Codeforces Round #168 (Div. 2)-A. Lights Out_html/css_WEB-ITnose

Lights Out

time limit per test

memory limit per test

input

output

Lenny is playing a game on a 3?×?3 grid of lights. In the beginning of the game all lights are switched on. Pressing any of the lights will toggle it and all side-adjacent lights. The goal of the game is to switch all the lights off. We consider the toggling as follows: if the light was switched on then it will be switched off, if it was switched off then it will be switched on.

Lenny has spent some time playing with the grid and by now he has pressed each light a certain number of times. Given the number of times each light is pressed, you have to print the current state of each light.

Input

The input consists of three rows. Each row contains three integers each between 0 to 100 inclusive. The j-th number in the i-th row is the number of times the j-th light of the i-th row of the grid is pressed.

Output

Print three lines, each containing three characters. The j-th character of the i-th line is "1" if and only if the corresponding light is switched on, otherwise it's "0".

Sample test(s)

input

1 0 00 0 00 0 1

output

001010100

input

1 0 18 8 82 0 3

output

010011100

题目大意：现有3*3个开关，初始全为开着。切换（开变成关，关变成开）每个开关的时候，与它直接相邻的四个方向上的开关也会切换，给出每个开关的切换次数，问最后各个开关的状态。

解题思路：我们只需要判断每个开关到最后总共被切换了多少次，直接判断次数的奇偶即可判断某个开关最后的状态。直接遍历每个开关，但是如果直接在原来的开关次数上加，会影响对后来的计算，所以，我们开了两个数组，A[][]和B[][]，A是输入的每个开关的切换次数，B是最后每个开关切换的总次数。最后在扫一遍B即可，若B[i][j]是奇数，则状态为0（关），否则状态为1（开）。

AC代码：

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define INF 0x7fffffffint a[4][4], b[4][4];int main(){    #ifdef sxk        freopen("in.txt","r",stdin);    #endif    int n;        for(int i=0; i<3; i++)            for(int j=0; j<3; j++)                scanf("%d", &a[i][j]);        memset(b,0,sizeof(b));        for(int i=0; i<3; i++)            for(int j=0; j<3; j++){                if(a[i][j]){                b[i][j] += a[i][j];                if(i > 0) b[i-1][j] += a[i][j];                if(i < 2) b[i+1][j] += a[i][j];                if(j > 0) b[i][j-1] += a[i][j];                if(j < 2) b[i][j+1] += a[i][j];                }            }        for(int i=0; i<3; i++){            for(int j=0; j<3; j++){                printf("%d", b[i][j]&1^1);            }            printf("\n");    }    return 0;}