Problem:


Score : 400 points

Problem Statement

There is a knight - the chess piece - at the origin (0, 0) of a two-dimensional grid.

When the knight is at the square (i, j), it can be moved to either (i+1,j+2) or (i+2, j+1).

In how many ways can the knight reach the square (X, Y)?

Find the number of ways modulo 10^9 + 7.

Constraints

1 \leq X \leq 10^6

1 \leq Y \leq 10^6

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X Y

Output

Print the number of ways for the knight to reach (X, Y) from (0, 0), modulo 10^9 + 7.

Sample Input 1

3 3

Sample Output 1

2

There are two ways: (0,0) \to (1,2) \to (3,3) and (0,0) \to (2,1) \to (3,3).

Sample Input 2

2 2

Sample Output 2

0

The knight cannot reach (2,2).

Sample Input 3

999999 999999

Sample Output 3

151840682

Print the number of ways modulo 10^9 + 7.



Sample input:
3 3


Sample output:
2

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
    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]

    # Graph: https://en.wikipedia.org/wiki/Directed_graph
    # 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]

    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array, use [[0] * N for _ in range(N)]
    # 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 = lambda: list(map(int, r().split()))
    Rmap = lambda: map(int, r().split())
    Rfloat = lambda: list(map(float, r().split()))
    Rtuple = 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+Y)/3 != (X+Y)//3:
        print(0)
        return
    tmp = (X+Y)//3
    if X-tmp < 0 or Y-tmp < 0:
        print(0)
        return
    X -= tmp
    Y -= tmp
    print(combinations_count(X+Y,X)%(10**9+7))
    
    # to finish code, use return instead of exit()
    # 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:
X,Y = list(map(int,input().split()))

from operator import mul
from functools import reduce

def cmb(n, r):
    if n - r < r: r = n - r
    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

if (X+Y)%3 != 0:
    print(0)
elif (X//2)>Y or (Y//2)>X:
    print(0)
else:
    dx = X-(X+Y)//3
    dy = Y-(X+Y)//3
    print (cmb(dx+dy,dx)%int(1e9+7))

Submission status: Accepted

Solution code:
# -*- coding: utf-8 -*-

X, Y = map(int, input().split())
mod = 10 ** 9 + 7

times = (X + Y) // 3
if (X + Y) % 3:
    print('0')
    exit()

x = X - times
y = Y - times


def egcd(a, b):
    if a == 0:
        return b, 0, 1
    else:
        g, y, x = egcd(b % a, a)
        return g, x - (b // a) * y, y


def modinv(a, m):
    g, x, y = egcd(a, m)
    if g != 1:
        raise Exception('modular inverse does not exist')
    else:
        return x % m


def combination(n, r, mod):
    r = min(r, n-r)
    res = 1
    for i in range(r):
        res = res * (n - i) * modinv(i+1, mod) % mod
    return res


if y < 0 or x < 0:
    print('0')
    exit()


print(combination(times, y, mod))

Submission status: Accepted

Solution code:
X, Y = map(int, input().split())
MOD = 10 ** 9 + 7


if (X + Y) % 3 != 0:
    print(0)
    exit()


X, Y = min(X, Y), max(X, Y)
ALL_MOVE = (X + Y) // 3
for m1 in range(ALL_MOVE):
    m2 = ALL_MOVE - m1

    if (m1 * 2 + m2 == X) and (m1 + m2 * 2 == Y):
        move_1_cnt = m1
        move_2_cnt = m2
        break
else:
    print(0)
    exit()


factorial = [1]
inverse = [1]
for i in range(1, ALL_MOVE + 2):
    factorial.append(factorial[-1] * i % MOD)
    inverse.append(pow(factorial[-1], MOD - 2, MOD))


# 組み合わせ計算
def nCr(n, r):
    if n < r or r < 0:
        return 0
    elif r == 0:
        return 1
    return factorial[n] * inverse[r] * inverse[n - r] % MOD


print(nCr(ALL_MOVE, move_1_cnt) % MOD)

Submission status: Accepted

Solution code:
#
import sys
import math
import numpy as np
import itertools
# いくつか入力
x,y = (int(i) for i in input().split())  

MOD = 1000000007

if (x+y) % 3 != 0:
    print(0)
    exit()

if (x > y*2) or (y > x*2):
    print(0)
    exit()


a = x - (x+y) // 3
b = y - (x+y) // 3

upper=1
lower=1

for i in range(1,a+b+1,1):
    upper *= i
    upper %= MOD

for i in range(1,a+1,1):
    lower *= i
    lower %= MOD

for i in range(1,b+1,1):
    lower *= i
    lower %= MOD

# karimono
# https://qiita.com/Yaruki00/items/fd1fc269ff7fe40d09a6
def mul(a, b):
    return ((a % MOD) * (b % MOD)) % MOD

def power(x, y):
    if   y == 0     : return 1
    elif y == 1     : return x % MOD
    elif y % 2 == 0 : return power(x, y//2)**2 % MOD
    else            : return power(x, y//2)**2 * x % MOD

def div(a, b):
    return mul(a, power(b, MOD-2))

answer = div(upper,lower)

print(answer)

