题解报告：poj 1426 Find The Multiple（bfs、dfs）

Description

Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only the digits 0 and 1. You may assume that n is not greater than 200 and there is a corresponding m containing no more than 100 decimal digits.

Input

The input file may contain multiple test cases. Each line contains a value of n (1 <= n <= 200). A line containing a zero terminates the input.

Output

For each value of n in the input print a line containing the corresponding value of m. The decimal representation of m must not contain more than 100 digits. If there are multiple solutions for a given value of n, any one of them is acceptable.

Sample Input

2
6
19
0

Sample Output

10
100100100100100100
111111111111111111
解题思路：题目的意思就是找一个不小于n的10进制数x，x只由1和0组成，并且满足x%n==0，输出这个x即可。由于x只由0和1组成，因此容易想到构建一棵平衡二叉树，从根节点值为1开始，每个左节点的值都为父节点值的10倍，每个右节点的值都为父节点值的10倍加1，这样只要找到x%n==0即可。亲测了一下x的值在long long范围内，于是bfs或dfs暴力即可。
AC代码之bfs：

 #include<iostream>
 #include<queue>
 #include<cstdio>
 using namespace std;
 typedef long long LL;LL n;
 queue<LL> que;
 void bfs(LL n){//在long long的范围内查找，按层遍历
     while(!que.empty())que.pop();//清空
     que.push(1LL);//模拟一棵平衡二叉树，先将入根节点的值1入队
     while(!que.empty()){
         LL x=que.front();que.pop();
         if(x%n==){cout<<x<<endl;return;}//只要满足条件，立即退出
         que.push(x*);//左节点的值为父节点值的10倍
         que.push(x*+);//右节点的值为父节点值的10倍加1
     }
 }
 int main(){
     while(cin>>n&&n){bfs(n);}
     return ;
 }

AC代码之dfs：

 #include<iostream>
 #include<queue>
 #include<cstdio>
 using namespace std;
 typedef long long LL;LL n;bool flag;
 void dfs(int num,LL x,LL n){
     if(num>||flag)return;//如果x的位数num超过19位即溢出了long long 范围，直接返回到上一层；如果已找到即flag为true则一直回退，直到栈空，表示不再深搜下去
     if(x%n==){cout<<x<<endl;flag=true;return;}
     dfs(num+,x*,n);//左递归
     dfs(num+,x*+,n);//右递归
 }
 int main(){
     while(cin>>n&&n){flag=false;dfs(,,n);}//从位数1，根节点值为1开始深搜
     return ;
 }