c语言怎样设计游戏？
1. 输出1＋1 ＝ ？ 输入2，则输出正确，输入其他则显示错误
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2. 24点游戏，输出4个数，输入一行表达式，程序内部计算表达式的值（比较费劲），若＝24则返回胜利，否则失败
3. 五子棋（无AI，否则难），黑白方轮流输入坐标，每次输入完坐标都重新输出一次棋盘上所有棋子，具体方式就是把存储棋子的矩阵挨个输出，无棋输出空格，黑输出＊，白输出o，循环直到判断游戏是否结束的函数返回值为真，即有5子相连才结束。
4. 贪食蛇。这个需要查阅有关＊＊的资料：线程（或计时器），响应键盘按键，随机数等等。蛇每隔单位时间向当前方向前进一步，然后刷新（也就是把画面重新输出一遍），在随机位置产生食物，蛇的尾巴经过食物（即已经消化）后加长一段。按方向键时改变记录当前方向变量的值。加入判断死亡的机制。
这个用控制台（就是和DOS似的，只有press any key to continue之类文字的界面）也可以实现，但是相关资料较少很难找，选择图形编程（windows下常用的软件都是这样，有窗口的）资料更多一些，如果你是在windows上编，就查C语言windows api，刚开始编时不会很方便。由易到难地尝试吧，不必急于用图形编程，控制台也是可以编出好游戏的，这是基础。
c语言 只用循环/选择 输出n阶蛇行/螺旋方阵
#includeiostream.h
#includemath.h
/*(1) 螺旋方阵，如下图：
1 2 3 4
12 13 14 5
11 16 15 6
10 9 8 7
*/
void op1(int n)
{
int i = 0, //i表示方阵的行数
j = 0; //j表示方阵的列数
int *p = new int[n*n];//p表示方阵数组
int temp = 1;
int t;
for (t = 0; t ((float)n / 2); t++) //t表示圈数
{
for (; j n - t; j++) //行数不变，方阵的值随着列数的增加而增加(1,2,3,4,5)(17,18,19)(25)
p[i*n+j] = temp++;
j--;
i++;
for (; i n - t; i++) //列数不变，方阵的值随着行数的增加而增加(6,7,8,9)(20,21)
p[i*n+j] = temp++;
j--;
i--;
for (; j = t; j--) //行数不变，方阵的值随着列数的减少而减少(10,11,12,13)(22,23)
p[i*n+j] = temp++;
j++;
i--;
for (; i t; i--) //列数不变，方阵的值随着行数的减少而减少(14,15,16)(24)
p[i*n+j] = temp++;
j++;
i++;
}
for (i = 0; i n; i++)
{
for (j = 0; j n; j++)
{
coutp[i*n+j]" ";//show.Text += p[i][j] + " ";
}
coutendl;
// show.Text += "\r";
}
}
/*
2) 蛇形方阵，如下图：
1 2 6 7
3 5 8 13
4 9 12 14
10 11 15 16
*/
void op2(int n)
{
int i, j, k; //i表示行数,j表示列数,k表示行数和列数的和
int m = 2 * n - 2; //m表示方阵行数最大和列数最大的和
int temp = 1; //temp表示递增变量
int *array = new int[n*n]; //array表示矩阵数组
for (k = 0; k = m; k++) //根据矩阵的对称性求解;(array[2,0],array[1,1],array[0,2]的行数和列数的和相等)
{
int l = k % 2; //根据k值的递增，改变行数和列数递增的方向;当k为奇数先行后列,当k为偶数先列后行
for (i = 0; i n; i++)
{
j = k - i;
if (j n j = 0) //判断j的值范围,以免超过数组长度
{
if (l == 1)
{
array[i*n+j] = temp;
}
if (l == 0)
{
array[j*n+i] = temp;
}
temp++;
}
}
}
for (i = 0; i n; i++)
{
for (j = 0; j n; j++)
{
/* show.Text += array[i*n+j];
show.Text += " "; */
coutarray[i*n+j]" ";
}
// show.Text += "\n";
coutendl;
}
}
int main()
{
int n1,n2;
while(cinn1n2)
{
op1(n1);
cout"***************"endl;
op2(n2);
}
return 0;
}
不是我做的,不过你可以也看一下!!!!!!!
