AYIT2017暑假集训第二周周三赛 A

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

26190

Sample Output

10100100100100100100111111111111111111

题目意思大概就是给你一个数，然后让你用0和1组成十进制的数去整除他.

这是一道搜索题，用简单的广搜就可以写出来，题目虽然说数很大，其实long long int就可以存下了，

因为只有0,1。所以直接让第一位为1，如果想加上0就直接乘10，加1的话直接乘十加1就行了。
#include<stdio.h>long long int p[10000000];int a,b,c,d,i,head,tail;void bfs(){   head=1,tail=2;//用数组模拟队列    p[1]=1;//设置第一个数为1    while(head<tail)//如果栈不为空    {        if(p[tail-1]%a==0)//如果第一个数就可以整除a就直接输出            printf("%lld",p[tail-1]);        else              //然后就进行分情况，只有加1和加0两种情况        {            for(i=1; i<=2; i++)            {                if(i==1)//当加0时直接乘10，进行存储                {                    p[tail]=p[head]*10;                    if(p[tail]%a==0)                    {                        b=1;                        return;                    }                    tail++;                }                else                {                    p[tail]=p[head]*10+1;//当加1时直接乘10+1,;                    if(p[tail]%a==0)                    {                        b=1;                        return;                    }                    tail++;//两种情况每次进行完就tail++；                }            }        }        head++;    }}int main(){    while(~scanf("%d",&a)&&a!=0)//输入要求解的数    {        bfs();//进行广搜        printf("%lld\n",p[tail]);    }    return 0;}