Binary Table (Easy Version) (Codeforces 1440C1)
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):
import math
inf = float('inf')
mod = 10 ** 9 + 7
''
for _ in range(int(input())):
    (m, n) = map(int, input().split())
    A = [[0] * (n + 1)] + [[0] + list(map(int, list(input()))) for _ in range(m)]
    ans = []
    for i in range(1, m - 1):
        for j in range(1, n + 1):
            if A[i][j]:
                ans.append([i, j, i + 1, j, i + 1, j + 1 if j < n else j - 1])
                A[i][j] ^= 1
                A[i + 1][j] ^= 1
                A[i + 1][j + 1 if j < n else j - 1] ^= 1
    for j in range(1, n - 1):
        if A[m - 1][j] and A[m][j]:
            ans.append([m - 1, j, m, j, m - 1, j + 1])
            A[m - 1][j] ^= 1
            A[m][j] ^= 1
            A[m - 1][j + 1] ^= 1
        elif A[m - 1][j]:
            ans.append([m - 1, j, m - 1, j + 1, m, j + 1])
            A[m - 1][j] ^= 1
            A[m - 1][j + 1] ^= 1
            A[m][j + 1] ^= 1
        elif A[m][j]:
            ans.append([m, j, m, j + 1, m - 1, j + 1])
            A[m][j] ^= 1
            A[m][j + 1] ^= 1
            A[m - 1][j + 1] ^= 1
    if A[m - 1][n - 1] ^ A[m - 1][n] ^ A[m][n - 1]:
        ans.append([m - 1, n - 1, m - 1, n, m, n - 1])
    if A[m - 1][n - 1] ^ A[m][n - 1] ^ A[m][n]:
        ans.append([m - 1, n - 1, m, n - 1, m, n])
    if A[m][n - 1] ^ A[m][n] ^ A[m - 1][n]:
        ans.append([m, n - 1, m, n, m - 1, n])
    if A[m][n] ^ A[m - 1][n] ^ A[m - 1][n - 1]:
        ans.append([m, n, m - 1, n, m - 1, n - 1])
    print(len(ans))
    for a in ans:
        print(*a)

Solution (Python):
def read_ints():
    return list(map(int, input().split()))
d = {'0': '1', '1': '0'}

def change(rest, a, b, c):
    v = list(rest)
    for i in (a, b, c):
        v[i] = d[v[i]]
    return ''.join(v)

def get_moves(*args):
    result = []
    for i in args:
        result.extend([i % 2, i // 2])
    return [result]
d_rest = {'0000': []}
while len(d_rest) < 16:
    known = list(d_rest)
    for rest in known:
        new = change(rest, 0, 1, 2)
        if new not in d_rest:
            d_rest[new] = d_rest[rest] + get_moves(0, 1, 2)
        new = change(rest, 0, 1, 3)
        if new not in d_rest:
            d_rest[new] = d_rest[rest] + get_moves(0, 1, 3)
        new = change(rest, 0, 2, 3)
        if new not in d_rest:
            d_rest[new] = d_rest[rest] + get_moves(0, 2, 3)
        new = change(rest, 1, 2, 3)
        if new not in d_rest:
            d_rest[new] = d_rest[rest] + get_moves(1, 2, 3)

def remove_row(v):
    y = len(v) - 1
    row = v[-1]
    row1 = v[-2]
    result = []
    x = len(row) - 1
    while x > 0:
        if row[x] == '0':
            x -= 1
            continue
        if row[x - 1] == '1':
            result.append([x, y, x - 1, y, x, y - 1])
            row1[x] = d[row1[x]]
            x -= 2
            continue
        result.append([x, y, x, y - 1, x - 1, y - 1])
        row1[x] = d[row1[x]]
        row1[x - 1] = d[row1[x - 1]]
        x -= 2
    if x == 0 and row[0] == '1':
        result.append([0, y, 0, y - 1, 1, y - 1])
        row1[0] = d[row1[0]]
        row1[1] = d[row1[1]]
    del v[-1]
    return result

