枚举_百炼 2811 熄灯问题 (美妙的枚举函数)

Posted 2020-10-22 mapreduce

 1 #define _CRT_SECURE_NO_WARNINGS  
 2 #include <stdio.h>
 3 #include <math.h>
 4 #include <algorithm>
 5 #include <stdlib.h>
 6 #include <vector>
 7 #include <map>
 8 #include <queue>
 9 #include <string>
10 #include <iostream>
11 #include <ctype.h>
12 #include <string.h>
13 #include <set>
14 #include <stack>
15 #include<functional>
16 using namespace std;
17 #define Size 110
18 #define maxn  1<<30
19 #define minn  1e-6
20 int a[6][8];
21 int press[6][8];
22 bool guest(){
23     for (int i = 1; i < 5; i++){
24         for (int j = 1; j <= 6; j++){
25             press[i + 1][j] = (a[i][j] + press[i][j] + press[i - 1][j] + press[i][j - 1] + press[i][j + 1]) % 2;
26         }
27         /*
28         按不按[4][3]  取决于[3][3]的状态, 等于号后面的是现在(3,3)的状态
29         */
30     }
31     for (int i = 1; i <= 6; i++){
32         if (a[5][i] != (press[5][i] + press[4][i] + press[5][i - 1] + press[5][i + 1]) % 2) return false;
33     }
34     return true;
35 }
36 void solve(){
37     int c;
38     while (guest() == false){
39         c = 1;
40         press[1][1]++;
41         while (press[1][c] > 1){
42             press[1][c] = 0;
43             c++;
44             press[1][c]++;
45         }
46     }
47     /*
48     美妙的枚举函数
49     0
50     1
51     01
52     11
53     001
54     101
55     011
56     111
57     0001
58     */
59     for (int i = 1; i <= 5; i++){
60         for (int j = 1; j <= 6; j++)
61             if (j == 1) cout << press[i][j];
62             else cout << " " << press[i][j];
63             cout << endl;
64     }
65 
66 }
67 int main(){
68     for (int i = 1; i <= 5; i++)
69         for (int j = 1; j <= 6; j++)
70             cin >> a[i][j];
71     solve();
72     system("pause");
73 }