Problem:


Score : 200 points

Problem Statement

There are some animals in a garden. Each of them is a crane with two legs or a turtle with four legs.

Takahashi says: "there are X animals in total in the garden, and they have Y legs in total." Determine whether there is a combination of numbers of cranes and turtles in which this statement is correct.

Constraints

1 \leq X \leq 100

1 \leq Y \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X Y

Output

If there is a combination of numbers of cranes and turtles in which the statement is correct, print Yes; otherwise, print No.

Sample Input 1

3 8

Sample Output 1

Yes

The statement "there are 3 animals in total in the garden, and they have 8 legs in total" is correct if there are two cranes and one turtle. Thus, there is a combination of numbers of cranes and turtles in which the statement is correct.

Sample Input 2

2 100

Sample Output 2

No

There is no combination of numbers of cranes and turtles in which this statement is correct.

Sample Input 3

1 2

Sample Output 3

Yes

We also consider the case in which there are only cranes or only turtles.



Sample input:
3 8


Sample output:
Yes

Submission status: Accepted

Solution code:
import sys
input=sys.stdin.readline
x,y=map(int,input().split())
z=y-2*x
if z%2!=0:
    print('No')
else:
    k=z//2
    t=x-k
    if t>=0 and k>=0:
        print('Yes')
    else:
        print('No')

Submission status: Accepted

Solution code:
import math


x,y=map(int,input().split())
for i in range(x+1):

    if 2*i+4*(x-i)==y:
        print("Yes")
        exit()

print("No")

Submission status: Accepted

Solution code:
X,Y = map(int, input().split())
if Y>=2*X and Y<=4*X and Y%2==0:
    print("Yes")
else:
    print("No")

Submission status: Accepted

Solution code:
X, Y = map(int, input().split())

if(2 * X <= Y <= 4 * X and Y % 2 == 0):
  print("Yes")
else:
  print("No")

Submission status: Accepted

Solution code:
def main():
    x, y = list(map(int, input().split()))
    for a in range(x + 1):
        b = x - a
        if 2 * a + 4 * b == y:
            print('Yes')
            break
    else:
        print('No')

if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
x,y = map(int,input().split())
if (4*x-y)//2 == (4*x-y)/2 and  0 <= (4*x-y)//2 <= x:
    print("Yes")
else:
    print("No")

Submission status: Accepted

Solution code:
X, Y = map(int, input().split())

for a in range(X + 1) :
  b = X - a
  if 2 * a + 4 * b == Y :
    print('Yes')
    break
else :
  print('No')

Submission status: Accepted

Solution code:
x, y = list(map(int, input().split()))

flag = False
for i in range(x + 1):
  kame = i
  turu = x - i
  
  if kame * 4 + turu * 2 == y:
    flag = True
    break
print("Yes" if flag else "No")

Submission status: Accepted

Solution code:
x,y = map(int, input().split())
if 2*x <= y <= 4*x and y%2==0:
    print("Yes")
else:
    print("No")

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 # 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)
    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
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    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 # Deprecated since version 3.5: Use math.gcd() instead.
    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)
    
    # first create factorial_list
    # fac_list = mod_factorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    AtoZ = list('abcdefghijklmnopqrsruvwxyz') # 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))

    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
        >>  else:
        >>      do something
        >>      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]

    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)
        """
        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)
        
    # ----------------------------------
    # main
    #%%
    X, Y = R()
    if X * 2 <= Y <= X * 4 and Y%2 == 0:
        print('Yes')
    else:
        print('No')
    
    #%%
    # 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
    # to make binary digits text
    >>> a = 5
    >>> bin_str_a = format(a, '#06b')
    >>> print(bin_str_a)
    0b0101 # first 2 strings (='0b') indicates it is binary
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
import sys
input = sys.stdin.readline

'''

'''

X, Y = map(int, input().split())
if Y % 2 == 0 and 2*X <= Y <= 4*X:
    print("Yes")
else:
    print("No")

Submission status: Accepted

Solution code:
from sys import stdin
import sys

X, Y = [int(x) for x in stdin.readline().rstrip().split()]
for i in range(0,X+1):
    if i*2+(X-i)*4==Y:
        print('Yes')
        sys.exit()
print('No')

Submission status: Accepted

Solution code:
import sys
sys.setrecursionlimit(10000000)
MOD = 10 ** 9 + 7
INF = 10 ** 15

def main():
    X,Y = map(int,input().split())
    B = Y - 2*X
    if B >= 0 and B%2 == 0 and 2*X - B >= 0:
        print('Yes')
    else:
        print('No')
if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
x,y = map(int, input().split())

for i in range(x+1):
    if y == i*2 + (x-i)*4:
        print('Yes')
        exit()
print("No")

Submission status: Accepted

Solution code:
X, Y = map(int, input().split())
if Y%2 == 0:
    Y //= 2
    if X <= Y and Y <= 2*X:
        print("Yes")
    else:
        print("No")
else:
    print("No")

Submission status: Accepted

Solution code:
X, Y = map(int, input().split())
ans = "No"
for i in range(100):
    for j in range(100):
        if i * 2 + j * 4 == Y and i + j == X:
            ans = "Yes"
print(ans)

Submission status: Accepted

Solution code:
from functools import reduce
from fractions import gcd
import math
import bisect
import itertools
import sys
sys.setrecursionlimit(10**7)
input = sys.stdin.readline
INF = float("inf")


# 処理内容
def main():
    X, Y = map(int, input().split())

    ans = "No"
    for a in range(X+1):
        for b in range(X+1):
            if a + b == X and a*2 + b*4 == Y:
                ans = "Yes"
    
    print(ans)


if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
X,Y = map(int,input().split())
a,b = (4*X-Y)/2,(Y-2*X)/2
if a>=0 and b >= 0 and int(a) == a and int(b) == b:
    print("Yes")
else:
    print("No")

