POJ-1426-Find The Multiple【BFS】

1426-Find The Multiple

                Time Limit:1000MS     Memory Limit:10000KB

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

题目链接：POJ-1426

题目大意：给出一个数n，求一个数m，只由0和1组成，且是n的倍数（存在多个可任意输出）

题目思路：直接BFS即可。数字大小不会超过19位，用long long可以存下。
G++可过。

以下是代码：

//
// E.cpp
// 搜索
//
// Created by pro on 16/3/21.
// Copyright (c) 2016年 pro. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>
#include<iomanip>
using namespace std;
queue<long long> que;
long long bfs(int n)
{
    que.push(1);
    while(!que.empty())
    {
        long long front = que.front();
        que.pop();
        if (front % n == 0) return front;
        que.push(front * 10);
        que.push(front * 10 + 1);
    }
    return 0;
}
int main()
{
    int n;
    while(cin >> n && n)
    {
        while (!que.empty()) {  //注意清空
            que.pop();
        }
        cout << bfs(n) << endl;
    }
}