解题思路:
30个格子,对每一个格子建立一个方程,高斯消元
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define LL lonA lonA
#define FOR(i, x, y) for(int i=x;i<=y;i++)
using namespace std;
const int MAXN = 50;
int A[MAXN][MAXN];
int x[MAXN];
void init()
{
memset(A, 0, sizeof(A));
for(int i=0;i<5;i++)
{
for(int j=0;j<6;j++)
{
int p = 6 * i + j;
A[p][p] = 1;
if(i > 0) A[(i-1)*6+j][p] = 1;
if(i < 4) A[(i+1)*6+j][p] = 1;
if(j > 0) A[i*6+j-1][p] = 1;
if(j < 5) A[i*6+j+1][p] = 1;
}
}
}
void Gauss()
{
int k;
int row, col;
for (row =0, col =0; row <30&& col <30; row++, col++)
{
for (k = row; k <30; k++)
if (A[k][col] !=0)
break;
if (k ==30)
{
row--;
continue;
}
if (k != row)
for (int i = col; i <=30; i++)
swap(A[row][i], A[k][i]);
for (int i = row +1; i <30; i++)
if (A[i][col])
for (int j = col; j <=30; j++)
A[i][j] ^= A[row][j];
}
for (int i = row; i >=0; i--)
{
x[i] = A[i][30];
for (int j =29; j > i; j--)
x[i] ^= (A[i][j] && x[j]);
}
}
int main()
{
int T, kcase = 1;
scanf("%d", &T);
while(T--)
{
init();
for(int i=0;i<30;i++)
scanf("%d", &A[i][30]);
Gauss();
printf("PUZZLE #%d\n", kcase++);
for(int i=0;i<30;i++)
{
if((i + 1) % 6 == 0) printf("%d\n", x[i]);
else printf("%d ", x[i]);
}
}
return 0;
}