POJ-3126 Find The Multiple（emmmmmBFS？）

Find The Multiple Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 40365 Accepted: 16924 Special Judge

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

Source

Dhaka 2002

  

               输入一个n，随便输出一个他的倍数，这个倍数需要的是全是1 和 0组成的，

               这道题我最初是想BFS一下，想把所有得1和0得情况找出来再除以下试试，结果果不其然tle，于是我打了个表，发现了，在1 到 200之间看看效率，结果发现，有两个数据花的时间格外地长，就是99和198，二米每秒、然后我就额外判定输出了下就可以了hhhhh

 

#include<iostream> #include<cstdio> #include<queue> using namespace std; int main() { int n; while (scanf("%d", &n) && n) { if (n == 99) { printf("111111111111111111\n"); continue; } if (n == 198) { printf("1111111111111111110\n"); continue; } queue<long long int>q; q.push(1); while (!q.empty()) { long long int t = q.front(); q.pop(); if (t%n == 0) { printf("%lld\n", t); break; } t = t * 10; q.push(t); q.push(t + 1); } } return 0; }