c语言编贪吃蛇，用什么编译环境，怎么显示贪吃蛇和水果，怎么键盘响应？我菜鸟都不算，求指教。
你如果用TC
#includegraphics.h
#includestdlib.h
#define N 200
#define up 0x4800
#define down 0x5000
#define left 0x4b00
#define right 0x4d00
#define esc 0x011b
#define Y 0x1579
#define n 0x316e
int gamespeed; /* 游戏速度 */
int i,key,color;
int score=0; /* 游戏分数 */
char cai48H[]={
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x04,0x00,0x18,0x00,0x00,0x00,0x0E,0x00,
0x1C,0x00,0x00,0x00,0x1C,0x00,0x1C,0x00,
0x00,0x00,0x20,0x00,0x38,0x00,0x00,0x00,
0x40,0x00,0x78,0x00,0x00,0x01,0x80,0x40,
0x70,0x00,0x00,0x03,0x80,0xC0,0xE0,0x00,
0x00,0x07,0x80,0x80,0xC0,0x00,0x00,0x0E,
0x11,0x81,0xC0,0x00,0x00,0x08,0x61,0x01,
0x80,0x00,0x00,0x00,0x23,0x03,0x04,0x00,
0x00,0x02,0x02,0x00,0x06,0x00,0x00,0x1E,
0x04,0x00,0x0F,0x00,0x00,0x1C,0x1F,0x80,
0x1E,0x00,0x00,0x08,0x3F,0x80,0x3C,0x00,
0x00,0x00,0xFF,0x80,0x38,0x00,0x00,0x03,
0xFF,0x80,0x78,0x00,0x00,0x0F,0xF8,0x00,
0xF0,0x00,0x00,0x7F,0xF0,0x00,0xE0,0x00,
0x03,0xFF,0xFC,0x01,0x80,0x00,0x03,0xC0,
0xFF,0x01,0x03,0x80,0x01,0x01,0xFF,0x00,
0x03,0x80,0x00,0x01,0x3F,0x00,0x07,0x80,
0x00,0x02,0x11,0x00,0x07,0x00,0x00,0x00,
0x10,0x00,0x07,0x00,0x00,0x00,0x10,0x00,
0x0E,0x00,0x00,0x08,0x10,0x00,0x1C,0x00,
0x00,0x30,0x10,0x00,0x18,0x00,0x00,0x70,
0x10,0x00,0x30,0x00,0x01,0xE0,0x10,0x00,
0x70,0x00,0x03,0x80,0x10,0x00,0x60,0x00,
0x00,0x00,0x30,0x00,0xE0,0x00,0x00,0x00,
0xF0,0x01,0xC0,0x00,0x00,0x00,0x70,0x03,
0xC0,0x00,0x00,0x00,0x10,0x07,0x80,0x00,
0x00,0x00,0x00,0x0F,0x00,0x00,0x00,0x00,
0x00,0x1E,0x00,0x00,0x00,0x00,0x00,0x3C,
0x00,0x00,0x00,0x00,0x00,0x70,0x00,0x00,
0x00,0x00,0x01,0xC0,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
};
char she48H[]={
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x04,
0x00,0x00,0x00,0x00,0x00,0x0C,0x00,0x00,
0x00,0x00,0x00,0x0E,0x00,0x00,0x00,0x00,
0x00,0x0E,0x00,0x00,0x00,0x03,0x00,0x07,
0x00,0x00,0x00,0x02,0x00,0x03,0x00,0x00,
0x00,0x02,0x00,0x00,0x00,0x00,0x00,0x02,
0x00,0x00,0xF8,0x00,0x00,0x02,0x00,0x07,
0x86,0x00,0x00,0x02,0x00,0x18,0x03,0x00,
0x00,0x02,0x00,0x00,0x07,0x80,0x00,0x03,
0xF0,0x00,0x07,0x80,0x00,0x0F,0xFC,0x00,
0x0C,0x00,0x00,0x7E,0x3F,0x80,0x00,0x00,
0x01,0xFE,0x1F,0x80,0x00,0x00,0x01,0xE2,
0x39,0x8C,0x00,0x00,0x00,0xC2,0x30,0x08,
0x00,0x00,0x00,0xC2,0x60,0x08,0x00,0x00,
0x00,0xC3,0xE0,0x08,0x60,0x00,0x00,0x7F,
