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
其实一开始我是没读懂题意的，可爱善良的伙伴给我说的：找到任意一由0,1组成的十进制数，满足是n的倍数，输出就行。
怎么做呢，还是while，queue。
#include<iostream> #include<queue> using namespace std; int n; long long bfs(long long sum) { queue <long long> Q; // 定义队列 Q.push(1); // 把1进队 while(1) { long long q = Q.front(); // 取数 if(q % n == 0) return q; // 满足题意结束函数，返回满足题意要求数 Q.pop(); Q.push(q*10); // 不满足，继续找下一个0,1数，进队 Q.push(q*10+1); // 10,11.，100,101,110,111 } } int main() { while(cin >> n, n) { cout << bfs(1) << endl; // 0,1数字最小是1，so，从1开始 } return 0; }