poj-1426-Find The Multiple

最新推荐文章于 2022-02-19 09:15:00 发布

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原文链接：http://www.cnblogs.com/w-y-1/p/6729838.html

本文提供了一种解决POJ 1426问题的有效算法。该问题要求找到一个正整数n的非零倍数m，且m仅由0和1组成。通过使用广度优先搜索策略，程序能够快速找到符合条件的最小m值。

http://poj.org/problem?id=1426

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 30731		Accepted: 12788		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

Sample Output

10
100100100100100100
111111111111111111

Source

Dhaka 2002

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<math.h>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>

using namespace std;
const int maxn=1000006;
int n;

long long bfs()
{
    queue<long long>Q;
    Q.push(1);

    while(!Q.empty())
    {
        long long q=Q.front();
        Q.pop();
        if(q%n==0)return q;

        Q.push(q*10);
        Q.push(q*10+1);
    }
    return -1;
}
int main()
{
    while(scanf("%d", &n), n)
    {
        long long ans=bfs();
        printf("%lld\n", ans);
    }
    return 0;
}