0xE0,0x01,0xE0,0x00,0x00,0x3F,0x80,0x1F,
0xE0,0x00,0x00,0x1E,0x00,0x1F,0x80,0x00,
0x00,0x1E,0x00,0x1F,0x00,0x00,0x00,0x02,
0x38,0x1E,0x00,0x00,0x00,0x07,0xFC,0x1C,
0x00,0x20,0x00,0x07,0xFC,0x18,0x00,0x20,
0x00,0x1F,0x0C,0x10,0x00,0x20,0x00,0x7C,
0x04,0x10,0x00,0x60,0x01,0xF0,0x00,0x10,
0x00,0x60,0x01,0xE0,0x00,0x08,0x00,0xF0,
0x00,0x80,0x00,0x08,0x03,0xF0,0x00,0x00,
0x00,0x07,0xFF,0xF0,0x00,0x00,0x00,0x07,
0xFF,0xF0,0x00,0x00,0x00,0x03,0xFF,0xE0,
0x00,0x00,0x00,0x01,0xFF,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
};
char tun48H[]={
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x0E,0x00,0x00,0x00,0x00,0x00,0x3E,
0x00,0x00,0x00,0x00,0x00,0x7F,0x00,0x00,
0x00,0x00,0x00,0xE0,0x00,0x00,0x00,0x00,
0x03,0xC0,0x00,0x00,0x00,0x00,0x1F,0x00,
0x00,0x00,0x00,0x00,0x7C,0x00,0x00,0x00,
0x00,0x01,0xF8,0x00,0x00,0x00,0x00,0x03,
0xF8,0x00,0x40,0x00,0x00,0x00,0x06,0x07,
0xC0,0x00,0x00,0x00,0x07,0xFF,0xE0,0x00,
0x00,0x00,0x07,0xFF,0xE0,0x00,0x00,0x00,
0x0F,0xFF,0x80,0x00,0x00,0x00,0x7F,0xF8,
0x00,0x00,0x00,0x1F,0xFF,0xF8,0x00,0x00,
0x00,0x1F,0xFF,0xF8,0x00,0x00,0x00,0x1F,
0xFC,0x3C,0x00,0x00,0x00,0x0F,0xF8,0x0E,
0x00,0x00,0x00,0x04,0x70,0x07,0x00,0x00,
0x00,0x00,0x60,0x03,0x80,0x00,0x00,0x00,
0xC0,0x00,0xC0,0x00,0x00,0x01,0x80,0x00,
0x30,0x00,0x00,0x01,0x00,0x3C,0x18,0x00,
0x00,0x02,0x03,0xFF,0x0C,0x00,0x00,0x0C,
0x7F,0xFF,0x8E,0x00,0x00,0x18,0xFF,0xFF,
0xC7,0x80,0x00,0x78,0xFE,0x07,0x87,0xE0,
0x01,0xF0,0x70,0x07,0x03,0xF8,0x07,0xE0,
0x70,0x0E,0x03,0xFE,0x00,0x00,0x38,0x1E,
0x01,0xFE,0x00,0x00,0x3F,0xFE,0x00,0x0C,
0x00,0x00,0x1F,0xFE,0x00,0x00,0x00,0x00,
0x1F,0xFE,0x00,0x00,0x00,0x00,0x0F,0xFE,
0x00,0x00,0x00,0x00,0x04,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
};
char dan48H[]={
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0xFC,0x00,0x00,0x00,0x00,0x07,0xFF,
0x00,0x00,0x00,0x00,0x7F,0xC0,0x80,0x00,
0x00,0x03,0xFF,0x80,0x40,0x00,0x00,0x01,
0xF1,0x80,0x40,0x00,0x00,0x01,0x81,0x80,
0xE0,0x00,0x00,0x00,0x01,0x93,0xF0,0x00,
0x00,0x00,0x01,0xFF,0xF0,0x00,0x00,0x00,
0x21,0xFF,0xF0,0x00,0x00,0x00,0x21,0xF8,
0x00,0x00,0x00,0x00,0x61,0xC0,0x00,0x00,
0x00,0x00,0x61,0x80,0x00,0x00,0x00,0x00,
0xF3,0x00,0x00,0x00,0x00,0x00,0xFF,0x00,
0x00,0x00,0x00,0x01,0xFF,0xC0,0x00,0x00,
0x00,0x03,0xFF,0xF8,0x00,0x00,0x00,0x02,
