Find The Multiple 《DFS》

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


最初想用队列搞一下，就像抓住那头牛一样，结果没搞过去，然后，DFS跑一遍能A掉。
可以的话可以用二进制码搞一下，感觉应该能过

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <string>
#include <cstring>
#include <cstdlib>
#include <map>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
using namespace std;
#define ull unsigned long long
#define lli long long
#define pq priority_queue<int>
#define pql priority_queue<ll>
#define pqn priority_queue<node>
#define v vector<int>
#define vl vector<ll>
#define read(x) scanf("%d",&x)
#define lread(x) scanf("%lld",&x);
#define pt(x) printf("%d
",(x))
#define YES printf("YES
");
#define NO printf("NO
");
#define gcd __gcd
#define out(x) cout<<x<<endl;
#define over cout<<endl;
#define rep(j,k) for (int i = (int)(j); i <= (int)(k); i++)
#define input(k) for (int i = 1; i <= (int)(k); i++)  {scanf("%d",&a[i]) ; }
#define mem(s,t) memset(s,t,sizeof(s))
#define ok return 0;
#define TLE std::ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define mod(x) ((x)%9973)
#define test cout<<"     ++++++      "<<endl;
//二叉树
#define lson rt<<1, l, m
#define rson rt<<1|1, m+1, r
//线段树
#define ls now<<1
#define rs now<<1|1
const int MAXN = 2e5+5;
//int dir[6][3] = {0,0,1,0,0,-1,1,0,0,-1,0,0,0,1,0,0,-1,0};
//int dir[4][2] = {1,0,-1,0,0,1,0,-1}; //单位移动
//int dir[8][2] = {2,1,2,-1,-2,1,-2,-1,1,2,1,-2,-1,2,-1,-2};
int t,n,m,x,y,col,ex,ey,ans,ly,flag;
struct node
{
    int id;
    int l,r;
} dp[7000+5];
int cmp(node a,node b)
{
    return a.l<b.l;
}
int cmd(node a,node b)
{
    return a.r>b.r;
}
void DFS(ull k,int cnt)
{
    if(flag==1) return ;
    if(k%n==0)
    {

        cout<<k<<endl;
        flag = 1;
        return ;
    }
    if(cnt==19) return ;
    DFS( k*10 ,cnt+1);
    DFS( k*10+1 ,cnt+1);
}
int main()
{
    TLE;
    while(cin>>n&&n)      //这里的&&n 要有，不然的话会RE，原因是除0了
    {
        flag=0;
        DFS(1,0);
    }
    ok;
}


/*
int main()
{
    while(cin>>n)
    {
        ans=0;
        for(int i=0; i<n; i++)
            cin>>dp[i].l;
        for(int i=0; i<n; i++)
            cin>>dp[i].r;
        sort(dp,dp+n,cmp);

        for(int i=0; i<n; i++)
            cout<<dp[i].l;

        for(int i=0; i<n; i++)
        {
            if(dp[i].l==dp[i+1].l || dp[i].l==dp[i-1].l)
            {
                ans += dp[i].r;
                dp[i].r=0;
            }
        }
        sort(dp,dp+n,cmd);
        cout<<ans+dp[0].r<<endl;
    }
}
*/