POJ 1426 Find The Multiple

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 33059 Accepted: 13834 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：只由0和1组成的十进制数，是否含有一个数整除所给的数

dfs回溯去查找太浪费时间，而且递归的终止条件很难去找，不如bfs搜索

#include <iostream>#include <stack>#include <queue>#include <map>using namespace std;typedef long long ll;queue <ll>q;ll a;void bfs() {   ll t = 1;   q.push(t);//从1开始找      while(!q.empty()) {       ll front = q.front();       if(front % a == 0) {//这里为了判断a为1           cout << front << endl;           return;       }       q.pop();//栈顶元素       for(int i = 0;i < 2;i++) {           ll x = front * 10 + i;           if(x % a == 0) {               cout << x << endl;               return;           }           q.push(x);//将接下来两个入队列       }   }}int main(int argc, char *argv[]) {    while(cin >> a && a) {        while(!q.empty()) {            q.pop();        }//清空队列，防止不同步相互影响        bfs();    }    return 0;}