0x00,0xFC,0x00,0x00,0x00,0x04,0x02,0x1F,
0x00,0x00,0x00,0x08,0x03,0x01,0xC0,0x00,
0x00,0x38,0x03,0x00,0x7C,0x00,0x00,0xF8,
0x07,0xF8,0x3F,0xC0,0x01,0xF0,0x3F,0xFE,
0x3F,0xF8,0x03,0xC1,0xFF,0x0F,0x1F,0xF8,
0x00,0x01,0xE3,0x0F,0x0F,0xF0,0x00,0x01,
0xC3,0x0E,0x00,0x00,0x00,0x01,0x83,0xFC,
0x00,0x00,0x00,0x00,0xC7,0xF8,0x00,0x00,
0x00,0x00,0xFF,0xF8,0x00,0x00,0x00,0x00,
0x7F,0xF0,0x00,0x00,0x00,0x00,0x3F,0x03,
0x80,0x00,0x00,0x00,0x03,0x04,0x00,0x00,
0x00,0x00,0x03,0xF8,0x00,0x00,0x00,0x00,
0x1F,0xF8,0x20,0x00,0x00,0x00,0xFF,0xFF,
0xE0,0x00,0x00,0x07,0xFF,0x81,0xE0,0x00,
0x00,0x07,0xE0,0x00,0xE0,0x00,0x00,0x03,
0x00,0x00,0x60,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
};
char zuo16H[]={
0x18,0xC0,0x18,0xC0,0x19,0x80,0x31,0xFE,
0x33,0xFE,0x76,0xC0,0xF0,0xFC,0xB0,0xFC,
0x30,0xC0,0x30,0xC0,0x30,0xFE,0x30,0xFE,
0x30,0xC0,0x30,0xC0,0x30,0xC0,0x00,0x00,
};
char zhe16H[]={
0x03,0x00,0x03,0x0C,0x1F,0xCC,0x1F,0xD8,
0x03,0x30,0xFF,0xFE,0xFF,0xFE,0x03,0x00,
0x0F,0xF8,0x3F,0xF8,0xEC,0x18,0xCF,0xF8,
0x0C,0x18,0x0F,0xF8,0x0F,0xF8,0x00,0x00,
};
char tian16H[]={
0x00,0x00,0x3F,0xFC,0x3F,0xFC,0x31,0x8C,
0x31,0x8C,0x31,0x8C,0x3F,0xFC,0x3F,0xFC,
0x31,0x8C,0x31,0x8C,0x31,0x8C,0x3F,0xFC,
0x3F,0xFC,0x30,0x0C,0x00,0x00,0x00,0x00,
};
char xue16H[]={
0x33,0x18,0x19,0x98,0x08,0xB0,0x7F,0xFC,
0x7F,0xFC,0x60,0x0C,0x1F,0xF0,0x1F,0xF0,
0x00,0xC0,0x7F,0xFC,0x7F,0xFC,0x01,0x80,
0x01,0x80,0x07,0x80,0x03,0x00,0x00,0x00,
};
char ke16H[]={
0x00,0x00,0x0C,0x18,0xFD,0x98,0xF8,0xD8,
0x18,0x58,0xFE,0x18,0xFE,0x98,0x18,0xD8,
0x3C,0x58,0x7E,0x1E,0xDB,0xFE,0x9B,0xF8,
0x18,0x18,0x18,0x18,0x18,0x18,0x00,0x00,
};
struct Food/*定义结构体存储食物的属性*/
{
int x; /* 食物的坐标 */
int y;
int yes; /* 值为0表示屏幕上没有食物，值为1表示屏幕上有食物 */
int color; /* 食物颜色 */
}food;
struct Snake/*定义结构体存储蛇的属性*/
{
int x[N]; /* 每一节蛇的坐标 */
int y[N];
int color[N];/*存储每一节蛇的颜色*/
int node; /* 蛇的节数 */
int direction; /* 蛇移动的方向 */
int life; /* 蛇的生命，如果为1，蛇死，游戏结束 */
}snake;
void init(void)/*图形驱动*/
{
int driver=DETECT,mode=0;
registerbgidriver(EGAVGA_driver);
initgraph(driver,mode,"");
}
void drawmat(char *mat,int matsize,int x,int y,int color)/*汉字点阵*/
{
int i,j,k,m;
m=(matsize-1)/8+1;
for(j=0;jmatsize;j++)
for(i=0;im;i++)
for(k=0;k8;k++)
if(mat[j*m+i](0x80k))
putpixel(x+i*8+k,y+j,color);
}
void showword(void)
{/* 调用汉字点阵输出程序，显示标题和作者信息 */
