先确定窗口左上角可能出现的区域，方法是对每个点确定这样一个区域，然后求交。接下来枚举窗口左上角，计算密码序列，插入一个set中。最后按字典序输出这个set。
/**//*************************************************************************
Author: WHU_GCC
Created Time: 2007-10-21 10:24:05
File Name: pku3424.cpp
Description:
************************************************************************/
#include <iostream>
#include <string>
#include <set>
using namespace std;
#define out(x) (cout<<#x<<": "<<x<<endl)
const int maxint=0x7FFFFFFF;
typedef long long int64;
const int64 maxint64 = 0x7FFFFFFFFFFFFFFFLL;
template<class T>void show(T a, int n)
{for(int i=0; i<n; ++i) cout<<a[i]<<' '; cout<<endl;}
template<class T>void show(T a, int r, int l){for(int i=0; i<r; ++i)show(a[i],l);cout<<endl;}
int a[110][2];
set <string> ans;
int main()
{
int n;
while (scanf("%d", &n) != EOF)
{
for (int i = 0; i < n; i++)
{
scanf("%d%d", &a[i][0], &a[i][1]);
a[i][1] = -a[i][1];
}
ans.clear();
int x0l = -maxint, x0r = maxint, y0l = -maxint, y0r = maxint;
for (int i = 0; i < n; i++)
{
x0l >?= a[i][0] - 300;
x0r <?= a[i][0];
y0l >?= a[i][1] - 300;
y0r <?= a[i][1];
}
for (int x0 = x0r; x0 > x0l; x0--)
for (int y0 = y0r; y0 > y0l; y0--)
{
char t[110];
int len = 0;
int flag = 1;
for (int i = 0; i < n; i++)
{
int t1 = a[i][0] - x0, t2 = a[i][1] - y0;
if (t1 < 0 || t1 >= 300 || t2 < 0 || t2 >= 300)
{
flag = 0;
break;
}
int tt = t1 / 100 + t2 / 100 * 3;
t[len++] = '1' + tt;
}
if (flag)
{
t[len] = 0;
string ttt = t;
ans.insert(ttt);
}
}
if (ans.size() == 0)
{
printf ("NONE\n");
}
else
{
for (set <string> ::iterator it = ans.begin(); it != ans.end(); it++)
printf ("%s\n", it->c_str());
}
break;
}
return 0;
}