POJ1426,Find The Multiple,bfs...注意类型 5兆内存过了...

Find The Multiple

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

分析：

STL的queue还是很好用的，开始考虑用位运算移位的方法来生成下一个数，这样就涉及到同一个数在二进制和十进制下的转换，增加了代码量。

用类似递推的方法考虑的话，每一个数，在结尾添一个0或添一个1能生成两个数。不难看出这样也能生成所有满足条件的数，可能顺序会有点奇怪。

但根据题意，不许要保证得出的结果是不是最小的，这样以上的分析足够保证程序的正确性了。

code:

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>using namespace std;int n;void bfs(){    queue<long long> q;    while(!q.empty())        q.pop();    q.push(1);    while(1)    {        long long tmp=q.front();        if(tmp%n==0)        {            cout<<tmp<<'\n';            return;        }        q.pop();        q.push(tmp*10);        q.push(tmp*10+1);    }}int main(){    while(scanf("%d",&n),n)        bfs();    return 0;}