Problem:


Score : 300 points

Problem Statement

There is a grid with infinitely many rows and columns. In this grid, there is a rectangular region with consecutive N rows and M columns, and a card is placed in each square in this region.
The front and back sides of these cards can be distinguished, and initially every card faces up.

We will perform the following operation once for each square contains a card:

For each of the following nine squares, flip the card in it if it exists: the target square itself and the eight squares that shares a corner or a side with the target square.

It can be proved that, whether each card faces up or down after all the operations does not depend on the order the operations are performed.
Find the number of cards that face down after all the operations.

Constraints

1 \leq N,M \leq 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

N M

Output

Print the number of cards that face down after all the operations.

Sample Input 1

2 2

Sample Output 1

0

We will flip every card in any of the four operations. Thus, after all the operations, all cards face up.

Sample Input 2

1 7

Sample Output 2

5

After all the operations, all cards except at both ends face down.

Sample Input 3

314 1592

Sample Output 3

496080



Sample input:
2 2


Sample output:
0

Submission status: Accepted

Solution code:
def main(sample_file = ''):

    """ convenient functions
    # for i, a in enumerate(iterable)
    # q, mod = divmod(a, b)
    # divmod(x, y) returns the tuple (x//y, x%y)
    # Higher-order function: reduce(operator.mul, xyz_count, 1)
    # manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
    """

    """convenient decorator
    # @functools.lru_cache():
    # to facilitate use of recursive function
        # ex:
        # from functools import lru_cache
        # import sys
        # sys.setrecursionlimit(10**9)
        # @lru_cache(maxsize=None)
        # def fib(n):
        #     if n < 2:
        #         return n
        #     return fib(n-1) + fib(n-2)
        # print(fib(1000))
    """
    
    # import numpy as np
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product, combinations_with_replacement # https://docs.python.org/ja/3/library/itertools.html
    # accumulate() returns iterator! to get list: list(accumulate())
    from math import factorial, ceil, floor, sqrt
    def factorize(n):
        """return the factors of the Arg and count of each factor
        
        Args:
            n (long): number to be resolved into factors
        
        Returns:
            list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
        """
        fct = []  # prime factor
        b, e = 2, 0  # base, exponent
        while b * b <= n:
            while n % b == 0:
                n = n // b
                e = e + 1
            if e > 0:
                fct.append((b, e))
            b, e = b + 1, 0
        if n > 1:
            fct.append((n, 1))
        return fct
    def combinations_count(n, r):   
        """Return the number of selecting r pieces of items from n kinds of items.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        # TODO: How should I do when n - r is negative?
        if n < 0 or r < 0:
            raise Exception('combinations_count(n, r) not defined when n or r is negative')
        if n - r < r: r = n - r
        if r < 0: return 0
        if r == 0: return 1
        if r == 1: return n
        numerator = [n - r + k + 1 for k in range(r)]
        denominator = [k + 1 for k in range(r)]
        for p in range(2,r+1):
            pivot = denominator[p - 1]
            if pivot > 1:
                offset = (n - r) % p
                for k in range(p-1,r,p):
                    numerator[k - offset] /= pivot
                    denominator[k] /= pivot
        result = 1
        for k in range(r):
            if numerator[k] > 1:
                result *= int(numerator[k])
        return result
    def combinations_with_replacement_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        if n < 0 or r < 0:
            raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
        elif n == 0:
            return 1
        else:
            return combinations_count(n + r - 1, r)
    def permutations_count(n, r, MOD=None):
        ans = 1
        if MOD:            
            for i in range(r):
                ans = (ans * (n-i))%MOD
        else:
            for i in range(r):
                ans = (ans * (n-i))
        return ans
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    # from operator import add, mul
    from operator import itemgetter #sort 
    # ex1: List.sort(key=itemgetter(1))
    # ex2: sorted(tuples, key=itemgetter(1,2))
    from functools import reduce, lru_cache
    def chmin(x, y):
        """change minimum
        if x > y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current minimum value
            y (long): potential minimum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x > y:
            x = y
            return (x, True)
        else:
            return (x, False)
    def chmax(x, y):
        """change maximum
        if x < y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current maximum value
            y (long): potential maximum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x < y:
            x = y
            return (x, True)
        else:
            return (x, False)

    from fractions import gcd
    def gcd(a,b):
        # from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
        # => 最近の環境でfractions.gcdを使うとTLEする。
        if a<b:
            a, b = b, a
        while a%b !=0 :
            a, b = b, a%b
        return b

