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：
1 #include<iostream> 2 #include<queue> 3 #include<cstdio> 4 using namespace std; 5 typedef long long LL;LL n; 6 queue<LL> que; 7 void bfs(LL n){//在long long的范围内查找，按层遍历 8 while(!que.empty())que.pop();//清空 9 que.push(1LL);//模拟一棵平衡二叉树，先将入根节点的值1入队 10 while(!que.empty()){ 11 LL x=que.front();que.pop(); 12 if(x%n==0){cout<<x<<endl;return;}//只要满足条件，立即退出 13 que.push(x*10);//左节点的值为父节点值的10倍 14 que.push(x*10+1);//右节点的值为父节点值的10倍加1 15 } 16 } 17 int main(){ 18 while(cin>>n&&n){bfs(n);} 19 return 0; 20 }
AC代码之dfs：
1 #include<iostream> 2 #include<queue> 3 #include<cstdio> 4 using namespace std; 5 typedef long long LL;LL n;bool flag; 6 void dfs(int num,LL x,LL n){ 7 if(num>19||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 8 if(x%n==0){cout<<x<<endl;flag=true;return;} 9 dfs(num+1,x*10,n);//左递归 10 dfs(num+1,x*10+1,n);//右递归 11 } 12 int main(){ 13 while(cin>>n&&n){flag=false;dfs(1,1,n);}//从位数1，根节点值为1开始深搜 14 return 0; 15 }