Submission status: Accepted

Solution code:



def nCr(n,r):
  if n < 0 or r < 0 or n < r: return 0
  if n - r < r: r = n - r
  if r == 0: return 1
  if r == 1: return n
  
  # 分子のn*(n-1)*...がr個分続くやつ
  numerator = [n-r+k+1 for k in range(r)]
  # 分母：r!=r*(r-1)*...*3*2の要素
  denominator = [k+1 for k in range(r)]
  
  # 分母の要素で割れる部分を割っていく部分
  for p in range(2, r+1):
    # 分母は1,2,3,...rのようになっており、1は意味がないのでスキップした形か
    pivot = denominator[p-1]
    if pivot > 1:
      # 分子のX番目と分母のX-offset番目が共通の約数を持つということだと思う。piv分ずれているのだから、pivの倍数というところか
      offset = (n-r) % p
      for k in range(p-1, r, p):
        # 約分できる要素について割る
        numerator[k - offset]  /= pivot
        denominator[k] /= pivot
        
  ret = 1
  for k in range(r):
    if numerator[k] > 1: ret *= int(numerator[k])
      
  return ret


def main():
    X, Y = map(int, input().split())
    if (2*Y - X) % 3 != 0:
        print(0)
    else:
        A = (2*Y - X) // 3
        B = Y - 2*A
        print(nCr(A+B,A)%(10**9+7))


if __name__ == "__main__":
    main()

Submission status: Accepted

Solution code:
import sys
stdin = sys.stdin

sys.setrecursionlimit(10 ** 7)

def li(): return map(int, stdin.readline().split())
def li_(): return map(lambda x: int(x) - 1, stdin.readline().split())
def lf(): return map(float, stdin.readline().split())
def ls(): return stdin.readline().split()
def ns(): return stdin.readline().rstrip()
def lc(): return list(ns())
def ni(): return int(stdin.readline())
def nf(): return float(stdin.readline())

class CombTools(object):
    def __init__(self, cap: int, mod: int):
        self.cap = cap
        self.mod = mod
        self.inv = self._calc_inv()
        self.fac = self._calc_fac()
        self.fac_inv = self._calc_fac_inv()

    def _calc_inv(self):
        inv = [0, 1]
        for i in range(2, self.cap+1):
            inv.append((self.mod - (self.mod // i) * inv[self.mod % i]) % self.mod)

        return inv

    def _calc_fac(self):
        fac = [1]
        for i in range(1, self.cap+1):
            fac.append((i * fac[-1]) % self.mod)

        return fac

    def _calc_fac_inv(self):
        fac_inv = [1]
        for i in range(1, self.cap+1):
            fac_inv.append((self.inv[i] * fac_inv[-1]) % self.mod)

        return fac_inv

    def nCr(self, n: int, r: int):
        # validation
        if r > n:
            raise ValueError("n must be larger than r (n={}, r={})".format(n, r))

        # calculation
        return self.fac[n] * self.fac_inv[n-r] * self.fac_inv[r] % self.mod

x, y = li()
MOD = 10**9 + 7

ct = CombTools(10**6, MOD)
n = (x+y) // 3

if (x+y)%3 != 0:
    print(0)
elif x < n or x > 2*n:
    print(0)
else:
    print(ct.nCr((x+y) // 3, x - (x+y) // 3))

Submission status: Accepted

Solution code:
X,Y=map(int,input().split())
mod=10**9+7

if (2*Y-X)%3!=0 or (2*X-Y)%3!=0:
  print(0)
  exit()

p,q=(2*Y-X)//3,(2*X-Y)//3

def cmb(n,r,mod):
    if (r<0 or r>n):
      return 0
    r=min(r,n-r)
    return g1[n]*g2[r]*g2[n-r]%mod

g1=[1,1]
g2=[1,1]
inverse=[0,1]

for i in range(2,10**6+1):
    g1.append((g1[-1]*i)%mod)
    inverse.append((-inverse[mod%i]*(mod//i))%mod)
    g2.append((g2[-1]*inverse[-1])%mod)

print(cmb(p+q,p,mod))

