这是个汉诺塔程序，在调试的时候，输入的数字最好不要大于15，因为每大一个数所得的结果的步骤都会多一倍。如果你有耐心等待结果的话除外。汉诺塔是在欧洲流行的一种游戏，有a,b,c三个竿。a竿上有若干个由大到小的圆盘，大的在下面，小的在上面，b,c都是空杆，请你把a杆上的圆盘都倒到别的杆上，或b或c，在倒盘的过程中不可以大的压小的，实例程序如下：
#include <stdio.h>
#include <stdlib.h>
void move(char x,char y) {
printf("%c-->%c\n",x,y);
}
void hanoi (int n,char one ,char two,char three) {
if(n==1)
move (one ,three);
else {
hanoi (n-1,one,three,two);
move(one,three);
hanoi(n-1,two,one,three);
}
}
int main()
{
int m;
printf("input the number of diskes:");
scanf("%d",&m);
printf("the step to moving %3d diskes:\n",m);
hanoi(m,'A','B','C');
system("pause");
}
#include <stdio.h>
#include <conio.h>
int i=0;
void movedisc(unsigned n,char fromneedle,char toneedle,char usingneedle){
if(n>0){
movedisc(n-1,fromneedle,usingneedle,toneedle);
i++;
switch(fromneedle){
case 'a':
switch(toneedle){
case 'b':printf("\t[%d]:\t%2d------>%2d\n",i,n,n);
break;
case 'c':printf("\t[%d]:\t%2d------------->%2d\n",i,n,n);
break;
}
break;
case 'b':
switch(toneedle){
case 'a':printf("\t[%d]:\t%2d<----------%2d\n",i,n,n);
break;
case 'c':printf("\t[%d]:\t\t%2d------>%2d\n",i,n,n);
break;
}
break;
case 'c':
switch(toneedle){
case 'a':printf("\t[%d]:\t%2d<--------------%2d\n",i,n,n);
break;
case 'b':printf("\t[%d]:\t\t%2d<--------%2d\n",i,n,n);
break;
}
break;
}
movedisc(n-1,usingneedle,toneedle,fromneedle);
}
}
int main(){
unsigned n;
printf("Please enter the number of discs: ");
scanf("%d",&n);
printf("\tneedle:\ta\t b\t c\n");
movedisc(n,'a','c','b');
printf("\t Total: %d\n",i);
getch();
}
#include <stdio.h>
#define width (rings+1)
int main(){
int rings, last, next, x, z[500], s[3];
printf("how many rings? "); scanf("%d",&rings);
for(x=1; x<=rings; x++) /* put rings on first peg */
z[x]=width-x;
for(x=0; x<=2*width; x+=width) /* set base for each peg */
z[x]=1000;
/* if even number of rings, put first ring on second peg; if odd, on third */
if(rings%2==0){
last=1; s[2]=0; s[1]=1;
z[width+1]=z[rings];
}else{
last=2; s[1]=0; s[2]=1;
z[2*width+1]=z[rings];
}
printf("from 1 to %d\n",last+1);
s[0]=rings-1;
while(s[0]+s[1]){ /* while first and second pegs aren't empty */
/* next ring to move is smaller of rings on the two pegs not moved onto last */
if(last==0)
next=z[width+s[1]]<z[2*width+s[2]]?1:2;
if(last==1)
next=z[s[0]]<z[2*width+s[2]]?0:2;
if(last==2)
next=z[s[0]]<z[width+s[1]]?0:1;
/* top ring of 'to' peg must be larger and an even 'distance' away */
if((z[next*width+s[next]])>(z[last*width+s[last]])||((z[last*width+s[last]]-z[next*width+s[next]])%2==0))
last=3-next-last;
printf("from %d to %d\n",next+1,last+1);
s[next]=s[next]-1; s[last]=s[last]+1; /* move from 'next' to 'last' peg */
z[last*width+s[last]]=z[next*width+s[next]+1];
}
}
#include <stdio.h>
//--------------------------------------------------------
// 打印搬运动作
//--------------------------------------------------------
int MoveIt(int x,int Source,int Target)
{
printf("Move %d From %d to %d\n",x,Source,Target);
return 0;
}
//--------------------------------------------------------
// 用 4 根柱子移动盘子
// n 是盘子个数，编号从1 到 n
// First 是源柱子号
// Second Third 是两根过渡柱
// Fourth 是目标柱
//--------------------------------------------------------
int MoveHanoi(int n,int First,int Second,int Third,int Fourth)
{
if (n<1) return 0; // 如果没有盘子就返回
if (n==1) // 如果只有一个盘子
{
MoveIt(n,First,Fourth); // 就直接从源柱子移到目标柱子上
return 0;
}
if (n==2) // 如果有两个盘子
{
MoveIt(n-1,First,Second); // 把上面的那片移到一个过渡柱上
MoveIt(n,First,Fourth); // 把下面的那片移到目标柱上
MoveIt(n-1,Second,Fourth); // 再把第 1 片从过渡柱移到目标柱上
return 0;
}
if (n==3) // 如果有 3 片盘子
{
MoveIt(n-2,First,Second); // 把最小的盘子移到一个过渡柱上
MoveIt(n-1,First,Third); // 把中间盘子移到另一过渡柱上
MoveIt(n,First,Fourth); // 把最大的盘子移到目标柱上
MoveIt(n-1,Third,Fourth); // 把中间盘子移到目标柱上
MoveIt(n-2,Second,Fourth); // 把最小的盘子移到目标柱上
return 0;
}
// 递归地把上面 n-2 盘子移到一个过渡柱上
// 留下最大的两个盘子
MoveHanoi(n-2,First,Third,Fourth,Second);
MoveIt(n-1,First,Third); // 把倒数第 2 个盘子移到另一个过渡柱上
MoveIt(n,First,Fourth); // 把最底下的盘子移到目标柱上
MoveIt(n-1,Third,Fourth); // 把倒数第 2 个盘子移到目标柱上
// 递归地把 n-2 个盘子从过渡柱上移到目标柱上
MoveHanoi(n-2,Second,First,Third,Fourth);
return 0;
}
int main()
{
int num;
scanf("%d",&num);
MoveHanoi(num,1,2,3,4);
return 0;
}
#include <stdio.h>
#include <math.h>
#define MAX_N 1000
int main(){
double f[MAX_N];
int n,i,j;
double maxvalue,curvalue;
printf("Please input count:");
scanf("%d",&n);
if(n>=MAX_N){
printf("Out of range\n");
return -1;
}
f[1]=1.0,f[2]=3.0;
for(j=3;j<=n;j++){
maxvalue=pow(2,j);
for(i=1;i<j;i++){
curvalue=2*f[i]+pow(2,j-i)-1;
if(curvalue<maxvalue)
maxvalue=curvalue;
else
break;
}
f[j]=maxvalue;
}
for(i=1;i<=n;i++)
printf("m[%d]=%f\n",i,f[i]);
return 0;
}
不得不佩服递归魅力，非递归实现方式实现起来让人有点费解。