    # def gcd(p,q):
    #     while q != 0:
    #         r = p % q
    #         p = q
    #         q = r
    #     return p

    def gcds(numbers):
        return reduce(gcd, numbers)
    def lcm(x, y):
        return (x * y) // gcd(x, y)
    def lcms(numbers):
        return reduce(lcm, numbers, 1)
    def make_divisors(n, reversed=False):
        """create list of divisors
        
        Args:
            number (int): number from which list of divisors is created
            reversed (bool, optional): ascending order if False. Defaults to False.
        
        Returns:
            list: list of divisors
        """
        divisors = set()
        for i in range(1, int(n**0.5)+1):
            if n % i == 0:
                divisors.add(i)
                divisors.add(n//i)
        return sorted(list(divisors),reverse=reversed)

    # Eratosthenesの篩により、整数end未満の素数を列挙する
    def sieve_of_eratosthenes(end, typecode="L"):
        # len(sieve_of_eratosthenes(10**7)) => 664579; 514ms in pypy
        # len(sieve_of_eratosthenes(10**6)) => 78498; 134ms in pypy

        # https://hamukichi.hatenablog.jp/entry/2016/02/17/215948#Eratosthenes%E3%81%AE%E7%AF%A9%E3%81%A8%E5%8C%BA%E9%96%93%E7%AF%A9
        import array
        import textwrap
        assert end > 1
        # 整数iが素数であるかをis_prime[i]が示す
        # 最初はすべてTrueで初期化しておく
        # 最終的にprimesではなくこれを返してもよい
        is_prime = array.array("B", (True for i in range(end)))
        # 0, 1はいずれも素数ではない
        is_prime[0] = False
        is_prime[1] = False
        # 素数を格納する配列
        primes = array.array(typecode)
        # 篩う
        for i in range(2, end):
            if is_prime[i]:
                primes.append(i)
                for j in range(2 * i, end, i):
                    is_prime[j] = False  # 素数ではないため除外する
        return primes

    class PrimeFactorization:
        '''初期化にO(NloglogN)/ 因数分解クエリをO(log(k))/ 素数判定クエリをO(1)
        >>> pf = PrimeFactorization(100)
        >>> pf.query(100)
        ((2, 2), (5, 2))
        
        '''
        # 自分で実装した。
        # https://atcoder.jp/contests/abc177/editorial/82
        # 高速素因数分解
        # 問題： A 以下の数が N 個与えられる。全て素因数分解せよ。
        # 前計算としてエラトステネスの篩を行い、「その数をふるい落とした素数」を配列 D に記録します。
        # 例えば D[4]=D[6]=2,D[35]=5 です。x が素数のときは D[x]=x としておきます。この配列はエラトステネスの篩と同様 O(AloglogA) で構築できます。
        # D[x] は x を割り切る最小の素数なので、この配列 D を利用すると素因数分解を行うときに「試し割り」をする必要がなくなり(D[x]で割ればよい)、1つの数の素因数分解が素因数の個数である O(logA) でできるようになります。
        def __init__(self, Num=10**6):
            from collections import defaultdict
            from math import floor
            self.Num = Num
            self.D = [-1] * (Num+1)
            for i in range(2, Num+1):
                if self.D[i] != -1:
                    continue
                self.D[i] = i
                for j in range(1, floor(Num/i)+1):
                    if self.D[j*i] != -1:
                        continue
                    else:
                        self.D[j*i] = i
        
        def query(self, k):
            ans = defaultdict(int)
            tmp = k
            while self.D[tmp] != -1:
                div = self.D[tmp]
                ans[div] += 1
                tmp = tmp//div
            return tuple(ans.items())
        
        def is_prime(self, k):
            return k == self.D[k]