Submission status: Accepted

Solution code:
x, y = map(int, input().split())

for turu in range(x+1):
  kame = x - turu
  if turu*2 + kame*4 == y:
    print("Yes")
    break
else:
  print("No")

Submission status: Accepted

Solution code:
X,Y=map(int,input().split())
for i in range(X+1):
    if((i*2)+(X-i)*4==Y):
        print("Yes")
        exit()
print("No")

Submission status: Accepted

Solution code:
# coding: utf-8


def main():
    X, Y = map(int, input().split())

    for n_turu in range(X+1):
        n_kame = X - n_turu
        if n_turu * 2 + n_kame * 4 == Y:
            return "Yes"

    return "No"


print(main())

Submission status: Accepted

Solution code:
def main():
    x, y = map(int, input().split())
    if y % 2 != 0:
        print("No")
    else:
        if x * 2 <= y <= x * 4:
            print("Yes")
        else:
            print("No")


if __name__ == "__main__":
    main()

Submission status: Accepted

Solution code:
x, y = [int(i) for i in input().split()]

flag = 0
for i in range(x+1):
    j = x - i
    if(2 * i + 4 * j == y):
        flag = 1

if (flag == 1):
    print("Yes")
else:
    print("No")

Submission status: Accepted

Solution code:
i = input().split()
i = [int(j) for j in i]
 
flg = 0
for a in range(i[0]+1):
  if 2*a+4*(i[0]-a)==i[1]:
    print("Yes")
    flg = 1
    
if flg==0:
  print("No")

Submission status: Accepted

Solution code:
import sys

sys.setrecursionlimit(1000000)

import bisect
import math
import collections
import heapq


def solve(X, Y):
    for k in range(X + 1):
        t = X - k
        if Y == (k * 4) + (t * 2):
            return "Yes"
    return "No"


if __name__ == "__main__":
    X, Y = tuple(map(int, input().split(" ")))
    print(solve(X, Y))

Submission status: Accepted

Solution code:
def get_ints():
    return list(map(int, input().split()))

x, y = get_ints()
ok = False
for i in range(x + 1):
    ok |= 4 * i + 2 * (x - i) == y

print("Yes" if ok else "No")

Submission status: Accepted

Solution code:
X, Y = list(map(int, input().split()))

if Y % 2 == 0 and Y // 2 >= X and 2 * X >= Y // 2:
    print("Yes")
else:
    print("No")

Submission status: Accepted

Solution code:
def resolve():
    X,Y = map(int,input().split())

    for turu in range(X+1):
        kame = X-turu
        if turu*2 + kame *4 == Y:
            print("Yes")
            return
    else:
        print("No")
resolve()

Submission status: Accepted

Solution code:
x, y = map(int, input().split())

f = 0
for i in range(x+1):
    if i * 2 + (x-i) * 4 == y:
        f = 1
        break

if f:
    print('Yes')
else:
    print('No')

Submission status: Accepted

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

YES = "Yes"  # type: str
NO = "No"  # type: str


def solve(X: int, Y: int):
    n = (Y - 2*X) / 2
    if (int(n) == n) and (0<=n<=X):
        print(YES)
    else:
        print(NO)
    return


# Generated by 1.1.7.1 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
def main():
    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    X = int(next(tokens))  # type: int
    Y = int(next(tokens))  # type: int
    solve(X, Y)

if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
import math,itertools,fractions,heapq,collections,bisect,sys,queue,copy

sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
dd=[(-1,0),(0,1),(1,0),(0,-1)]
ddn=[(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]

def LI(): return [int(x) for x in sys.stdin.readline().split()]
# def LF(): return [float(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def LS(): return sys.stdin.readline().split()
def S(): return input()

def main():
  x,y=LI()

  for i in range(x+1):
    j=x-i
    z=i*2+j*4

    if z==y:
      return 'Yes'

  return 'No'

# main()
print(main())

Submission status: Accepted

Solution code:
X, Y = map(int, input().split())

for i in range(X+1):
    j = X - i
    if 2*i + 4*j == Y:
        print("Yes")
        exit(0)
print("No")

Submission status: Accepted

Solution code:
import sys
sys.setrecursionlimit(10 ** 8)

def Z(): return int(input())
def ZZ(): return [int(_) for _ in input().split()]

def main():
    X, Y = ZZ()
    for n in range(101):
        for m in range(101):
            if n+m == X and 2*n+4*m == Y:
                print('Yes')
                break
        else: continue
        break
    else: print('No')

    return

if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
def solve():    
    x,y=map(int,input().split())
    
    for i in range(x+1):
        if 2*i+4*(x-i) ==y:
            print("Yes")
            return
    print("No")
solve()

Submission status: Accepted

Solution code:
x,y=map(int,input().split())
for i in range(x+1):
  if 2*i+4*(x-i) == y:
    print('Yes')
    exit()
print('No')

Submission status: Accepted

Solution code:
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
# import numpy as np
sys.setrecursionlimit(int(pow(10,6)))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = int(pow(10, 9) + 7)
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]



x,y=l()
for a in range(1000):
    for b in range(1000):
        if a+b==x and 2*a+4*b==y:
            print("Yes")
            exit()
print("No")
exit()

Submission status: Accepted

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

if n*2<=m and m<=n*4 and m%2==0:
    print("Yes")
    exit()


print("No")

Submission status: Accepted

Solution code:
x,y = map(int, input().split())

if y % 2 == 1:
    print("No")
elif x * 2 <= y <= x * 4:
    print("Yes")
else:
    print("No")