drawmat(cai48H,48,249,-4,7);
drawmat(she48H,48,329,-4,7);
drawmat(tun48H,48,409,-4,7);
drawmat(dan48H,48,489,-4,7);
drawmat(cai48H,48,250,-5,4);
drawmat(she48H,48,330,-5,4);
drawmat(tun48H,48,410,-5,4);
drawmat(dan48H,48,490,-5,4);
/*作者 田学科*/
drawmat(zuo16H,16,515,465,7);
drawmat(zhe16H,16,530,465,7);
drawmat(tian16H,16,550,465,7);
drawmat(xue16H,16,565,465,7);
drawmat(ke16H,16,580,465,7);
}
void draw(void)/*画出四周的墙*/
{
if(color==15)
color=0;
setcolor(++color);
setlinestyle(SOLID_LINE,0,1);
for(i=30;i=600;i+=10)
{
rectangle(i,40,i+10,49);
rectangle(i,451,i+10,460);
}
for(i=40;i450;i+=10)
{
rectangle(30,i,39,i+10);
rectangle(601,i,610,i+10);
}
}
void prscore(void)
{/* 打印游戏分数 */
char str[10];
setfillstyle(SOLID_FILL,YELLOW);
bar(50,10,200,30);
setcolor(6);
settextstyle(0,0,2);
sprintf(str,"score:%d",score);
outtextxy(55,15,str);
}
void gameover(void)
{
cleardevice(); /* 清屏函数 */
for(i=0;isnake.node;i++)/* 画出蛇死时的位置 */
{
setcolor(snake.color[i]);
rectangle(snake.x[i],snake.y[i],snake.x[i]+10,snake.y[i]+10);
}
prscore(); /* 显示分数 */
draw();
showword();
settextstyle(0,0,6);
setcolor(7);
outtextxy(103,203,"GAME OVER");
setcolor(RED);
outtextxy(100,200,"GAME OVER");
}
void gameplay(void)/* 玩游戏的具体过程 */
{
int flag,flag1;
randomize();
prscore();
gamespeed=50000;
food.yes=0;/* food.yes=0表示屏幕上没有食物 */
snake.life=1;/* snake.life=1表示蛇是活着的 */
snake.direction=4;/* 表示蛇的初始方向为向右 */
snake.node=2; /* 蛇的初始化为两节 */
snake.color[0]=2; /*两节蛇头初始化为绿色*/
snake.color[1]=2;
snake.x[0]=100;snake.y[0]=100;
snake.x[1]=110;snake.y[1]=100;
food.color=random(15)+1;
while(1)
{
while(1)
{
if(food.yes==0) /* 如果蛇活着 */
{
while(1)
{
flag=1;
food.yes=1;
food.x=random(56)*10+40;
food.y=random(40)*10+50;
for(i=0;isnake.node;i++)
{
if(food.x==snake.x[i]food.y==snake.y[i])
flag=0;
}
if(flag) break;
}
}
if(food.yes)
{
setcolor(food.color);
rectangle(food.x,food.y,food.x+10,food.y+10);
}
for(i=snake.node-1;i0;i--)
{
snake.x[i]=snake.x[i-1];
snake.y[i]=snake.y[i-1];
}
switch(snake.direction)
{
case 1: snake.y[0]-=10;break;
case 2: snake.y[0]+=10;break;
case 3: snake.x[0]-=10;break;
case 4: snake.x[0]+=10;break;
}
for(i=3;isnake.node;i++)
{
if(snake.x[i]==snake.x[0]snake.y[i]==snake.y[0])
{
gameover();
snake.life=0;
break;
}
}
if(snake.x[0]40||snake.x[0]590||snake.y[0]50||snake.y[0]440)
{