    # first create factorial_list
    # fac_list = modfactorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    AtoZ = list('abcdefghijklmnopqrstuvwxyz') # 26
    modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
    def modinv(a, p = MOD):
        # evaluate reciprocal using Fermat's little theorem:
        # a**(p-1) is identical to 1 (mod p) when a and p is coprime
        return modpow(a, p-2, p)
    def modinv_list(n, p = MOD):
        if n <= 1:
            return [0,1][:n+1]
        else:
            inv_t = [0,1]
            for i in range(2, n+1):
                inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
            return inv_t
    def modfactorial_list(n, p = MOD):
        if n == 0:
            return [1]
        else:
            l = [0] * (n+1)
            tmp = 1
            for i in range(1, n+1):
                tmp = tmp * i % p
                l[i] = tmp
            return l
    def modcomb(n, k, fac_list = [], p = MOD):
        # fac_list = modfactorial_list(100)
        # print(modcomb(100, 5, modfactorial_list(100)))
        from math import factorial
        if n < 0 or k < 0 or n < k: return 0
        if n == 0 or k == 0: return 1
        if len(fac_list) <= n:
            a = factorial(n) % p
            b = factorial(k) % p
            c = factorial(n-k) % p
        else:
            a = fac_list[n]
            b = fac_list[k]
            c = fac_list[n-k]
        return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
    def modadd(a, b, p = MOD):
        return (a + b) % MOD
    def modsub(a, b, p = MOD):
        return (a - b) % p
    def modmul(a, b, p = MOD):
        return ((a % p) * (b % p)) % p
    def moddiv(a, b, p = MOD):
        return modmul(a, modpow(b, p-2, p))

    # ************ Data Structure ************
    from bisect import bisect_left, bisect_right
    from collections import deque, Counter, defaultdict # https://docs.python.org/ja/3/library/collections.html#collections.deque
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
    
    class UnionFindTree:
        """union find tree class
        TODO: fix this description...
        how to use (example):
        >>  uf = UnionFindTree(N) 
        >>  if uf.find_root(a) == uf.find_root(b):
        >>      do something like:
        >>      pass
        >>  else:
        >>      do something like:
        >>      L[find_root(a)] = L[find_root(a)] + L[find_root(b)]
        >>      L[find_root(b)] = L[find_root(a)]
        >>      
        >>      uf.unite(a, b)
        """
        def __init__(self, N):
            self.root = [-1] * (N+1)
            self.rank   = [0]  * (N+1)
            self.connected_num = [1] * (N+1)
    
        def find_root(self,x):
            root = self.root
            while root[x] != -1:
                x = root[x]
            return x
    
        def unite(self,x,y):
            root = self.root
            rank = self.rank
            connected_num = self.connected_num
            find_root = self.find_root
    
            rx = find_root(x)
            ry = find_root(y)
            if rx != ry:
                if rank[rx] < rank[ry]:
                    root[rx] = ry
                    rx,ry = ry,rx
                else:
                    if rank[rx] == rank[ry]:
                        rank[rx] += 1
                    root[ry] = rx       
                connected_num[rx] += connected_num[ry]

    # c++ set
    # https://nagiss.hateblo.jp/entry/2020/09/08/203701
    
    class SegTree:
        """
        ref: https://qiita.com/takayg1/items/c811bd07c21923d7ec69
        init(init_val, ide_ele): 配列init_valで初期化 O(N)
        update(k, x): k番目の値をxに更新 O(logN)
        query(l, r): 区間[l, r)をsegfuncしたものを返す O(logN)
        
        操作	    segfunc	        単位元
        最小値	    min(x, y)	    float('inf')
        最大値	    max(x, y)	    -float('inf')
        区間和	    x + y	        0
        区間積	    x * y	        1
        最大公約数	math.gcd(x, y)	0
        
        ex: seg = SegTree([14, 5, 9, 13], min, float('inf'))
        """
        def __init__(self, init_val, segfunc, ide_ele):
            """
            init_val: 配列の初期値
            segfunc: 区間にしたい操作
            ide_ele: 単位元
            n: 要素数
            num: n以上の最小の2のべき乗
            tree: セグメント木(1-index)
            """
            n = len(init_val)
            self.segfunc = segfunc
            self.ide_ele = ide_ele
            self.num = 1 << (n - 1).bit_length()
            self.tree = [ide_ele] * 2 * self.num
            # 配列の値を葉にセット
            for i in range(n):
                self.tree[self.num + i] = init_val[i]
            # 構築していく
            for i in range(self.num - 1, 0, -1):
                self.tree[i] = self.segfunc(self.tree[2 * i], self.tree[2 * i + 1])

        def update(self, k, x):
            """
            k番目の値をxに更新
            k: index(0-index)
            x: update value
            """
            k += self.num
            self.tree[k] = x
            while k > 1:
                self.tree[k >> 1] = self.segfunc(self.tree[k], self.tree[k ^ 1])
                k >>= 1

        def query(self, l, r):
            """
            [l, r)のsegfuncしたものを得る
            l: index(0-index)
            r: index(0-index)
            """
            res = self.ide_ele
            l += self.num
            r += self.num
            while l < r:
                if l & 1:
                    res = self.segfunc(res, self.tree[l])
                    l += 1
                if r & 1:
                    res = self.segfunc(res, self.tree[r - 1])
                l >>= 1
                r >>= 1
            return res

