阶梯问题的递归解法

#include <stdio.h>

#define N 10

int steped[N];
int i=0;

void steping(int n){                                  \\走楼梯
       if(n==0){                                        \\已走完
            for(int j=0;j<i;j++){
                  printf(" %d ",steped[j]);        \\打印
            }
      printf("\n");
      }
      if(n>=1){                                       \\若剩下的阶梯数大等于1
            steped[i++]=1;                         \\迈1个阶梯
            steping(n-1);                             \\走剩下的阶梯
            i--;                                           \\退一个阶梯,寻找其它上法
      }
      if(n>=2){                                       \\若剩下的阶梯数大等于2
            steped[i++]=2;待                      \\迈两个阶梯
            steping(n-2);                            \\走剩下的阶梯
            i--;                                          \\退两个阶梯,寻找其它上法
     }
     if(n>=3){                                      \\同上.....略
           steped[i++]=3;
           steping(n-3);
           i--;
     }
}

void main(){
      int n;
      n=N;
      steping(n);
}