POJ 1426 Find The Multiple

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 12564		Accepted: 5175		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



BFS就可以解决。但是用自己编的队列94ms，用STL居然超时。

STL代码（TLE）：

#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
typedef long long LL;
LL m,n,ans;

queue<LL>Q;
void BFS()
{
    Q.push(1);
    while(!Q.empty())
    {
        m=Q.front();
        Q.pop();
        for(int i=0;i<2;++i)
        {
            if(i==0)
            {
                m*=10;
            }
            else
            {
                m+=1;
            }
            if(m%n==0)
            {
                ans=m;
                return ;
            }
            else
            {
                Q.push(m);
            }
        }
    }
}
int main()
{
    while(scanf("%lld",&n)&&n)
    {
        while(!Q.empty())
        {
                    Q.pop();
        }
        BFS();
        printf("%lld\n",ans);
    }
}

自己编的队列实现（94ms）：

#include<iostream>
#include<cstdio>
using namespace std;
typedef long long LL;
LL m,n,ans;
LL Q[1000000];
LL Size=1000000-1;


void BFS()
{
    int rear=1;
    int front=1;
    Q[rear++]=1;
    while(true)
    {
        m=Q[front];
        front++;
        for(int i=0;i<2;++i)
        {
            if(i==0) m*=10;
            else     m+=1;
            if(m%n==0)
            {
                ans=m;
                return ;
            }
            else
            {
                Q[rear++]=m;
            }
        }
    }
}
int main()
{
    while(scanf("%lld",&n)&&n)
    {
        BFS();
        printf("%lld\n",ans);
    }
}