    # ************ Graph ************
    # Graph: https://en.wikipedia.org/wiki/Directed_graph
    # 最短経路問題: https://ja.wikipedia.org/wiki/%E6%9C%80%E7%9F%AD%E7%B5%8C%E8%B7%AF%E5%95%8F%E9%A1%8C
    
    # Bellman-Ford: O(|V||E|). Use this if there exists an edge with negative length in the graph
    # After N steps, the shortest path has converded if there doesn't exist an cycle of edges with negative
    # Watch out: d[N] == d[2*N] doesn't necessarily mean the graph doesn't have negative cycle
    # ref: https://www.youtube.com/watch?v=1Z6ofKN03_Y

    def BellmanFord(N, M, ABC, vertex_start, vertex_end, value_if_inf = -1, find_shortest = False):
        """to calculate furthest or shortest length between vertex_start and vertex_end using BellmanFord algorithm
        
        Args:
            N (int): number of vertices
            M (int): number of edges
            ABC (list): [(ai, bi, ci) for _ in range(N)] where i-th edge is directed from vertex ai to vertex bi and the length is ci 
            vertex_start (int): start vertex. usually use 0.
            vertex_end (int): end vertex. usually use N-1.
            value_if_inf (int or string as you like, optional): value you want when the furthest (or shortest) distance is infinite (or -infinite). Defaults to -1.
            find_shortest (bool, optional): choose False to find furthest path. Defaults to False.
        
        Returns:
            int or string: normally int (but can be str if you set value_if_inf to str)
            
        Example:
                N, M, P = R()
                ABC = [R() for _ in range(M)]
                ABC = [(a-1, b-1, c-P) for a, b, c in ABC]
                print(BellmanFord(N, M, ABC, 0, N-1, value_if_inf = 'inf'))

        """

        def make_reachable_list(N, M, ABC, vertex_start, vertex_end):
            reachable_to_direct = defaultdict(list)
            reachable_from_direct = defaultdict(list)
            reachable_from_start = [False] * N
            reachable_to_end = [False] * N
            reachable_from_start[vertex_start] = True
            reachable_to_end[vertex_end] = True
            reachable_from_both_sides = [False] * N
            dfs_from_start = []
            dfs_to_end = []
            for a, b, c in ABC:
                reachable_to_direct[a].append(b)
                reachable_from_direct[b].append(a)
                if a == vertex_start:
                    dfs_from_start.append(b)
                    reachable_from_start[b] = True
                if b == vertex_end:
                    dfs_to_end.append(a)
                    reachable_to_end[a] = True
            while dfs_from_start:
                v = dfs_from_start.pop()
                for i in reachable_to_direct[v]:
                    if not reachable_from_start[i]:
                        reachable_from_start[i] = True
                        dfs_from_start.append(i)
            while dfs_to_end:
                v = dfs_to_end.pop()
                for i in reachable_from_direct[v]:
                    if not reachable_to_end[i]:
                        reachable_to_end[i] = True
                        dfs_to_end.append(i)
            for i in range(N):
                reachable_from_both_sides[i] = reachable_from_start[i] and reachable_to_end[i]
            return reachable_from_both_sides
        
        reachable_from_both_sides = make_reachable_list(N, M, ABC, vertex_start, vertex_end)
        
        if find_shortest:
            dist = [INF for i in range(N)]
        else:
            dist = [-INF for i in range(N)]
        
        dist[vertex_start] = 0
        for i in range(N):
            updated = False
            for a, b, c in ABC:
                if not reachable_from_both_sides[a]:
                    continue
                elif find_shortest:
                    update_condition = dist[a] + c < dist[b]
                else:
                    update_condition = dist[a] + c > dist[b]
                if dist[a] != INF and update_condition:
                    dist[b] = dist[a] + c
                    updated = True
                    if i == N-1:
                        return value_if_inf
            if not updated:
                break
        return dist[vertex_end]

