Description: Towers of Hanoi was brought to France from Thailand by French M. Claus (Lucas) in 1883. Hanoi was the capital of North Vietnam during the Vietnam War, which is now Ho Chi Minh City; in 1883, French mathematician Edouard Lucas This story has been mentioned. It is said that at the time of Genesis, Benares had a Polo tower, which was supported by three diamond rods (Pag). At the beginning, God placed 64 on the first rod, from top to bottom, from small to large. Arrange the gold discs (Disc), and order the monks to move all the gold discs from the first stone rod to the third stone rod, and observe the principle of the large plate under the small plate during the transportation process, if only one plate is moved a day , then when all the plates have been carried, the tower will be destroyed, and the end of the world will come.
Solution: If the column is marked as ABC, it needs to be moved from A to C. When there is only one plate, it is directly moved to C. When there are two plates, B is used as an auxiliary column. If the number of plates exceeds 2, it is very simple to cover the plates below the third, and process two plates at a time, that is, the three steps of A->B, A->C, B->C , and the covered part is actually the recursive processing that enters the program. In fact, if there are n disks, the number of times required to complete the move is 2^n -1, so when the number of disks is 64, the required number of times is: 264- 1 =18446744073709551615 is 5.05390248594782e+16 years, also It is about 5000 centuries, if you have no idea about this number, assuming that one plate is moved every second, it will take about 585 billion years.
Standard Edition
#include <stdio.h>
void hanoi(int n,char A, char B, char C){
if(n == 1){
printf("Move sheet %d from %cto %cn", n, A, C);
}
else{
hanoi(n-1, A, C, B);
printf("Move sheet %d from %c to %cn", n, A, C);
hanoi(n-1, B, A, C);
}
}
int main(){
int n;
printf("请输入盘数：");
scanf("%d", &n);
hanoi(n, 'A', 'B', 'C');
return 0;
}Enhanced Edition
#include <windows.h>
#include <stdio.h>
//常量定义区
#define MAX_VALUE 100//定义数组最大值
#define DATA_1 1
#define DATA_2 2
#define DATA_3 3
#define SPEED 800000
//全局变量定义区
int data_1[MAX_VALUE][MAX_VALUE];
int data_2[MAX_VALUE][MAX_VALUE];
int data_3[MAX_VALUE][MAX_VALUE];
int data_l_1=0;
int data_l_2=0;
int data_l_3=0;
int max_n;
int move_num=0;
//函数定义区
void hanoi(int n, char A, char B, char C);
int int_input(int i,int j);
void program_introduction();
void initialization_data();
void initialization_data_1(int n);
void show_data(int n,int K);
void record_data(int n,char A,char C);
void sleep_order();
int main()
{
int n;
program_introduction();
printf("请输入盘数：");
n=int_input(1,100);
max_n=n;
initialization_data_1(n);
show_data(n,DATA_1);
system("cls");
hanoi(n, 'A', 'B', 'C');
system("pause");
return 0;
}
void hanoi(int n, char A, char B, char C)
{
if(n == 1){
printf("Move sheet %2d from %c to %c , move number %4dn", n, A, C,++move_num);
