都没有棋子 */ for(i=0;i<19;i++) for(j=0;j<19;j++) gChessBoard[j]=CHESSNULL; /*光标初始位置*/ gCursor.x=gCursor.y=0; /*画棋盘*/ textmode(C40); DrawMap(); /*显示操作键说明*/ i=0; textcolor(BROWN); while(Msg!=NULL) { gotoxy(25,3+i); cputs(Msg); i++; } /*显示当前行棋方*/ ShowOrderMsg(gPlayOrder); /*光标移至棋盘的左上角点处*/ gotoxy(gCursor.x+MAPXOFT,gCursor.y+MAPYOFT); } /*画棋盘*/ void DrawMap(void) { int i,j; clrscr(); for(i=0;i<19;i++) for(j=0;j<19;j++) DrawCross(i,j); } /*画棋盘上的交叉点*/ void DrawCross(int x,int y) { gotoxy(x+MAPXOFT,y+MAPYOFT); /*交叉点上是一号玩家的棋子*/ if(gChessBoard[x][y]==CHESS1) { textcolor(LIGHTBLUE); putch(CHESS1); return; } /*交叉点上是二号玩家的棋子*/ if(gChessBoard[x][y]==CHESS2) { textcolor(LIGHTBLUE); putch(CHESS2); return; } textcolor(GREEN); /*左上角交叉点*/ if(x==0&&y==0) { putch(CROSSLU); return; } /*左下角交叉点*/ if(x==0&&y==18) { putch(CROSSLD); return; } /*右上角交叉点*/ if(x==18&&y==0) { putch(CROSSRU); return; } /*右下角交叉点*/ if(x==18&&y==18) { putch(CROSSRD); return; } /*左边界交叉点*/ if(x==0) { putch(CROSSL); return; } /*右边界交叉点*/ if(x==18) { putch(CROSSR); return; } /*上边界交叉点*/ if(y==0) { putch(CROSSU); return; } /*下边界交叉点*/ if(y==18) { putch(CROSSD); return; } /*棋盘中间的交叉点*/ putch(CROSS); } /*交换行棋方*/ int ChangeOrder(void) { if(gPlayOrder==CHESS1) gPlayOrder=CHESS2; else gPlayOrder=CHESS1; return(gPlayOrder); } /*获取按键值*/ int GetKey(void) { char lowbyte; int press; while (bioskey(1) == 0) ;/*如果用户没有按键，空循环*/ press=bioskey(0); lowbyte=press&0xff; press=press&0xff00 + toupper(lowbyte); return(press); } /*落子错误处理*/ void DoError(void) { sound(1200); delay(50); nosound(); } /*赢棋处理*/ void DoWin(int Order) { sound(1500);delay(100); sound(0); delay(50); sound(800); delay(100); sound(0); delay(50); sound(1500);delay(100); sound(0); delay(50); sound(800); delay(100); sound(0); delay(50); nosound(); textcolor(RED+BLINK); gotoxy(25,20); if(Order==CHESS1) cputs("PLAYER1 WIN!"); else cputs("PLAYER2 WIN!"); gotoxy(25,21); cputs(" \<^+^>/"); getch(); } /*走棋*/ int ChessGo(int Order,struct point Cursor) { /*判断交叉点上有无棋子*/ if(gChessBoard[Cursor.x][Cursor.y]==CHESSNULL) { /*若没有棋子, 则可以落子*/ gotoxy(Cursor.x+MAPXOFT,Cursor.y+MAPYOFT); textcolor(LIGHTBLUE); putch(Order); gotoxy(Cursor.x+MAPXOFT,Cursor.y+MAPYOFT); gChessBoard[Cursor.x][Cursor.y]=Order; return TRUE; } else return FALSE; } /*判断当前行棋方落子后是否赢棋*/ int JudgeWin(int Order,struct point Cursor) { int i; for(i=0;i<4;i++) /*判断在指定方向上是否有连续5个行棋方的棋子*/ if(JudgeWinLine(Order,Cursor,i)) return TRUE; return FALSE; } /*判断在指定方向上是否有连续5个行棋方的棋子*/ int JudgeWinLine(int Order,struct point Cursor,int direction) { int i; struct point pos,dpos; const int testnum = 5; int count; switch(direction) { case 0:/*在水平方向*/ pos.x=Cursor.x-(testnum-1); pos.y=Cursor.y; dpos.x=1; dpos.y=0; break; case 1:/*在垂直方向*/ pos.x=Cursor.x; pos.y=Cursor.y-(testnum-1); dpos.x=0; dpos.y=1; break; case 2:/*在左下至右上的斜方向*/ pos.x=Cursor.x-(testnum-1); pos.y=Cursor.y+(testnum-1); dpos.x=1; dpos.y=-1; break; case 3:/*在左上至右下的斜方向*/ pos.x=Cursor.x-(testnum-1); pos.y=Cursor.y-(testnum-1); dpos.x=1; dpos.y=1; break; } count=0; for(i=0;i=0&&pos.x<=18&&pos.y>=0&&pos.y<=18) { if(gChessBoard[pos.x][pos.y]==Order) { count++; if(count>=testnum) return TRUE; } else count=0; } pos.x+=dpos.x; pos.y+=dpos.y; } return FALSE; } /*移动光标*/ void MoveCursor(int Order,int press) { switch(press) { case PLAY1UP: if(Order==CHESS1&&gCursor.y>0) gCursor.y--; break; case PLAY1DOWN: if(Order==CHESS1&&gCursor.y<18) gCursor.y++; break; case PLAY1LEFT: if(Order==CHESS1&&gCursor.x>0) gCursor.x--; break; case PLAY1RIGHT: if(Order==CHESS1&&gCursor.x<18) gCursor.x++; break; case PLAY2UP: if(Order==CHESS2&&gCursor.y>0) gCursor.y--; break; case PLAY2DOWN: if(Order==CHESS2&&gCursor.y<18) gCursor.y++; break; case PLAY2LEFT: if(Order==CHESS2&&gCursor.x>0) gCursor.x--; break; case PLAY2RIGHT: if(Order==CHESS2&&gCursor.x<18) gCursor.x++; break; } gotoxy(gCursor.x+MAPXOFT,gC