    # Warshall Floyd O(V**3)
    def warshall_floyd(number_vertex, XYD, shift_one_for_vertex=False):
        # ref: https://juppy.hatenablog.com/entry/2018/11/01/%E8%9F%BB%E6%9C%AC_python_%E5%85%A8%E7%82%B9%E5%AF%BE%E6%9C%80%E7%9F%AD%E7%B5%8C%E8%B7%AF%E6%B3%95%EF%BC%88%E3%83%AF%E3%83%BC%E3%82%B7%E3%83%A3%E3%83%AB%E3%83%95%E3%83%AD%E3%82%A4%E3%83%89%E6%B3%95
        # number_vertex: len(vertex)
        # XYD: list of distance from x to y
        # shift_one_for_vertex: XYDの値が1~Nの時(0~N-1でない時)に調整するフラグ。
        # d[i][j]: iからjへの最短距離
        d = [[float("inf")]*number_vertex for i in range(number_vertex)]
        for x, y, dist in XYD:
            if shift_one_for_vertex:
                x = x - 1
                y = y - 1
            d[x][y] = dist
            d[y][x] = dist
        for i in range(number_vertex):
            d[i][i] = 0 # distance of same vertex is 0
        for k in range(number_vertex):
            for i in range(number_vertex):
                for j in range(number_vertex):
                    d[i][j] = min(d[i][j],d[i][k] + d[k][j])
        return d


    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array:
            # ex) [[0] * N for _ in range(N)]
            # ex2) dp = [[0] * (N+1) for _ in range(W*2)]
    # set inputs
        # put inputs between specific values (almost as quickly)
        # ex) S = [-INF] + [int(r()) for _ in range(A)] + [INF]
        # open(0).read() is sometimes useful:
        # ex) n, m, *x = map(int, open(0).read().split())
        #     min(x[::2]) - max(x[1::2])
        # ex2) *x, = map(int, open(0).read().split())
        #     don't forget to add comma after *x if only one variable is used
    # preprocessing
        # transpose = [x for x in zip(*data)]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
        # flat = [flatten for inner in data for flatten in inner]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [1, 2, 3, 4, 5, 6, 7, 8, 9]
    # calculate and output
        # output pattern
        # ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
    """

    # functions to read input
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    r_float = lambda: float(r())
    R = lambda: list(map(int, r().split()))
    # R_str = lambda: r().split()
    R_map = lambda: map(int, r().split())
    R_float = lambda: list(map(float, r().split()))
    R_tuple = lambda: tuple(map(int, r().split()))

    """ how to treat input
    # single int: int(r())
    # single string: r()
    # single float: float(r())
    # line int: R()
    # line string: r().split()
    # line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
    # lines int: [R() for _ in range(n)]
    """

    # for test
    if sample_file:
        sys.stdin = open(sample_file)
    
    # for creating random input
    import random
    
    # ----------------------------------
    # main
    
    # 問題の解き方
    # https://kimiyuki.net/blog/2016/06/21/how-to-solve-problems-in-competitive-programming/
    # https://algo-logic.info/how-to-think-cp/
    
    # OEIS
    # https://kabukimining.hateblo.jp/entry/atcoder-OEIS
    
    #%%
    # while True:
    #     N = random.randint(1, 2*10**2)
    #     A = sorted([random.randint(1, N) for _ in range(N)])
    #     B = sorted([random.randint(1, N) for _ in range(N)])
    #     print(A)
    #     print(B)
    
    H, W = R()
    if H == 1 or W == 1:
        if H*W == 2:
            print(0)
        elif H*W == 1:
            print(1)
        else:
            print(H*W-2)
    else:
        print((H-2)*(W-2))

    
    #%%
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # three methods for loop: keys(), values(), items()
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    https://atcoder.jp/contests/abc173/submissions/14966961
    for i in range(1<<h):
	    if (i >> x) & 1:
            hoge
    (1 >> x) & 1 でNビット目のビット値を取得している
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
n,m = map(int, input().split())

if n > m:
    n,m = m,n

if n==1 and m==1:
    print(1) 
elif n==1:
    print(m-2)
elif n==2 or m==2:
    print(0)
else:
    print((m-2)*(n-2))

Submission status: Accepted

Solution code:
n, m = map(int, input().split())