Submission status: Accepted

Solution code:
def cmb(n, r, mod):
    if ( r<0 or r>n ):
        return 0
    r = min(r, n-r)
    return g1[n] * g2[r] * g2[n-r] % mod
mod = 10**9+7 #出力の制限
N = 10**6
g1 = [1, 1] # 元テーブル
g2 = [1, 1] #逆元テーブル
inverse = [0, 1] #逆元テーブル計算用テーブル

for i in range( 2, N + 1 ):
    g1.append( ( g1[-1] * i ) % mod )
    inverse.append( ( -inverse[mod % i] * (mod//i) ) % mod )
    g2.append( (g2[-1] * inverse[-1]) % mod )
    
X, Y = list(map(int, input().split()))
u = [1, 2]
r = [2, 1]
flag = True
for i in range(Y//2+1):
    if i*1+2*(Y-i*2)==X:
        flag = False
        N = i+Y-i*2
        print(cmb(N, i, 10**9+7))
        
if flag:
    print(0)

Submission status: Accepted

Solution code:
import sys
import itertools
import math
from math import factorial
from heapq import heapify, heappop, heappush
MAX_INT = int(10e15)
MIN_INT = -MAX_INT
mod = 1000000007
sys.setrecursionlimit(1000000)
def IL(): return list(map(int,input().split()))
def SL(): return input().split()
def I(): return int(sys.stdin.readline())
def S(): return input()

def cmb(n, r):
    if n - r < r: r = n - r
    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]) %mod

    return result

X, Y = IL()

num = 0
n = 0
f = 0
for i in range(X+1): #(1, 2)を取る回数
  x = X - i
  y = Y - 2*i
  if x == 2*y:
    n = i
    num = i + y
    f = 1
    break
if f == 0:
  print(0)
else:
  print(cmb(num, n) %mod)

Submission status: Accepted

Solution code:
def cmb(n, r):
    if n - r < r: r = n - r
    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


X,Y=map(int,input().split())
mod=10**9+7

if  X>Y*2 or Y>X*2:
    print(0)
    exit()
if (X+Y)%3!=0:
    print(0)
    exit()

count=0
_count=0
while X>0 and Y>0:
    if X>Y:
        X-=2
        Y-=1

        _count+=1
    else:

        X-=1
        Y-=2
    #print(X,Y)
    count+=1

#print(count,_count)
print(cmb(count,_count)%mod)

Submission status: Accepted

Solution code:
class BigCombination(object):
    __slots__ = ["mod", "factorial", "inverse"]

    def __init__(self, mod: int = 10**9+7, max_n: int = 10**6):
        fac, inv = [1], []
        fac_append, inv_append = fac.append, inv.append

        for i in range(1, max_n+1):
            fac_append(fac[-1] * i % mod)

        inv_append(pow(fac[-1], mod-2, mod))

        for i in range(max_n, 0, -1):
            inv_append(inv[-1] * i % mod)

        self.mod, self.factorial, self.inverse = mod, fac, inv[::-1]

    def get_combination(self, n, r):
        return self.factorial[n] * self.inverse[r] * self.inverse[n-r] % self.mod

    def get_permutation(self, n, r):
        return self.factorial[n] * self.inverse[n-r] % self.mod

import math
Big = BigCombination()
n, d = map(int, input().split())
def cal(a,b):
    m = min(a,b)
    M = max(a,b)
    res = 0
    T = (m+M)//3
    if (a+b)%3 == 0 and M<=m*2:
        res = 2*M - m
        res = res// 3
        k = Big.get_combination(T, res)
        print(k)
        
    else:
        print(0)

cal(n,d)

Submission status: Accepted

Solution code:
Mod = 10**9+7
X, Y = map(int, input().split())
ans1 = (2*X-Y)//3
ans2 = (2*Y-X)//3
if (2*X-Y)%3!=0 or (2*Y-X)%3!=0:
  print(0)
  exit()
fac = [1, 1]
finv = [1, 1]
inv = [0, 1]
def COMinit():
    for i in range(2, ans1+ans2+111):
        fac.append(fac[-1]*i%Mod)
        inv.append((-inv[Mod%i] * (Mod//i)) % Mod)
        finv.append(finv[-1] * inv[-1] % Mod)
    

def COM(n, k):
    if n < 0 or k < 0 or n < k:
        return 0
    return fac[n] * (finv[k] * finv[n-k] % Mod) % Mod

COMinit()
print(COM(ans1+ans2, ans1))

Submission status: Accepted

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

mod = 10**9+7 #出力の制限
N = 10**6
g1 = [1, 1] # 元テーブル
g2 = [1, 1] #逆元テーブル
inverse = [0, 1] #逆元テーブル計算用テーブル

for i in range(2, N + 1 ):
    g1.append( ( g1[-1] * i ) % mod )
    inverse.append( ( -inverse[mod % i] * (mod//i) ) % mod )
    g2.append( (g2[-1] * inverse[-1]) % mod )

def cmb(n, r, mod):
    if ( r<0 or r>n ):
        return 0
    r = min(r, n-r)
    return g1[n] * g2[r] * g2[n-r] % mod

if (2*X-Y) % 3 == 0 and (2*Y-X) % 3 == 0:
    s = (2*X-Y)//3
    t = (2*Y-X)//3
    print(cmb(s+t,s, mod))
else:
    print(0)

