[POJ]Find The Multiple(DFS)

题目链接

http://poj.org/problem?id=1426

题意

输入一个数n，输出任意一个 只含0、1且能被n整除的数m。保证n<=200,m最多100位。

题解

DFS/BFS都能做。
这里使用DFS。然后T了。就变成了打表A了。emm ,m最多会19位，这点我也很迷==。

其他

使用了Java大数，get了打开方式。

todo

• 为啥m19？？
• BFS法
• 0/1背包法
• 和鸽笼原理有什么关系？

代码

打表A版

import java.math.BigInteger;
import java.util.Scanner;

public class Main {
    
    static int num;
    static boolean tag;
    
    static BigInteger[] arr=new BigInteger[201];
    static String[] ans= {"0","1","10","1000000000000000011","100","10","1000000000000000110","1000000000000000111","1000","1000000000011111111","10","100000000000000001","1000000000000001100","1000000000000001","10000000000000010","1000000000000000110","10000","10000000000000101","1000000000111111110","1000000000000001101","100","10000000000000101","1000000000000000010","1000000000000010111","1000000000000011000","100","10000000000000010","1000000000101111111","100000000000000100","1000000000000000111","1000000000000000110","1000000000000110011","100000","1000000000000101111","100000000000001010","10000000000000010","1000000001111111100","100000000000000011","10000000000000110","10000000000000101","1000","1000000000000010111","100000000000001010","1000000000000010101","100000000000000100","1000000000111111110","100000000000001110","1000000000000101","1000000000000110000","1000000000000011101","100","10000000000000101","100000000000000100","1000000000000000011","1000000001101111110","1000000000000000010","1000000000000001000","1000000000000011","1000000000000001010","10000000000000011","1000000000000001100","1000000000000011","100000000001110010","1000000001011101111","1000000","10000000000000010","1000000000000111110","100000000000011","1000000000000010100","100000000000011","10000000000000010","1000000000000100101","1000000011111111000","1000000000001","1000000000000000110","1000000000000001100","100000000000001100","1000000000000001","100000000000001010","100000000000110001","10000","1000000000111111101","1000000000000110110","100000000000001101","1000000000000010100","100000000000001010","100000000000010010","100000000000000101","1000000000000001000","1000000000000011011","1000000000111111110","1000000000000001","1000000000000011100","1000000000001111001","10000000000001010","10000000000000110","1000000000001100000","100000000001111","1000000000000111110","1101111111111111111","100","1000000000000000001","100000000000001010","100000000000000001","1000000000000001000","100000000000001010","100000000010000110","10000000000001001","1000000001111111100","1000000000001000011","1000000000000000010","100000000000000011","100000000000110000","1000000000000111101","10000000000000110","100000000000001110","10000000000000100","1000000001101111101","100000000000000110","10000000000000101","1000000000000011000","1000000000000101111","10000000000000110","1000000001100011001","1000000000011100100","1000","1000000001111011110","1000000000000110101","10000000","10000000000001001","10000000000000010","10000000001101","1000000000010111100","1000000000001010001","1000000000000110","1000000001101111110","100000000000111000","1000000000001","1000000000000110","1000000000001001111","100000000000000100","1000000000000101","1000000000011110","1000000000000001","1000000111111110000","1000000000000001010","10000000000010","100000000000011111","1000000000000010100","10000000000001111","1000000000000001100","1000000000010111001","1000000000000011000","1000000000111111101","10000000000000010","100000000001110010","1000000000000010100","1000000000000101","1000000000001100010","1000000000000000011","100000","1000000000001101001","100000000111111110","1000000000011011101","1000000000001110100","1000000000000111110","1000000000000011010","1000000000001001101","10000000000011000","1000000000001100111","100000000000001010","1000000000111111011","1000000000000100100","1000000000000010111","1000000000000001010","100000000000000100","100000000000010000","10000000000000011","100000000000001110","1000000000000100111","1000000001111111100","10000000000000111","10000000000000010","1000000000000011","100000000000011000","1000000000000000110","10000000001101110","100000000001001001","100000000000010100","1000000010111010111","10000000000000110","1000000000011111101","1000000000011000000","1000000010000100001","1000000000011110","100000000000001010","100000000010010100","100000000000111001","1111111111111111110","1000000000101111001","1000"};
    
    public static void main(String args[]) {
//      //打表
//      for(int i=1;i<201;++i) {
//          tag=false;
//          dfs(BigInteger.valueOf(1),i,1);
//          System.out.print("\"");
//          System.out.print(arr[i]);
//          System.out.print("\"");
//          System.out.print(",");
//      }
        
        Scanner in=new Scanner(System.in);
        while(in.hasNext()) {
            int num=in.nextInt();
            if(num==0) {break;}
            System.out.println(ans[num]);
        }
    }
    
    public static void dfs(BigInteger x,int num,int layer) {
        if(tag) {return;}//
        if(x.mod(BigInteger.valueOf(num))==BigInteger.ZERO) {
            //System.out.println(x);
            arr[num]=x;
            tag=!tag;
            return;
        }
        if(layer==19) {return;}
        dfs(x.multiply(BigInteger.valueOf(10)),num,layer+1);
        dfs(x.multiply(BigInteger.valueOf(10)).add(BigInteger.ONE),num,layer+1);
    }
}

DFS T版，估计是JAVA大数T的？C++用unsigned long long这么写感觉不会T？

import java.math.BigInteger;
import java.util.Scanner;

public class Main {
    
    static int num;
    static boolean tag;
    
    public static void main(String args[]) {
        Scanner in=new Scanner(System.in);
        while(in.hasNext()) {
            int num=in.nextInt();
            if(num==0) {break;}
            tag=false;
            dfs(BigInteger.valueOf(1),num,1);
        }
    }
    
    public static void dfs(BigInteger x,int num,int layer) {
        if(tag) {return;}//
        if(x.mod(BigInteger.valueOf(num))==BigInteger.ZERO&&!tag) {
            System.out.println(x);
            tag=!tag;
            return;
        }
        if(layer==19) {return;}
        dfs(x.multiply(BigInteger.valueOf(10)),num,layer+1);
        dfs(x.multiply(BigInteger.valueOf(10)).add(BigInteger.ONE),num,layer+1);
    }
}