【其他】【USACO】Party Lamps

To brighten up the gala dinner of the IOI'98 we have a set of N (10 <= N <= 100) colored lamps numbered from 1 to N.

The lamps are connected to four buttons:

Button 1: When this button is pressed, all the lamps change their state: those that are ON are turned OFF and those that are OFF are turned ON.

Button 2: Changes the state of all the odd numbered lamps.

Button 3: Changes the state of all the even numbered lamps.

Button 4: Changes the state of the lamps whose number is of the form 3xK+1 (with K>=0), i.e., 1,4,7,...

A counter C records the total number of button presses.

When the party starts, all the lamps are ON and the counter C is set to zero.

You are given the value of counter C (0 <= C <= 10000) and the final state of some of the lamps after some operations have been executed. Write a program to determine all the possible final configurations of the N lamps that are consistent with the given information, without repetitions.

PROGRAM NAME: lamps

INPUT FORMAT

No lamp will be listed twice in the input.

Line 1:	N
Line 2:	Final value of C
Line 3:	Some lamp numbers ON in the final configuration, separated by one space and terminated by the integer -1.
Line 4:	Some lamp numbers OFF in the final configuration, separated by one space and terminated by the integer -1.

SAMPLE INPUT (file lamps.in)

10
1
-1
7 -1

In this case, there are 10 lamps and only one button has been pressed. Lamp 7 is OFF in the final configuration.

OUTPUT FORMAT

Lines with all the possible final configurations (without repetitions) of all the lamps. Each line has N characters, where the first character represents the state of lamp 1 and the last character represents the state of lamp N. A 0 (zero) stands for a lamp that is OFF, and a 1 (one) stands for a lamp that is ON. The lines must be ordered from least to largest (as binary numbers).

If there are no possible configurations, output a single line with the single word `IMPOSSIBLE'

SAMPLE OUTPUT (file lamps.out)

0000000000
0101010101
0110110110

In this case, there are three possible final configurations:

All lamps are OFF

Lamps 1, 4, 7, 10 are OFF and lamps 2, 3, 5, 6, 8, 9 are ON.

Lamps 1, 3, 5, 7, 9 are OFF and lamps 2, 4, 6, 8, 10 are ON.

描述

在IOI98的节日宴会上，我们有N(10<=N<=100)盏彩色灯，他们分别从1到N被标上号码。这些灯都连接到四个按钮：

按钮1：当按下此按钮，将改变所有的灯：本来亮着的灯就熄灭，本来是关着的灯被点亮。
按钮2：当按下此按钮，将改变所有奇数号的灯。
按钮3：当按下此按钮，将改变所有偶数号的灯。
按钮4：当按下此按钮，将改变所有序号是3*K+1(K>=0)的灯。例如：1,4,7...

一个计数器C记录按钮被按下的次数。当宴会开始，所有的灯都亮着，此时计数器C为0。

你将得到计数器C(0<=C<=10000)上的数值和经过若干操作后某些灯的状态。写一个程序去找出所有灯最后可能的与所给出信息相符的状态，并且没有重复。

格式

PROGRAM NAME: lamps

INPUT FORMAT:

(file lamps.in)

不会有灯会在输入中出现两次。

第一行: N。

第二行: C最后显示的数值。

第三行: 最后亮着的灯,用一个空格分开，以-1为结束。

第四行: 最后关着的灯,用一个空格分开，以-1为结束。

OUTPUT FORMAT:

(file lamps.out)

每一行是所有灯可能的最后状态(没有重复)。每一行有N个字符，第1个字符表示1号灯，最后一个字符表示N号灯。0表示关闭，1表示亮着。这些行必须从小到大排列（看作是二进制数）。

如果没有可能的状态，则输出一行'IMPOSSIBLE'。

SAMPLE INPUT

10
1
-1
7 -1

在这个样例中，有10盏灯，只有1个按钮被按下。最后7号灯是关着的。

SAMPLE OUTPUT

0000000000
0101010101
0110110110

在这个样例中，有三种可能的状态：

所有灯都关着

1,4,7,10号灯关着，2,3,5,6,8,9亮着。

1,3,5,7,9号灯关着，2, 4, 6, 8, 10亮着。

来自"http://www.nocow.cn/index.php/Translate:USACO/lamps"

找规律 然后打个表 然后枚举

/*
ID:sephiro5
LANG:C++
PROG:lamps
*/
#include<stdio.h>
int li[10][10]={
{},
{0,0,0,0,0,0,0},
{0,0,0,1,1,1,0},
{0,0,1,0,1,0,1},
{0,0,1,1,0,1,1},
{0,1,0,0,1,0,0},
{0,1,0,1,0,1,0},
{0,1,1,0,0,0,1},
{0,1,1,1,1,1,1},
};
int minnum[50]={0,1,2,1,1,2,1,2,0};
int v[1000];
int n,c;
int main(){
freopen("lamps.in","r",stdin);
freopen("lamps.out","w",stdout);
scanf("%d%d",&n,&c);
for (int i=1;i<=n;++i)
v[i]=-1;
int temp;
while (1){
scanf("%d",&temp);
if (temp==-1)break;
else v[temp]=1;
}
while (1){
scanf("%d",&temp);
if (temp==-1)break;
else v[temp]=0;
}
int t,p=0,q=0;
for (int k=1;k<=8;++k){
p=1;
for (int i=1;i<=n;++i)
{
if (v[i]==-1) continue;
t=i%6; if (t==0) t=6;
if (v[i]!=li[k][t])
{
p=0;
break;
}
}
if (p==1&&c>=minnum[k])
{
q=1;
for (int j=1;j<=n;++j)
{
t=j%6;
if (t==0) t=6;
printf("%d",li[k][t]);
}
printf("/n");
}
}
if (q==0) printf("IMPOSSIBLE/n");
return 0;
}