Binary Table (Easy Version) (Codeforces 1439A1)
Tags: Others, implementation

This is the easy version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.You are given a binary table of size $$$n \times m$$$. This table consists of symbols $$$0$$$ and $$$1$$$.You can make such operation: select $$$3$$$ different cells that belong to one $$$2 \times 2$$$ square and change the symbols in these cells (change $$$0$$$ to $$$1$$$ and $$$1$$$ to $$$0$$$).Your task is to make all symbols in the table equal to $$$0$$$. You are allowed to make at most $$$3nm$$$ operations. You don't need to minimize the number of operations.It can be proved that it is always possible.

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 5000$$$) — the number of test cases. The next lines contain descriptions of test cases.The first line of the description of each test case contains two integers $$$n$$$, $$$m$$$ ($$$2 \leq n, m \leq 100$$$).Each of the next $$$n$$$ lines contains a binary string of length $$$m$$$, describing the symbols of the next row of the table.It is guaranteed that the sum of $$$nm$$$ for all test cases does not exceed $$$20000$$$.

Output: For each test case print the integer $$$k$$$ ($$$0 \leq k \leq 3nm$$$) — the number of operations.In the each of the next $$$k$$$ lines print $$$6$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3$$$ ($$$1 \leq x_1, x_2, x_3 \leq n, 1 \leq y_1, y_2, y_3 \leq m$$$) describing the next operation. This operation will be made with three cells $$$(x_1, y_1)$$$, $$$(x_2, y_2)$$$, $$$(x_3, y_3)$$$. These three cells should be different. These three cells should belong into some $$$2 \times 2$$$ square.

Note: In the first test case, it is possible to make only one operation with cells $$$(1, 1)$$$, $$$(2, 1)$$$, $$$(2, 2)$$$. After that, all symbols will be equal to $$$0$$$.In the second test case:  operation with cells $$$(2, 1)$$$, $$$(3, 1)$$$, $$$(3, 2)$$$. After it the table will be: 011001000  operation with cells $$$(1, 2)$$$, $$$(1, 3)$$$, $$$(2, 3)$$$. After it the table will be: 000000000 In the fifth test case:  operation with cells $$$(1, 3)$$$, $$$(2, 2)$$$, $$$(2, 3)$$$. After it the table will be: 010110  operation with cells $$$(1, 2)$$$, $$$(2, 1)$$$, $$$(2, 2)$$$. After it the table will be: 000000 

Sample Inputs:
5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101

Sample Outputs:
1
1 1 2 1 2 2
2 
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2 
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2 
1 4 1 5 2 5 
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2

Solution (Python):
t = int(input())
for _ in range(t):
    (n, m) = map(int, input().split())
    a = [list(map(int, input())) for i in range(n)]

    def add(p):
        a[p[0]][p[1]] ^= 1
        a[p[2]][p[3]] ^= 1
        a[p[4]][p[5]] ^= 1
        ans.append(' '.join(map(lambda x: str(x + 1), p)))
    ans = []
    for i in range(n - 1, 1, -1):
        for j in range(m):
            if a[i][j]:
                add((i, j, i - 1, j, i - 1, j + (1 if j < m - 1 else -1)))
    for j in range(m - 1, 1, -1):
        for i in (0, 1):
            if a[i][j]:
                add((i, j, 0, j - 1, 1, j - 1))
    for i in (0, 1):
        for j in (0, 1):
            if (a[0][0] + a[0][1] + a[1][0] + a[1][1] + a[i][j]) % 2:
                add((i ^ 1, j, i, j ^ 1, i ^ 1, j ^ 1))
    print(len(ans))
    print('\n'.join(ans))

Solution (Python):
import bisect
import collections
import random
import sys
p2D = lambda x: print(*x, sep='\n')

def II():
    return int(sys.stdin.buffer.readline())

def MI():
    return map(int, sys.stdin.buffer.readline().split())

def LI():
    return list(map(int, sys.stdin.buffer.readline().split()))

def LLI(rows_number):
    return [LI() for _ in range(rows_number)]

def BI():
    return sys.stdin.buffer.readline().rstrip()

def SI():
    return sys.stdin.buffer.readline().rstrip().decode()

def li():
    return [int(i) for i in input().split()]

def lli(rows):
    return [li() for _ in range(rows)]

def si():
    return input()

def ii():
    return int(input())

def ins():
    return input().split()

def drank(d, processing, da, rank):
    tmp = 10 ** 9
    if len(d[da]) == 1:
        return 1
    for di in d[da]:
        if processing[di - 1] == 0:
            processing[di - 1] = 1
            tmp = min(tmp, drank(d, processing, di, rank))
            processing[di - 1] = 0
    rank[da - 1] = tmp + 1
    return tmp + 1

def binary_search(n, a):
    l = len(a)
    low = 0
    high = l - 1
    while high >= low:
        mid = (high + low) // 2
        if a[mid] == n:
            return mid
        elif a[mid] > n:
            high = mid - 1
        else:
            low = mid + 1
    return -1
import math

def invert(matrix, x1, y1, x2, y2, x3, y3):
    matrix[x1 - 1][y1 - 1] = 1 - matrix[x1 - 1][y1 - 1]
    matrix[x2 - 1][y2 - 1] = 1 - matrix[x2 - 1][y2 - 1]
    matrix[x3 - 1][y3 - 1] = 1 - matrix[x3 - 1][y3 - 1]

