带分数 隔板dfs

​​题目地址​​

思路：
先枚举数字1~9可能出现的组合形式，再双重循环分离出a,b,c，满足题目要求a + b /c == n.
慎用java的String 转 Integer函数,易超时.
因为是9个数字,复杂度为O（9!）

// package ch01;

import java.util.Scanner;

public class Main {
    public static long gcd(long fm, long fm2) 
    {return fm2 == 0 ? fm : gcd (fm2, fm % fm2);}
    public static int n, cnt;
    public static int vis[] = new int[10 + 5];
    public static int a[] = new int[10 + 5];
    
    public static int trans(int l , int r) {
      int ans = 0;
      for(int i = l;i <= r;i++)
      ans = ans * 10 + a[i];
        return ans;
    }
    
    public static void dfs(int k) {
      
      if(k == 9) {
//        for(int i = 1;i <= 9;i++) {
//          System.out.print(a[i] + " ");
//        }
//        System.out.println();
//        
        for(int i = 1;i <= k - 2;i++)
          for(int j = i + 1;j <= k - 1;j++)
          {
            int a = trans(1, i);
            int b = trans(i + 1, j);
            int c = trans(j + 1, k);
            
            if(b % c == 0 && a + b /c == n)cnt++;
          
          }
      } else {
        for(int i = 1;i <= 9;i++) {
          if(vis[i] == 0)
          {
            a[k + 1] = i;
            vis[i] = 1;
            dfs(k + 1);
            vis[i] = 0;
          }
        }
      }
    }
    
    public static void main(String[] args) {
      
      Scanner cin = new Scanner(System.in);
      n = cin.nextInt();
      
      dfs(0);
      System.out.println(cnt);
    }
    
    
}