# flip odd or even
# check concatinated mass
# if that odd, it will back. if even, be opposite.
# ans=0
# for i in range(n):
#     for j in range(m):
#         c=0
#         for h in range(-1,2):
#             for w in range(-1,2):
#                 if 0<i+h<n-1 and 0<j+w<m-1:
#                     c+=1
#         if c%2==1:
#             ans+=1
# print(ans, flush=True)
# TLE


ans=0
if n==1 and m==1:
    print(1, flush=True)
elif n==1 or m==1:
    print(max(n-2, m-2), flush=True)
else:
    print((n-2)*(m-2))

Submission status: Accepted

Solution code:
N, M = map(int, input().split())

if N == 1 and M == 1:
  print(1)
elif N == 1 or M == 1:
  print(N*M-2)
else:
  print((N-2) * (M-2))

Submission status: Accepted

Solution code:
# -*- coding: utf-8 -*-
"""
Created on Wed Sep 23 14:04:30 2020

@author: liang
"""

N, M = map(int,input().split())

if N == 1 and M == 1:
    ans = 1
elif N == 1:
    ans = M - 2
elif M == 1:
    ans = N - 2
else:
    ans = (N-2)*(M-2)
print(ans)

Submission status: Accepted

Solution code:
n,m = map(int, input().split())

if n == 1 and m == 1: print(1)
elif n != 1 and m == 1: print(n-2)
elif n == 1 and m != 1: print(m-2)
else: print((n-2)*(m-2))

Submission status: Accepted

Solution code:
import math
from math import gcd,pi,sqrt
INF = float("inf")
MOD = 10**9 + 7

import sys
sys.setrecursionlimit(10**6)
import itertools
import bisect
from collections import Counter,deque
def i_input(): return int(input())
def i_map(): return map(int, input().split())
def i_list(): return list(i_map())
def i_row(N): return [i_input() for _ in range(N)]
def i_row_list(N): return [i_list() for _ in range(N)]
def s_input(): return input()
def s_map(): return input().split()
def s_list(): return list(s_map())
def s_row(N): return [s_input for _ in range(N)]
def s_row_str(N): return [s_list() for _ in range(N)]
def s_row_list(N): return [list(s_input()) for _ in range(N)]


def main():
    n,m = i_map()
    if n == 1 or m == 1:
        print(abs(n*m - 2))
        exit()

    t = n*2 + (m-2)*2
    print(n*m - t)

if __name__=="__main__":
    main()

Submission status: Accepted

Solution code:
n, m = map(int, input().split())
if n > m:
  n, m = m, n
if n == 1:
  if m == 1:
    ans = 1
  else:
    ans = m - 2
else:
  ans = (n - 2) * (m - 2)
print(ans)

Submission status: Accepted

Solution code:
n,m=map(int,input().split())
if n >2 and m >2:
    ans = (n-2)*(m-2)
elif n ==2 or m ==2:
    ans = 0
elif n == 1 and m == 1:
    ans = 1
elif n == 1:
    ans = m-2
else:
    ans = n-2

print(ans)

Submission status: Accepted

Solution code:
a,b=map(int,input().split())
print(abs((a-2)*(b-2)))

Submission status: Accepted

Solution code:
n,m=map(int,input().split())
if n==1 and m==1:print(1)
elif n==1:print(m-2)
elif m==1:print(n-2)
else:print((n-2)*(m-2))

Submission status: Accepted

Solution code:

def main():
    N, M = map(int, input().split())
    if N >= 2 and M >= 2:
        ans = (N-2)*(M-2)
    else:
        if N == M:
            ans = 1
        else:
            ans = max(N, M) - 2
    print(ans)

main()

Submission status: Accepted

Solution code:
N,M = map(int,input().split())

if N == M == 1:
  print(1)
elif N == 1:
  print(M-2)
elif M == 1:
  print(N-2)
else:
  print((N-2)*(M-2))

Submission status: Accepted

Solution code:
l = sorted(list(map(int,input().split())))
#N<=Mとして一般性を失わない
N=l[0]
M=l[1]

if N==M==1:
    print(1)
if N==1 and M>=2:
    print(M-2)
if N==2 and M>=2:
    print(0)
if N>=3 and M>=3:
    print((N-2)*(M-2))

