数论 - 同余 + BFS (Find The Multiple)

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 16995		Accepted: 6921		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

【题目大意】

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。

【题目分析】

数论 +bfs

用到了同余定理，用bfs搜索当前位，每位都只可能是0或1,所以这是双入口的bfs，同时还涉及到了大数的知识。

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cmath>

#include<algorithm>

using namespace std;

int mod[];

int main(int i)

{

    int n;

    while(cin>>n)

    {

        if(!n)

            break;

        mod[]=%n;

        for(i=;mod[i-]!=;i++)

            mod[i]=(mod[i/]*+i%)%n;

        i--;

        int pm=;

        while(i)

        {

            mod[pm++]=i%;

            i/=;

        }

        while(pm)

            cout<<mod[--pm];  //倒序输出

        cout<<endl;

    }

    return ;

}

巴特西

数论 - 同余 + BFS (Find The Multiple)

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