Submission status: Accepted

Solution code:
x,y=[int(j) for j in input().split()]
U = 2*10**6
MOD = 10**9+7
 
fact = [1]*(U+1)
fact_inv = [1]*(U+1)
 
for i in range(1,U+1):
  fact[i] = (fact[i-1]*i)%MOD
fact_inv[U] = pow(fact[U],MOD-2,MOD)
 
for i in range(U,0,-1):
	fact_inv[i-1] = (fact_inv[i]*i)%MOD
 
def comb(n,k):
  if k < 0 or k > n:
    return 0
  x = fact[n]
  x *= fact_inv[k]
  x %= MOD
  x *= fact_inv[n-k]
  x %= MOD
  return x

if (2*y-x)//3==(2*y-x)/3 and (2*x-y)//3==(2*x-y)/3:
    i=(2*y-x)//3
    j=(2*x-y)//3
    print(comb(i+j,min(i,j)))
else:
    print(0)

Submission status: Accepted

Solution code:
M = 10 ** 9 + 7

X, Y = map(int, input().split())
N = max(X, Y)

def calc_inverse(n, law):
	if n == 1:
		return 1
	
	pre_r = law
	pre_x = 0
	pre_y = 1
	cur_r = n
	cur_x = 1
	cur_y = 0
	
	while True:
		d = pre_r // cur_r
		r = pre_r % cur_r
		
		nxt_r = r
		nxt_x = pre_x - d * cur_x
		nxt_y = pre_y - d * cur_y
		
		if nxt_r == 1:
			ret = nxt_x
			while ret < 0:
				ret += law;
			return ret;
		elif nxt_r == 0:
			return None
		
		pre_r = cur_r;
		pre_x = cur_x;
		pre_y = cur_y;
		cur_r = nxt_r;
		cur_x = nxt_x;
		cur_y = nxt_y;

fact_list = [1, 1]
inv_fact_list = [1, 1]
f = 1
i_f = 1
for i in range(2, N + 1):
	f = (f * i) % M
	fact_list.append(f)
	i_f = (i_f * calc_inverse(i, M)) % M
	inv_fact_list.append(i_f)

total = 0

for i in range(X + 1):
	x = X - i
	y = Y - i * 2
	if x == y * 2:
		total = (total + fact_list[i + y] * inv_fact_list[i] * inv_fact_list[y]) % M

print(total)

Submission status: Accepted

Solution code:
x,y=map(int,input().split())
n=(2*x-y)/3
m=(2*y-x)/3
def cmb(n, r):
    if n - r < r: r = n - r
    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

if n>=0 and m>=0 and n%1==0 and m%1==0:
  n=int(n)
  m=int(m)
  print(cmb(n+m,n)%(10**9+7))
else:
  print(0)

Submission status: Accepted

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

MOD = 10**9+7

# コピペ

g1 = [1, 1] # 元テーブル
g2 = [1, 1] #逆元テーブル
inverse = [0, 1] #逆元テーブル計算用テーブル

for i in range(2, 10**6+1):
    g1.append((g1[-1] * i) % MOD)
    inverse.append((-inverse[MOD % i] * (MOD//i)) % MOD)
    g2.append((g2[-1] * inverse[-1]) % MOD)

def combi_mod(n, r):
    if (r<0 or r>n):
        return 0
    r = min(r, n-r)
    return g1[n] * g2[r] * g2[n-r] % MOD

# ====

sum_num = 0
for n in range(X+1):
    x, y = 1*n, 2*n
    xr, yr = X-x, Y-y
    if xr == yr*2:
        m = yr
        sum_num = (sum_num + combi_mod(n+m, m)) % MOD

print(sum_num)

Submission status: Accepted

Solution code:
from functools import reduce
X, Y = map(int, input().split())
MOD = 10**9 + 7

def cmb(n, r, mod):
    '''nCrを返却(modは素数であるべし)
    '''
    r = min(n-r, r)
    if r == 0: return 1
    over = reduce(lambda x, y : x*y % mod, range(n, n-r, -1))
    under = reduce(lambda x, y : x*y % mod, range(1, r+1))
    return (over * pow(under, mod-2, mod)) % mod

if 2*Y-X < 0 or 2*X-Y < 0 or (2*Y-X)%3 != 0 or (2*X-Y)%3 != 0:
    print(0)
else:
    a, b = (2*X-Y)//3, (2*Y-X)//3
    print(cmb(a+b, a, MOD))