POJ1426 Find The Multiple —— BFS

题目链接：http://poj.org/problem?id=1426

 

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 34218		Accepted: 14337		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

Dhaka 2002

 

题解:

题目说答案的位数不会超过100，所以以为是：高精度 + 模运算。这样也太丧心病狂了吧！

后来实在想不到其他方法，就看了题解。结果是一道普通的搜索题，答案用long long存就够了（题目不是说位数<=100），感觉被骗了。

再一次受到了启发：面对一些题目（人生也如此）时，如果想到的方法似乎不能解决问题，但除此之外又无其他办法，那就就要放开手脚试一试，不要畏手畏脚。如果想到的唯一方法都搁置不试，那就相当于交白卷了；试一试，或许能出现意想不到的结果。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#define ms(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e9+7;
const int MAXN = 200000+10;

queue<LL>q;
LL bfs(int n)
{
    while(!q.empty()) q.pop();
    q.push(1);

    while(!q.empty())
    {
        LL x = q.front();
        q.pop();
        if(x%n==0)
            return x;
        q.push(1LL*x*10);
        q.push(1LL*x*10+1);
    }
}

int main()
{
    int n;
    while(scanf("%d",&n) && n)
    {
        cout<< bfs(n) <<endl;
    }
}