随即生成五子棋问题:
题目描述如下:每次生成一组术随机数作为五子棋中的坐标位置,第二组数与第一组相邻,第三组、第四组、第五组分别为前两组组成的直线上的点,直到生成的五组坐标相邻,程序结束。
代码如下:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#define MAX 10
int chess[MAX][MAX]; //棋盘
int r[5][2]; //存储结果
int a,b; //行列 y = a*x + b
int len; //当前满足要求点的个数
//初始化棋盘
void init()
{
int i,j;
for(i=0; i<MAX; i++)
{
for(j=0; j<MAX; j++)
{
chess[i][j] = 0;
}
}
}
int Judge1(int c,int d)
{
int m,n;
// int flag=0;
m = r[0][0];
n = r[0][1];
// m = abs(c-r[0][0]);
// n = abs(d-r[0][1]);
if( (c==m-1&&d==n-1) || (c==m+1&&d==n+1) || (c==m+1&&d==n-1)
|| (c==m-1&&d==n+1) || (c==m&&d==n-1) || (c==m&&d==n+1)
||(c==m-1&&d==n) || (c==m+1&&d==n) )
return 0; //相邻
else return 1; //不相邻
}
//确定斜率参数
void direction()
{
int x1,x2,y1,y2;
x1=r[0][0];
y1=r[0][1];
x2=r[1][0];
y2=r[1][1];
if(x1==x2)
{
a=-100;
b=0;
}
else
{
a=(y1-y2)/(x1-x2);
b=y1-a*x1;
}
}
int Judge2(int c,int d)
{
int i,m,n;
if(chess[c][d] == 1)
return 1; //此处已经下过
else if(a!=-100 && a*c+b != d)
return 1; //不在一条直线上
else if(a==-100 && c != r[0][0])
return 1;
else
{
for(i=0; i<len; i++)
{
m = r[i][0];
n = r[i][1];
if( (c==m-1&&d==n-1) || (c==m+1&&d==n+1) || (c==m+1&&d==n-1)
|| (c==m-1&&d==n+1) || (c==m&&d==n-1) || (c==m&&d==n+1)
||(c==m-1&&d==n) || (c==m+1&&d==n) )
return 0; //相邻
}
}
return 1;
}
int main()
{
int m,n,i;
srand(time(NULL));
init();
m = rand() % MAX + 1;
n = rand() % MAX + 1;
r[0][0] = m;
r[0][1] = n;
chess[m][n] = 1; //标记改坐标已经下过棋子
len=1;
printf("(%d,%d)\n",m,n);
//判断第二组数是否与第一组数相邻
while(Judge1(m,n))
{
m = rand() % MAX + 1;
n = rand() % MAX + 1;
}
printf("(%d,%d)\n",m,n);
r[1][0] = m;
r[1][1] = n;
chess[m][n] = 1;
len=2;
direction();
//判断第二组数是否与第一组数相邻
for(i=0; i<3; i++)
{
m = rand() % MAX + 1;
n = rand() % MAX + 1;
while(Judge2(m,n))
{
m = rand() % MAX + 1;
n = rand() % MAX + 1;
}
len++;
chess[m][n] = 1;
printf("(%d,%d)\n",m,n);
r[i+2][0] = m;
r[i+2][1] = n;
}
return 0;
}