Find The Multiple

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 32537 Accepted: 13604 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

26190

Sample Output

10100100100100100100111111111111111111

题意：找出任意的一个由0和1组成的数字，并且是n的倍数

解题思路：两个路径的广搜（t*10）,(t*10+1);

代码：

#include<stdio.h>#include<queue>using namespace std;
void bfs(int n){    queue<long long>q;   //命名一个名称为q的队列    q.push(1);         //将1推入队列中    while(!q.empty())  //如果队列不为空    {        long long t;   //由于t可能比较大，定义为long long型        t=q.front();   //将队列的首元素赋给t        q.pop();       //将队列的首元素pop掉        if(t%n==0)        {            printf("%lld\n",t); //如果符合条件，就将t输出            return ;        }        q.push(t*10);             q.push(t*10+1);   //两个路径改变t的值    }}int main(){    int n;    while(~scanf("%d",&n)&&n)    {        bfs(n);        //调用n    }    return 0;}