wuziqi.c

linux环境下c语言写的不需要图形界面的五子棋

#include <stdio.h>

#define BUFSIZE 256
#define AREA 10

int qipan[AREA][AREA] = {};
void p_qipan(void);
void luozi(void);
int win(void);


int main(void)
{
	int flag;
	while (1) {
		p_qipan();
		luozi();
		if ((flag = win()))
			break;
	}

	if (flag == 1)
		printf("black win\n");
	else if (flag == 2) 
		printf("white win\n");
	else 
		printf("no win\n");
}


void p_qipan(void)
{
	int i,j,k;
	for (k = 0; k < AREA; k++){
		printf("%X ",k+1);
	}
	printf("\n");
//	printf("1 2 3 4 5 6 7 8 9 A\n"); 	
	for(i = 0; i < AREA; i++){
		for (j = 0; j < AREA ; j++){
			if (qipan[i][j] == 0){
				printf("+ ");
			} else if (qipan[i][j] == 1){
				printf("# ");
			} else {
				printf("O ");
			}
		}
		printf("%X ",i+1);
		printf("\n");
	}	
}

void luozi(void)
{
	char buf[BUFSIZE];
	static int flag = 1;
	int x, y;

	while (1) {
	printf("Please input zuobiao (like x,y)\n");
	fgets(buf, BUFSIZE, stdin);

	if (*buf == '\n') {
		printf("argument not enough\n");
		continue;
	}

	sscanf(buf, "%d,%d",&x,&y);
	if (x > AREA || y > AREA) {
		printf("number is too big\n");
		continue;
	}
	if (qipan[y-1][x-1] == 0){
		if (flag == 1)
			qipan[y-1][x-1] = 1;
		else
			qipan[y-1][x-1] = 2;
	} else {
		printf("qizi is exist\n");
		continue;
	}
	flag = -1 *flag;
	break;
	}
	
	return;
}


#if 1
int win(void)
{
	int i,j,k,num;
	for (i = 0; i < AREA; i++) {
		for (j = 0; j < AREA; j++) {

			if (qipan[i][j] == 0) {
				/* 空位，返回循环 */
				continue; 
			} else {
				/* 判断x方向是否存在5子 */
				for (k = 1; qipan[i][j] == qipan[i+k][j] && i < AREA - 4 ;) {
					k++;
					if (k == 5){
						return qipan[i][j];
					}					
				}
				/* 判断y方向是否存在5子 */
				for (k = 1; qipan[i][j] == qipan[i][j+k] && j < AREA - 4;) {
                                        k++;
                                        if (k == 5){
						return qipan[i][j];
                                        }
                                }

				/* 判断左上指向右下方向是否存在5子 */
				for (k = 1; qipan[i][j] == qipan[i+k][j+k] && i < AREA - 4 && j < AREA - 4;) {
                                        k++;
                                        if (k == 5){
						return qipan[i][j];
                                        }
                                }
				/* 判断右上指向左下方向是否存在5子 */
				for (k = 1; qipan[i][j] == qipan[i-k][j+k] && i > 3 && j < AREA - 4;) {
                                        k++;
                                        if (k == 5){
						return qipan[i][j];
                                        }
                                }
			}
		}
	}
	return 0;
}
#endif