【C语言】汉诺塔自动递归演示程序(源码)
【C语言】汉诺塔自动递归演示程序(源码)
尝试写出可视化的小算法
程序介绍: 【C语言】用C语言实现汉诺塔自动递归演示程序
/*------------------------------------
project : 汉诺塔演示软件;
language: C语言
author: 404name
叠64层需要开全窗口并且将字号调整为6
------------------------------------0-*/
#include <stdio.h>
#include <windows.h>
int len, width, left, mid, right, time;
char map[80][1000];
int next[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; //上0下1左2右3;
int turn[2][3] = {{0, 3, 1}, //a -> b b -> c a -> c 上右下
{0, 2, 1}}; //b -> a c -> b c -> a 上左下
void gotoxy(int x, int y)
{
HANDLE handle = GetStdHandle(STD_OUTPUT_HANDLE);
COORD pos;
pos.X = x;
pos.Y = y;
SetConsoleCursorPosition(handle, pos);
}
int init()
{
system("color 30");
printf("请输入你要递归的\n汉诺塔数目_");
int n;
scanf("%d", &n);
time = 50 / n;
left = 1, mid = 2 * (n + 1), right = mid + 2 * n + 1;
len = 3 * (2 * n + 1);
width = n + 1;
for (int i = 0; i <= width; i++)
{
for (int j = 0, num = i; j <= len; j++)
{
if (i == 0 || j == 0 || i == width || j == len)
map[i][j] = '#';
if ((i > 1 && i < width) && (j == mid - 1 || j == right - 1))
map[i][j] = '|';
if (num && (i >= 1 && i <= width - 1) && j % 2 != 0 && j < mid)
{
map[i][j] = -95;
map[i][j + 1] = -10; //打印 ■ ■占2个字节可以拆开来
num--;
}
}
}
gotoxy(0, 0); //容易出现东西卡顿
for (int i = 0; i <= width; i++)
{
for (int j = 0; j <= len; j++)
printf("%c", map[i][j]);
printf("\n");
}
return n;
}
void play(int x, int y)
{
int turn_0, n = 0, i, j, k, tx, ty, flag = 0;
if ((x == mid && y == right) || (x == left && y == mid) || (x == left && y == right))
turn_0 = 0; //往右
else if ((x == mid && y == left) || (x == right && y == left) || (x == right && y == mid))
turn_0 = 1; //往左
for (i = 1, j = y; i <= width; i++)
{ //目的地
if (map[i][j] != 0)
{
tx = i - 1;
ty = j;
break;
}
}
for (i = 1, j = x; i <= width; i++)
{ //出发点
if (map[i][j] != 0)
{
break;
}
}
while (1)
{
while ((i != 1 || j != x) && (i != 1 || j != y) && (i != tx || j != ty))
{
if (turn_0 == 0)
for (k = mid - 3; k >= 0; k--)
{
map[i + next[turn[turn_0][n]][0]][j + next[turn[turn_0][n]][1] + k] = map[i][j + k];
map[i][j + k] = 0;
}
else
for (k = 0; k < mid - 2; k++)
{
map[i + next[turn[turn_0][n]][0]][j + next[turn[turn_0][n]][1] + k] = map[i][j + k];
map[i][j + k] = 0;
}
gotoxy(j, i);
Sleep(time);
for (k = 0; k < mid - 2; k++)
{
printf(" ");
}
i = i + next[turn[turn_0][n]][0];
j = j + next[turn[turn_0][n]][1];
gotoxy(j, i);
Sleep(time);
for (k = 0; k < mid - 2; k++)
{
printf("%c", map[i][j + k]);
}
}
n++; //改变方向;
if (i == tx && j == ty)
return;
if (turn_0 == 0)
for (k = mid - 3; k >= 0; k--)
{
map[i + next[turn[turn_0][n]][0]][j + next[turn[turn_0][n]][1] + k] = map[i][j + k];
map[i][j + k] = 0;
}
else
for (k = 0; k < mid - 2; k++)
{
map[i + next[turn[turn_0][n]][0]][j + next[turn[turn_0][n]][1] + k] = map[i][j + k];
map[i][j + k] = 0;
}
gotoxy(j, i);
for (k = 0; k < mid - 2; k++)
{
printf(" ");
}
Sleep(time);
i = i + next[turn[turn_0][n]][0];
j = j + next[turn[turn_0][n]][1];
gotoxy(j, i);
for (k = 0; k < mid - 2; k++)
printf("%c", map[i][j + k]);
Sleep(time);
}
}
void move(int a, int b, int c, char aa, char bb, char cc, int n)
{
if (n == 1)
{
gotoxy(0, width + 1);
printf("from %c to %c", aa, cc);
play(a, c);
return;
}
move(a, c, b, aa, cc, bb, n - 1);
gotoxy(0, width + 1);
printf("from %c to %c", aa, cc);
play(a, c);
move(b, a, c, bb, aa, cc, n - 1);
gotoxy(0, width + 1);
}
int main()
{
int n = init();
move(left, mid, right, 'a', 'b', 'c', n);
return 0;
}