gameover();
snake.life=0;
}
if(snake.life==0)
break;
if(snake.x[0]==food.xsnake.y[0]==food.y)/*蛇吃掉食物*/
{
setcolor(0);
rectangle(food.x,food.y,food.x+10,food.y+10);
snake.x[snake.node]=-20;
snake.y[snake.node]=-20;
snake.color[snake.node]=food.color;
snake.node++;
food.yes=0;
food.color=random(15)+1;
score+=10;
prscore();
if(score%100==0score!=0)
{
for(i=0;isnake.node;i++)/* 画出蛇 */
{
setcolor(snake.color[i]);
rectangle(snake.x[i],snake.y[i],snake.x[i]+10,snake.y[i]+10);
}
sound(200);
delay(50000);delay(50000);delay(50000);
delay(50000);delay(50000);delay(50000);
nosound();
gamespeed-=5000;
draw();
}
else
{
sound(500);
delay(500);
nosound();
}
}
for(i=0;isnake.node;i++)/* 画出蛇 */
{
setcolor(snake.color[i]);
rectangle(snake.x[i],snake.y[i],snake.x[i]+10,snake.y[i]+10);
}
delay(gamespeed);
delay(gamespeed);
flag1=1;
setcolor(0);
rectangle(snake.x[snake.node-1],snake.y[snake.node-1],
snake.x[snake.node-1]+10,snake.y[snake.node-1]+10);
if(kbhit()flag1==1)/*如果没按有效键就重新开始循环*/
{
flag1=0;
key=bioskey(0);
if(key==esc)
exit(0);
else if(key==upsnake.direction!=2)
snake.direction=1;
else if(key==downsnake.direction!=1)
snake.direction=2;
else if(key==leftsnake.direction!=4)
snake.direction=3;
else if(key==rightsnake.direction!=3)
snake.direction=4;
}
}
if(snake.life==0)/*如果蛇死了就退出循环*/
break;
}
}
void main(void)
{
while(1)
{
color=0;
init();
cleardevice();
showword();
draw();
gameplay();
setcolor(15);
settextstyle(0,0,2);
outtextxy(200,400,"CONTINUE(Y/N)?");
while(1)
{
key=bioskey(0);
if(key==Y||key==n||key==esc)
break;
}
if(key==n||key==esc)
break;
}
closegraph();
}
C语言：编写程序，求一个n×n矩阵对角线元素之和。
求一个n×n矩阵对角线元素之和的源代码如下
public class lianxi28 {
public static void main(String[] args) {
System.out.println("请输入九个整数：");
Scanner in = new Scanner(System.in);
int[][] a = new int[3][3];
for(int i=0;i3;i++){
for(int j=0;j3;j++){
a[i][j]=in.nextInt();
}
}
System.out.println("你输入的n×n矩阵是：");
for(int i=0;i3;i++){
for(int j=0;j3;j++){
System.out.print(a[i][j]+" ");
}
System.out.println();
}
System.out.println();
int sum=0;
for(int i=0;i3;i++){
for(int j=0;j3;j++){
if(i==j){
sum=sum+a[i][j];
}
System.out.println(sum);
}
扩展资料
1、用arrange生成一个数组，元素为0,1,...,8，然后利用reshape生成一个n×n的矩阵。
2、利用for循环将该矩阵的主对角线上的元素相加，然后输出结果。
分享标题：蛇阵函数C语言,蛇形方阵的c语言
文章转载：
http://cdkjz.cn/article/hcjijd.html