有,前提是双方都不想赢棋,.如果先手方下必胜定式,那是摆不下去的.

Four sons chess and backgammon are two well-known game.To them based on,someone has invented a game of this three four sons.It's playing and the first two are almost identical,but the more interesting because people can take part in more.First of all players to use different color pens.We can observe one of the parties as to the other side to make the attack.Attacking side should have tried their best to move his line of four pieces composed a line,and Shou-side as long as they can to stop the attack until the attack on the successful board victory even under the full piece.Of course,such modifications can play chess,for example,to add a player or two for the attack one side for Shou-side,a lot of fun.
This piece has an interesting little time-consuming,very suitable for rest and friends entertainment.

My hobby
Everyone has his or her own hobby.Today I am going to talk about my hobby.My hobby is swimming.I like it very much,although I don't swim very well.And I go to swim with my friends every summer.My swimming coat is blue,because blue is my favourite color.I enjpy swimming in the sun,it's really cool.
I hope I can learn to swim well this summer.
你看看行不?

解题思路：根据A点的位置表示的坐标规律，结合五子棋中白棋已经有三个在一条直线上的情况，合理地选择黑棋的落点．

∵白棋已经有三个在一条直线上，
∴甲必须在（5，3）或（1，7）位置上落子，才不会让乙马上获胜．
故答案为：（5，3）或（1，7）．

点评：
本题考点： 坐标确定位置．

考点点评： 本题考查了坐标确定点的位置的方法．关键是根据题目所给的表示方法，结合图形确定黑棋的落点．

当然多看看高手的棋谱对提高很有帮助的. 你应该多做做VCF练习题,那个对计算力的提高帮助很大的.

图片就先不上了,实在太多,我手上有24开局的简谱,基本可以满足你的要求.
开局有：黑必胜、黑优、平衡、白优和白必胜五种
平衡局常见的是：斜月、银月、疏星、瑞星等.
黑大优：新月等?
活二：两子相连,未被阻挡.
活三：三子相连,未被阻挡
一打：一点
三手交换：黑走完第三手,白可以选择自行交换.
五手两打：黑第五手放两个点,白拿掉其中之一.
可以不走天元,只是天元的优势最大!