数论

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

2
6
19
0

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[524286]; 
int main(int i)
{
    int n;
    while(cin>>n)
    {
        if(!n)
            break;

        mod[1]=1%n;  
        for(i=2;mod[i-1]!=0;i++)  
            mod[i]=(mod[i/2]*10+i%2)%n;
        i--;
        int pm=0;
        while(i)
        {
            mod[pm++]=i%2;   
            i/=2;
        }
        while(pm)
            cout<<mod[--pm];  //倒序输出
        cout<<endl;
    }
    return 0;
}