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.
考试的时候一看200位直接就被吓到了，直接跳过，后来听说答案开long long 可过，就是普通宽搜，感觉亏了一个亿，既然每一位不是0就是1，首先第一位一定是1，然后每次这个数乘10，或者乘10加1，就是可以产生每一位都是1或者0的数字了。
代码：
1 #include<iostream> 2 #include<cstdio> 3 #include<queue> 4 #define ll long long 5 using namespace std; 6 int n; 7 queue<long long>q; 8 ll ans; 9 int main() 10 { 11 while(1) 12 { 13 cin>>n; 14 if(!n)break; 15 while(!q.empty())q.pop(); 16 q.push(1); 17 while(!q.empty()) 18 { 19 ll k=q.front();q.pop(); 20 if(k%n==0) 21 { 22 ans=k;break; 23 } 24 q.push(k*10),q.push(k*10+1); 25 } 26 printf("%lld ",ans); 27 } 28 29 return 0; 30 }