Code

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 1645	Accepted: 625

Description

KEY Inc., the leading company in security hardware, has developed a new kind of safe. To unlock it, you don't need a key but you are required to enter the correct n-digit code on a keypad (as if this were something new!). There are several models available, from toy safes for children (with a 2-digit code) to the military version (with a 6-digit code). 

The safe will open as soon as the last digit of the correct code is entered. There is no "enter" key. When you enter more than n digits, only the last n digits are significant. For example (in the 4-digit version), if the correct code is 4567, and you plan to enter the digit sequence 1234567890, the door will open as soon as you press the 7 key. 

The software to create this effect is rather simple. In the n-digit version the safe is always in one of 10^n-1 internal states. The current state of the safe simply represents the last n-1 digits that have been entered. One of these states (in the example above, state 456) is marked as the unlocked state. If the safe is in the unlocked state and then the right key (in the example above, 7) is pressed, the door opens. Otherwise the safe shifts to the corresponding new state. For example, if the safe is in state 456 and then you press 8, the safe goes into state 568. 

A trivial strategy to open the safe is to enter all possible codes one after the other. In the worst case, however, this will require n * 10ⁿ keystrokes. By choosing a good digit sequence it is possible to open the safe in at most 10ⁿ + n - 1 keystrokes. All you have to do is to find a digit sequence that contains all n-digit sequences exactly once. KEY Inc. claims that for the military version (n=6) the fastest computers available today would need billions of years to find such a sequence - but apparently they don't know what some programmers are capable of...

Input

The input contains several test cases. Every test case is specified by an integer n. You may assume that 1<=n<=6. The last test case is followed by a zero.

Output

For each test case specified by n output a line containing a sequence of 10ⁿ + n - 1 digits that contains each n-digit sequence exactly once.

Sample Input

1
2
0

Sample Output

0123456789
00102030405060708091121314151617181922324252627282933435363738394454647484955657585966768697787988990

Source

思路：要想最短，那麼每個相鄰的N位數前一個的後N-1位就是後一個的前N-1位。有要求序列字典序最小。因此使用棧從小到大搜。

#include<iostream>
#include<cstring>
using namespace std;
const int mm=1000109;
int g[mm],stak[mm],pos;///g[x]（N-1位）代表x出現了多少次
char ans[mm];
int n;
void see(int v,int m)///找出v為開頭最小後序
{ int w;
  while(g[v]<10)
  {
    w=v*10+g[v];++g[v];
    v=w%m;stak[pos++]=w;
  }
}
int main()
{ int m,v,z;
  while(cin>>n&&n)
  { m=1;pos=0;
    memset(g,0,sizeof(g));
    memset(stak,0,sizeof(stak));
    for(int i=1;i<n;i++)m*=10;
    v=0;
    see(v,m);
    z=0;
    while(pos)
    {
      v=stak[--pos];ans[z++]=v%10+'0';
      v/=10;see(v,m);
    }
    for(int i=1;i<n;i++)cout<<"0";
    for(int i=z-1;i>=0;i--)
      cout<<ans[i];
    cout<<"\n";
  }
}