http://poj.org/problem?id=1426
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
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 <iostream> #include <stdio.h> #include <string.h> #include <stdlib.h> 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<e) { t=q[s++]; if(t.x%n==0) { printf("%lld ",t.x); break; } f.x=t.x*10; q[e++]=f; f.x=t.x*10+1; q[e++]=f; } } int main() { while(scanf("%d",&n)!=EOF&&n!=0) { bfs(); } return 0; }
G++;
#include <iostream> #include <stdio.h> #include <string.h> #include <stdlib.h> #include <queue> using namespace std; int n; long long t; void bfs() { queue<long long >q; q.push(1); while(!q.empty()) { t=q.front(); q.pop(); if(t%n==0) { printf("%lld ",t); break; } q.push(t*10); q.push(t*10+1); } } int main() { while(scanf("%d",&n)!=EOF&&n!=0) { bfs(); } return 0; }