Description
Given a positive integer n,
给你正整数n
write a program to find out a nonzero multiple m of n whose decimal representation contains only the digits 0 and 1.
找出n的没有0的 非零倍数m ， m的十进制表示只有01
You may assume that n is not greater than 200
n <= 200
and there is a corresponding m containing no more than 100 decimal digits.
对应m 不超过100位数字
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
思路
2 的1倍 = 2 是01组成吗？ 2 的2倍 是01组成吗 。。。。很复杂。。
显然要逆向思考
num * 10 、 num * 10 +1 肯定是01 组成 == 末尾追加了0 或者 1
考虑所有由01组成的数，然后看这个数 是不是m的倍数
#include <iostream>
#include <queue>
#include <cmath>
using namespace std;
queue<long long> q;
long long bfs(int n) {
while(!q.empty()) {
q.pop();
}
long long father;
long long child;
q.push(1);
while(!q.empty()) {
father = q.front();
q.pop(); // 这个竟然忘了加！！多可恶啊！！
if (father % n == 0) {
return father;
}
q.push(father * 10);
q.push(father * 10 + 1);
}
}
int main() {
int n;
while(cin >> n) {
long long res = bfs(n);
printf("%lld\n",res);
}
}