汉诺塔问题的非递归实现.docx

汉诺塔问题的非递归实现#include #include #include /第0位置是柱子上的塔盘数目 int zhua100=0,zhub100=0,zhuc100=0; char charis(char x,int n) /左右字符出现顺序固定,且根据n值奇偶而不同 switch(x) case A: return (n%2=0)?C:B; case B: return (n%2=0)?A:C; case C: return (n%2=0)?B:A; default: return 0; void print(char lch,char rch) /打印字符 if(lch=A) switch(rch) case B: zhub0+; zhubzhub0=zhuazhua0; zhuazhua0=0; zhua0-; break; case C: zhuc0+; zhuczhuc0=zhuazhua0; zhuazhua0=0; zhua0-; break; default: break; if(lch=B) switch(rch) case A: zhua0+; zhuazhua0=zhubzhub0; zhubzhub0=0; zhub0-; break; case C: zhuc0+; zhuczhuc0=zhubzhub0; zhubzhub0=0; zhub0-; break; default: break; if(lch=C) switch(rch) case A: zhua0+; zhuazhua0=zhuczhuc0; zhuczhuc0=0; zhuc0-; break; case B: zhub0+; zhubzhub0=zhuczhuc0; zhuczhuc0=0; zhuc0-; break; default: break; printf(t); int i; printf(); for(i=1;i=zhua0;i+) printf( %d ,zhuai); printf(); printf(); for(i=1;i=zhub0;i+) printf( %d ,zhubi); printf(); printf(); for(i=1;i=zhuc0;i+) printf( %d ,zhuci); printf(); printf(n); void Hannuo(int n) /分配2n个空间 bool *isrev; isrev=(bool *)malloc(sizeof(bool)*(int)pow(2,n); for(int i=1;ipow(2,n);i+) isrevi=false; /循环计算是否逆序 for(int ci=2;ci=n;ci+) for(int zixh=(int)pow(2,ci-1);zixh%c,lch,rch); print(lch,rch); for(int k=2;k%c,rch,lch); print(rch,lch); else printf(%c-%c,lch,rch); print(lch,rch); int main() int n; printf(Input the num:); scanf(