def remove_col(v):
    x = len(v[0]) - 1
    col = v[0][-1] + v[1][-1]
    del v[0][-1]
    del v[1][-1]
    if col == '00':
        return []
    if col == '01':
        v[0][-1] = d[v[0][-1]]
        v[1][-1] = d[v[1][-1]]
        return [[x, 1, x - 1, 0, x - 1, 1]]
    if col == '10':
        v[0][-1] = d[v[0][-1]]
        v[1][-1] = d[v[1][-1]]
        return [[x, 0, x - 1, 0, x - 1, 1]]
    v[0][-1] = d[v[0][-1]]
    return [[x, 0, x - 1, 0, x, 1]]
(t,) = read_ints()
for i in range(t):
    (n, m) = read_ints()
    values = [list(input()) for j in range(n)]
    moves = []
    while len(values) > 2:
        moves.extend(remove_row(values))
    while len(values[0]) > 2:
        moves.extend(remove_col(values))
    rest = ''.join(values[0] + values[1])
    moves.extend(d_rest[rest])
    print(len(moves))
    for (a, b, c, d0, e, f) in moves:
        print(*(b + 1, a + 1, d0 + 1, c + 1, f + 1, e + 1))

Solution (Python):
def main():
    (n, m) = map(int, input().split())
    ig = [input() for _ in range(n)]
    g = [[False] * (m + 1) for _ in range(n + 1)]
    ans = []

    def swap(a, b):
        g[a][b] = not g[a][b]

    def out(a, b, c, d, e, f):
        ans.append(' '.join(map(str, [a, b, c, d, e, f])))
        swap(a, b)
        swap(c, d)
        swap(e, f)
    for i in range(n):
        for j in range(m):
            if ig[i][j] == '1':
                g[i + 1][j + 1] = True
    for i in range(1, n - 1):
        for j in range(1, m):
            if g[i][j]:
                if g[i][j + 1]:
                    out(i, j, i, j + 1, i + 1, j + 1)
                else:
                    out(i, j, i + 1, j, i + 1, j + 1)
            if g[i][j + 1]:
                out(i, j + 1, i + 1, j, i + 1, j + 1)
    for i in range(1, m - 1):
        if g[n - 1][i]:
            if g[n][i]:
                out(n - 1, i, n, i, n - 1, i + 1)
            else:
                out(n - 1, i, n, i + 1, n - 1, i + 1)
        elif g[n][i]:
            out(n, i, n - 1, i + 1, n, i + 1)
    rest = [[n - 1, m - 1], [n - 1, m], [n, m - 1], [n, m]]

    def three():
        cur = []
        for (cx, cy) in rest:
            if g[cx][cy]:
                cur.append(cx)
                cur.append(cy)
        out(*cur)

    def two():
        flag = 1
        cur = []
        for (cx, cy) in rest:
            if g[cx][cy]:
                if flag:
                    flag -= 1
                    cur.append(cx)
                    cur.append(cy)
            else:
                cur.append(cx)
                cur.append(cy)
        out(*cur)
        three()

    def one():
        flag = 2
        cur = []
        for (cx, cy) in rest:
            if not g[cx][cy]:
                if flag:
                    flag -= 1
                    cur.append(cx)
                    cur.append(cy)
            else:
                cur.append(cx)
                cur.append(cy)
        out(*cur)
        two()

    def four():
        flag = 3
        cur = []
        for (cx, cy) in rest:
            if flag:
                flag -= 1
                cur.append(cx)
                cur.append(cy)
        out(*cur)
        one()
    mx = 0
    for i in rest:
        if g[i[0]][i[1]]:
            mx += 1
    if mx == 1:
        one()
    elif mx == 2:
        two()
    elif mx == 3:
        three()
    elif mx == 4:
        four()
    print(len(ans))
    for i in ans:
        print(i)
t = int(input())
for _ in range(t):
    main()
''