Submission status: Accepted

Solution code:
n,m=map(int,input().split())
if n==1 and m==1:
  print(1)
  exit()
if n==1 or m==1:
  print(n*m-2)
  exit()
if n==2 or m==2:
  print(0)
  exit()
print((m-2)*(n-2))

Submission status: Accepted

Solution code:
n,m = map(int, input().split())
if n ==1 and m == 1:
    ans = 1
elif n == 1 or m == 1:
    ans = n*m-2
else:
    ans = (n-2)*(m-2)
print(ans)

Submission status: Accepted

Solution code:
# -*- coding: utf-8 -*-
import sys
from collections import deque
from collections import defaultdict
import heapq
import collections
import itertools
import bisect
import copy
import math
from itertools import accumulate
sys.setrecursionlimit(10**6)

# lis_of_lis = [[] for _ in range(N)]


def floor(a, b): return -(-a // b)


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


def z(): return int(sys.stdin.readline())


def S(): return sys.stdin.readline()[:-1]


def C(line): return [sys.stdin.readline() for _ in range(line)]


def is_kaibun(kaibun):
    for i in range(len(kaibun)//2):
        if kaibun[i] != kaibun[-i-1]:
            return False
    return True


N, M = zz()
next1, next2, next3, next4, next5, next6, next7, next8 = 0, 0, 0, 0, 0, 0, 0, 0
if (N == 1 and M == 1):
    print(1)
    exit()
elif (N == 1 or M == 1):
    next1 = 2
    next2 = (N*M)-2
elif (M == 2 or N == 2):
    next3 = 4
    next5 = (N*M)-4

else:
    next1 = 0
    next2 = 0
    next3 = 4
    next4 = 0
    next5 = (M-2)*2 + (N-2)*2
    next6 = 0
    next8 = (N-2)*(M-2)

# print(next1+next2 + next3 + next4 + next5 + next6 + next7+next8, N*M)
print(next2 + next4 + next6 + next8)

Submission status: Accepted

Solution code:
n, m = map(int, input().split())
r = (n - 2) * (m - 2)
if (r < 0):
    r *= -1
print(r)

Submission status: Accepted

Solution code:
N,M=map(int,input().split())
if N==1:
    if M==1:
        print(1)
    else:
        print(M-2)
elif M==1:
    print(N-2)
else:
    print((N-2)*(M-2))

Submission status: Accepted

Solution code:
n,m=map(int,input().split())
ans=0
if n==1 and m==1:
  ans=1
elif n==1 or m==1:
  ans=max(n,m)-2
else:
  ans=(n-2)*(m-2)
print(ans)

Submission status: Accepted

Solution code:
N, M = map(int, input().split())
if N == 1:
  if M == 1:
    print(1)
  elif M == 2:
    print(0)
  else:
    print(M - 2)
elif N == 2:
  print(0)
else:
  if M == 1:
    print(N - 2)
  elif M == 2:
    print(0)
  else:
    print((N - 2) * (M - 2))

Submission status: Accepted

Solution code:
def main():
    N, M = map(int, input().split())
    if N > M:
        N, M = M, N

    if M == 1:
        print(1)
    elif N == 1:
        print(M-2)
    elif N == 2:
        print(0)
    else:
        print((N-2) * (M-2))


if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
n, m = map(int, input().split())
if n == 1 or m == 1:
    if n == m:
        print(1)
    else:
        print(max(n, m)-2)
elif n == 2 or m == 2:
    print(0)
else:
    print((n-2)*(m-2))

Submission status: Accepted

Solution code:
n, m = map(int, input().split())

if n == 1 and m == 1:
    print(1)
elif n == 1:
    print(m-2)
elif m == 1:
    print(n-2)
else:
    print((n-2)*(m-2))

Submission status: Accepted

Solution code:
#!/usr/bin/env python3

n, m = list(map(int, input().split()))


if n == 1 and m == 1:
    corner = 1
    edge = 0
    other = 0
    print(1)
elif n == 1 or m == 1:
    corner = 2
    edge = n*m-corner
    other = 0
    print(edge)
else:
    corner = 4
    edge = n*2+m*2-corner*2
    other = n*m - edge-corner
    print(other)

Submission status: Accepted

Solution code:
N,M = map(int,input().split())
if N >= 3 and M >= 3:
    print((N-2)*(M-2))
elif N == 2 or M == 2:
    print(0)
else:
    print(max(1,N-2)*max(1,M-2))