33个盘子的非递归汉诺塔

// helloworld.cpp : Defines the entry point for the console application. // //by 陈墨仙 2019-07-18 //完全不用递归解汉诺塔 #include "stdafx.h" #include int h[34];//为了便于理解，0号元素不用,33个盘子 int a[4][34]; int b[4]; int jihao; int dijici; void printH() { //system("cls"); for(int i = 1;i<=34;i++) { printf("%d:%d\t",i,a[3][i]); } } bool jiancha(int * h)//检测是否在下面的都是编号小的 { int d[4]; d[1]=0; d[2]=0; d[3]=0; for(int j = 1;j<4;j++) { for(int i = 1;i<34;i++) { if(a[j][i]==0 || a[j][i]>a[j][i-1]) { } else { printf("error a[%d][%d] = %d a[%d][%d]=%d",j,i-1,a[j][i-1],j,i,a[j][i]); scanf("%d"); return false; } } } return true; } bool shunxu(int zhuzi) { for(int i = 1; i< b[zhuzi];i++) { if(a[zhuzi][i]==34-i) { } else { return false; } } return true; } bool chenggong(int n,int zhuzi)//n号盘是否都移到3 { int d = 0; int t = 34; for(int i = 33;i>33-n;i--) { if(a[zhuzi][34-i]!=i-n+1) { return false; } } return true; } int jc3()//检测3号柱的盘子有几个 { int d = 0; int t = 34; for(int i = 33;i>0;i--) { if(h[i]==3) { d=d+1; t = i; } } return d; } int jc1()//检测1号柱的盘子有几个 { int d = 0; int t = 34; for(int i = 33;i>0;i--) { if(h[i]==1) { d=d+1; t = i; } } return d; } void initH() { for(int i = 0;i<34;i++) { h[i]=1; a[1][i]=i; a[2][i]=0; a[3][i]=0; } b[1]=34; b[2]=1; b[3]=1; } int jc2()//检测2号柱从33往下盘子有几个 { int d = 0; int t = 34; for(int i = 33;i>0;i--) { if(h[i]==2) { d=d+1; t = i; } } return d; } int getTop(int zhuzi) { int d = 0; int t = 0; for(int i = 1;i<34;i++) { if(h[i]==zhuzi && i>t) { t = i; } } return t; } bool jiou(int s) { if(s % 2 == 0) { return true; } else { return false; } } void change(int i,int yuan,int mubiao) { if(h[i] == yuan) { h[i] = mubiao; a[mubiao][b[mubiao]]=i; b[mubiao]=b[mubiao]+1; a[yuan][b[yuan]]=0; b[yuan]=b[yuan]-1; printf("h[%d]:%d->%d",i,yuan,mubiao); } else { printf("Error h[%d] = %d",i,h[i]); scanf("%d"); } jiancha(h); } int FastLog2(int x) { float fx; unsigned long ix, exp; fx = (float)x; ix = *(unsigned long*)&fx; exp = (ix >> 23) & 0xFF; return exp - 127; } int chu2(int n ,int cishu) { for(int j = 1; j<= cishu;j++) { n=n/2; } return n; } void jihaopan(int n) { int i = 1; int yuan = n; while(1) { if(n%2==1) { jihao = i; dijici = chu2(yuan,i)+1; break; } n=n/2; i=i+1; } //printf("几号盘%d,第几次%d",jihao,dijici); } int main(int argc, char* argv[]) { printf("汉诺塔!\n"); int times = 0; initH(); int ji = 1; for(int i = 1;i<5559060534555523;i++) { jihaopan(i); if(jiou(jihao)==false) { int tmp; tmp = dijici%3; if(tmp==1) { change(getTop(1),1,3); } else if(tmp==2) { change(getTop(3),3,2); } else { change(getTop(2),2,1); } } else { int tmp; tmp = dijici%3; if(tmp==1) { change(getTop(1),1,2); } else if(tmp==2) { change(getTop(2),2,3); } else { change(getTop(3),3,1); } } if(chenggong(33,3)) { printf("sssss"); break; } } return 0; }