POJ1426 Find The Multiple （宽搜思想）

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 24768		Accepted: 10201		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

Source

 

题意：求n个一个倍数，但这个数必须是01串

其实就是利用宽搜的思想枚举： 第一个数是1 ， 然后下一个就是 10 (1 * 10) , 11( 1 * 10 + 1)，再下一个就是 100（10 * 10）， 101（10 * 10 + 1），110 （11 * 10），111（11 * 10 + 1） ...然后比较坑的就是只要是n个一个倍数就行，结果不用跟这个样例一样也可以，然后用long long也能过=_=

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