record_data(n,A,C);
show_data(max_n,1);
show_data(max_n,2);
show_data(max_n,3);
sleep_order();
system("cls");
}
else{
hanoi(n-1, A, C, B);
printf("Move sheet %2d from %c to %c , move number %4dn", n, A, C,++move_num);
record_data(n,A,C);
show_data(max_n,1);
show_data(max_n,2);
show_data(max_n,3);
sleep_order();
system("cls");
hanoi(n-1, B, A, C);
}
}
void program_introduction()
{
printf("Created At: 20170802n");
printf("Algorithm Gossip: 河内之塔n");
printf("广州航海学院 - 吴南辉 n");
printf("Referenced By: http://www.wunanhui.wang:81n");
printf("n");
printf("说明 河内之塔(Towers of Hanoi)是法国人M.Claus(Lucas)于1883年从泰国带至法国的，河内为越");
printf("战时北越的首都，即现在的胡志明市；1883年法国数学家 Edouard Lucas曾提及这个故事，据说");
printf("创世纪时Benares有一座波罗教塔，是由三支钻石棒（Pag）所支撑，开始时神在第一根棒上放置");
printf("64个由上至下依由小至大排列的金盘（Disc），并命令僧侣将所有的金盘从第一根石棒移至第三");
printf("根石棒，且搬运过程中遵守大盘子在小盘子之下的原则，若每日仅搬一个盘子，则当盘子全数");
printf("搬运完毕之时，此塔将毁损，而也就是世界末日来临之时。n");
printf("n");
printf("解法 如果柱子标为ABC，要由A搬至C，在只有一个盘子时，就将它直接搬至C，当有两个盘子，");
printf("就将B当作辅助柱。如果盘数超过2个，将第三个以下的盘子遮起来，就很简单了，每次处理两");
printf("个盘子，也就是：A->B、A ->C、B->C这三个步骤，而被遮住的部份，其实就是进入程式的递");
printf("回处理。事实上，若有n个盘子，则移动完毕所需之次数为2^n - 1，所以当盘数为64时，则所需");
printf("次数为：2 64 - 1 = 18446744073709551615为5.05390248594782e+16年，也就是约5000世纪，如果");
printf("对这数字没什幺概念，就假设每秒钟搬一个盘子好了，也要约5850亿年左右。n");
printf("n");
system("pause");
system("cls");
}
int int_input(int i,int j)//整形数据输入函数
{
int a=-100,k=0;
while(a>j||a<i){
if(k!=0){printf("数据错误，请重新输入n"); }
scanf("%d",&a);
k++;
}
return a;
}
void initialization_data(){
int i,j;
for(i=0;i<MAX_VALUE;i++){
for(j=0;j<MAX_VALUE;j++){
data_1[i][j]=0;
data_2[i][j]=0;
data_3[i][j]=0;
}
}
}
void initialization_data_1(int n){
int i,j;
for(i=0;i<n;i++){
for(j=0;j<=i;j++){data_1[i][j]=1;}
++data_l_1;
}
}
void show_data(int n,int K){
int i,j;
if(K==DATA_1){printf("A:n");}
if(K==DATA_2){printf("B:n");}
if(K==DATA_3){printf("C:n");}
for(i=0;i<n;i++){
printf("%2d | ",i+1);
for(j=n-1;j>=0;j--){
if(K==DATA_1){
if(data_1[i][j]==1){printf("■");}
else{printf(" ");}
}
if(K==DATA_2){
if(data_2[i][j]==1){ printf("■");}
else{printf(" ");}
}
if(K==DATA_3){
if(data_3[i][j]==1){ printf("■");}
else{printf(" "); }
}
}
for(j=0;j<n;j++){
if(K==DATA_1){
if(data_1[i][j]==1){printf("■");}
else{printf(" ");}
}
if(K==DATA_2){
if(data_2[i][j]==1){printf("■");}
else{printf(" "); }
}
if(K==DATA_3){
if(data_3[i][j]==1){printf("■");}
else{printf(" ");}
}
}
printf("n");
}
}
void record_data(int n,char A,char C)
{
int i,j;
if(A=='A'){
for(j=0;j<n;j++){data_1[max_n-data_l_1][j]=0;}
--data_l_1;
}
else if(A=='B'){
for(j=0;j<n;j++){ data_2[max_n-data_l_2][j]=0;}
--data_l_2;
}
else if(A=='C'){
for(j=0;j<n;j++){ data_3[max_n-data_l_3][j]=0;}
--data_l_3;
}
if(C=='A'){
for(j=0;j<n;j++){ data_1[max_n-data_l_1-1][j]=1;}
++data_l_1;
}
else if(C=='B'){
for(j=0;j<n;j++){ data_2[max_n-data_l_2-1][j]=1;}
++data_l_2;
}
else if(C=='C'){
for(j=0;j<n;j++){data_3[max_n-data_l_3-1][j]=1;}
++data_l_3;
}
}
void sleep_order(){
usleep(SPEED);
}