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
26190
Sample Output
10100100100100100100111111111111111111
大致题意：
给出一个整数n，(1 <= n <= 200)。求出任意一个它的倍数m，要求m必须只由十进制的'0'或'1'组成。
没看见是Special Judge 题解： 我感觉这题能够是投机取巧，输出结果根本就没有100位，之前以为是大数，一直没敢做，谁知是一个超级大坑题。 还有给的测试数据给的那么大，害我一看测试数据就不敢做了。还有为什么我用STL中的queue用C++交超时，而用G++就A了 ，而自己写的结构体用C++交就过了。 主要思想： 和二叉树差不多，1->10,11;10->100,101,11->110,111..... 就是q.push(t*10);q.push(t*10+1);
#include#include #include #include using namespace std;int n;struct node{ long long int x;} q[10000001];struct node t,f;void bfs(){ int s=0; int e=0; t.x=1; q[e++]=t; while(s
G++;
#include#include #include #include #include using namespace std;int n;long long t;void bfs(){ queue q; q.push(1); while(!q.empty()) { t=q.front(); q.pop(); if(t%n==0) { printf("%lld\n",t); break; } q.push(t*10); q.push(t*10+1); }}int main(){ while(scanf("%d",&n)!=EOF&&n!=0) { bfs(); } return 0;}