def solve():
    (n, m) = LI()
    grid = []
    for i in range(n):
        grid.append(list(map(int, list(SI()))))
    ans = []
    for row in range(n - 1):
        if grid[row][0] == 1:
            ans.append([row + 1, 1, row + 2, 1, row + 2, 2])
            invert(grid, row + 1, 1, row + 2, 1, row + 2, 2)
        for col in range(1, m):
            if grid[row][col] == 1:
                ans.append([row + 1, col + 1, row + 2, col, row + 2, col + 1])
                invert(grid, row + 1, col + 1, row + 2, col, row + 2, col + 1)
    for i in range(0, m - 1, 2):
        if grid[n - 1][i] == grid[n - 1][i + 1] == 1:
            ans.append([n - 1, i + 1, n - 1, i + 2, n, i + 1])
            invert(grid, n - 1, i + 1, n - 1, i + 2, n, i + 1)
            ans.append([n - 1, i + 1, n - 1, i + 2, n, i + 2])
            invert(grid, n - 1, i + 1, n - 1, i + 2, n, i + 2)
        elif grid[n - 1][i] != grid[n - 1][i + 1]:
            ans.append([n, i + 1, n, i + 2, n - 1, i + 1])
            invert(grid, n, i + 1, n, i + 2, n - 1, i + 1)
            if grid[n - 1][i] == 1:
                ans.append([n - 1, i + 2, n, i + 2, n, i + 1])
                invert(grid, n - 1, i + 2, n, i + 2, n, i + 1)
                ans.append([n - 1, i + 2, n, i + 2, n - 1, i + 1])
                invert(grid, n - 1, i + 2, n, i + 2, n - 1, i + 1)
            else:
                ans.append([n, i + 1, n - 1, i + 1, n - 1, i + 2])
                invert(grid, n, i + 1, n - 1, i + 1, n - 1, i + 2)
                ans.append([n - 1, i + 2, n, i + 1, n, i + 2])
                invert(grid, n - 1, i + 2, n, i + 1, n, i + 2)
    if m % 2 == 1 and grid[n - 1][m - 1] == 1:
        ans.append([n - 1, m, n, m - 1, n, m])
        invert(grid, n - 1, m, n, m - 1, n, m)
        ans.append([n - 1, m - 1, n, m, n - 1, m])
        invert(grid, n - 1, m - 1, n, m, n - 1, m)
        ans.append([n, m - 1, n, m, n - 1, m - 1])
        invert(grid, n, m - 1, n, m, n - 1, m - 1)
    print(len(ans))
    for i in range(len(ans)):
        print(' '.join(map(str, ans[i])))

def main():
    for _ in range(II()):
        solve()
if __name__ == '__main__':
    main()

Solution (Python):
import sys

def minp():
    return sys.stdin.readline().strip()

def mint():
    return int(minp())

def mints():
    return map(int, minp().split())

def solve():
    (n, m) = mints()
    a = [list(map(int, minp())) for i in range(n)]
    ans = []
    for i in range(n - 2):
        l = a[i]
        ll = a[i + 1]
        for j in range(m - 1):
            if l[j] == 1:
                if l[j + 1] == 1:
                    ans.append((i, j, i, j + 1, i + 1, j))
                    l[j + 1] ^= 1
                    ll[j] ^= 1
                else:
                    ans.append((i, j, i + 1, j, i + 1, j + 1))
                    ll[j] ^= 1
                    ll[j + 1] ^= 1
            elif l[j + 1] == 1:
                ans.append((i, j + 1, i + 1, j, i + 1, j + 1))
                l[j + 1] ^= 1
                ll[j] ^= 1
                ll[j + 1] ^= 1
    aa = a[n - 2]
    bb = a[n - 1]
    for j in range(m - 2):
        if aa[j] == 1:
            if bb[j] == 1:
                ans.append((n - 2, j, n - 1, j, n - 2, j + 1))
                aa[j + 1] ^= 1
            else:
                ans.append((n - 2, j, n - 2, j + 1, n - 1, j + 1))
                aa[j + 1] ^= 1
                bb[j + 1] ^= 1
        elif bb[j] == 1:
            ans.append((n - 1, j, n - 2, j + 1, n - 1, j + 1))
            aa[j + 1] ^= 1
            bb[j + 1] ^= 1
    c = [(n - 2, m - 2), (n - 2, m - 1), (n - 1, m - 2), (n - 1, m - 1)]
    while True:
        ss = 0
        for (x, y) in c:
            ss += a[x][y]
        if ss == 0:
            break
        if ss == 1 or ss == 3:
            for i in range(4):
                (x, y) = c[i]
                if a[x][y] == 0:
                    break
        if ss == 2:
            for i in range(4):
                (x, y) = c[i]
                if a[x][y] == 1:
                    break
        if ss == 4:
            i = 0
        d = c.copy()
        d.pop(i)
        tmp = []
        for (x, y) in d:
            a[x][y] ^= 1
            tmp.append(x)
            tmp.append(y)
        ans.append(tmp)
    print(len(ans))
    for l in ans:
        print(' '.join([str(i + 1) for i in l]))
for i in range(mint()):
    solve()