Problem:


Score : 500 points

Problem Statement

You are going to hold a competition of one-to-one game called AtCoder Janken. (Janken is the Japanese name for Rock-paper-scissors.)
N players will participate in this competition, and they are given distinct integers from 1 through N.
The arena has M playing fields for two players. You need to assign each playing field two distinct integers between 1 and N (inclusive).
You cannot assign the same integer to multiple playing fields.
The competition consists of N rounds, each of which proceeds as follows:

For each player, if there is a playing field that is assigned the player's integer, the player goes to that field and fight the other player who comes there.

Then, each player adds 1 to its integer. If it becomes N+1, change it to 1.

You want to ensure that no player fights the same opponent more than once during the N rounds.
Print an assignment of integers to the playing fields satisfying this condition.
It can be proved that such an assignment always exists under the constraints given.

Constraints

1 \leq M

M \times 2 +1 \leq N \leq 200000

Input

Input is given from Standard Input in the following format:

N M

Output

Print M lines in the format below.
The i-th line should contain the two integers a_i and b_i assigned to the i-th playing field.

a_1 b_1
a_2 b_2
:
a_M b_M

Sample Input 1

4 1

Sample Output 1

2 3

Let us call the four players A, B, C, and D, and assume that they are initially given the integers 1, 2, 3, and 4, respectively.

The 1-st round is fought by B and C, who has the integers 2 and 3, respectively. After this round, A, B, C, and D have the integers 2, 3, 4, and 1, respectively.

The 2-nd round is fought by A and B, who has the integers 2 and 3, respectively. After this round, A, B, C, and D have the integers 3, 4, 1, and 2, respectively.

The 3-rd round is fought by D and A, who has the integers 2 and 3, respectively. After this round, A, B, C, and D have the integers 4, 1, 2, and 3, respectively.

The 4-th round is fought by C and D, who has the integers 2 and 3, respectively. After this round, A, B, C, and D have the integers 1, 2, 3, and 4, respectively.

No player fights the same opponent more than once during the four rounds, so this solution will be accepted.

Sample Input 2

7 3

Sample Output 2

1 6
2 5
3 4



Sample input:
4 1


Sample output:
2 3

Submission status: Accepted

Solution code:
# 君指先跃动の光は、私の一生不変の信仰に、唯私の超电磁炮永生き！
from sys import stdin, stdout
import math
import bisect
from collections import defaultdict
import datetime
import random

# n = int(stdin.readline().strip())
# arr = list(map(int, stdin.readline().strip().split()))

n, m = map(int, stdin.readline().strip().split())

if m == 1:
    stdout.writelines('1 2\n')
else:
    m1 = m // 2
    m2 = m - m1
    a = m2
    b = a + 1
    for i in range(m2):
        stdout.writelines('%d %d\n' % (a, b))
        a -= 1
        b += 1
    a = b
    b = a + m1 * 2
    for i in range(m1):
        stdout.writelines('%d %d\n' % (a, b))
        a += 1
        b -= 1

Submission status: Accepted

Solution code:
N, M = map(int, input().split())
 
if M % 2 == 0:
  l = 1
  for i in range(M//2):
    L = []
    L.append(l)
    L.append(2*(M//2-i)+l)
    print(*L)
    l += 1
  l = 2*(M//2)+2
  for i in range(M//2):
    L = []
    L.append(l)
    L.append(2*(M//2-i)+l-1)
    print(*L)
    l += 1
  
  
elif M % 2 == 1:
  l = 1
  for i in range(M//2):
    L = []
    L.append(l)
    L.append(2*(M//2-i)+l)
    print(*L)
    l += 1
  l = 2*(M//2)+2
  for i in range(M//2+1):
    L = []
    L.append(l)
    L.append(2*(M//2+1-i)+l-1)
    print(*L)
    l += 1

Submission status: Accepted

Solution code:
n, m = map(int, input().split())
if n%2 == 1:
    l = 1
    r = n-1
    for i in range(m):
        print(l, r)
        l += 1
        r -= 1
else:
    l = 1
    r = n-1
    flag = False
    for i in range(m):
        if (not flag) and r-l <= n//2:
            r -= 1
            flag = True
        print(l, r)
        l += 1
        r -= 1

Submission status: Accepted

Solution code:
import math

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

start = 1
end = n

wq = n // 4
for i in range(m):
    if i == wq and n % 2 == 0:
        start += 1
    print(start, end)
    start += 1
    end -= 1

Submission status: Accepted

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

x=list(range(1,N+1,1))

if N%2==1:
    for i in range(M):
        print(x.pop(0),x.pop(-1))
        
else:
    x_f=x[1:int(N/2+1)]
    x_l=x[int(N/2+1):]
    
    count=0
    
    for i in range(M):
        if len(x_f)>1:
            print(x_f.pop(0),x_f.pop(-1))
            count+=1
            
        if count==M:
            break
        
        if len(x_l)>1:
            print(x_l.pop(0),x_l.pop(-1))
            count+=1
            
        if count==M:
            break

Submission status: Accepted

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


result=[]
if m%2==1:
    j=1
    for i in range(m//2):
        result.append((j,2*(m//2-i)+j))
        j+=1
    j=2*(m//2)+2
    for i in range(m//2+1):
        result.append((j,2*(m//2+1-i)+j-1))
        j+=1

else:
    j=1
    for i in range(m//2):
        result.append((j,2*(m//2-i)+j))
        j+=1
    j=2*(m//2)+2
    for i in range(m//2):
        result.append((j,2*(m//2-i)+j-1))
        j+=1

    
for item in result:
    print(item[0],item[1])

Submission status: Accepted

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

# 奇数のときは最後の数だけを最初に飛ばした形で先頭、末尾から作っていけばok
if N % 2 == 1:
    for i in range(1, M+1):
        print(i, N-i)
# 偶数のときは奇数と同じように作ると対戦距離がNの半分になるものがあるので一周したときに重複する
# 対戦距離がNの半分になるものが出ないように、対戦距離が半分になった時点で1つずらす
# 対戦距離 = b - a
else:
    for i in range(1, M+1):
        if N-2*i <= N//2:
            print(i, N-i-1)
        else:
            print(i, N-i)

Submission status: Accepted

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

import math
ans = []
if M == 1:
    ans.append((1, 2))
else:
    m0 = (1, M+1)
    m1 = (M+2, 2*M+1)
    for _ in range(math.ceil(M/2)):
        if m0[0] < m0[1]:
            ans.append(m0)
        if m1[0] < m1[1]:
            ans.append(m1)
        m0 = (m0[0]+1, m0[1]-1)
        m1 = (m1[0]+1, m1[1]-1)

for a in ans:
    print(a[0], a[1])

Submission status: Accepted

Solution code:
import numpy as np
N,M=map(int,input().split())
A=np.arange(M)
B=2*M-1-A
B[:M//2]-=(2*M+1)
A%=N
B%=N
A+=1
B+=1
for i in range(M):
  print(A[i],B[i])

Submission status: Accepted

Solution code:
#!/usr/bin/env python3
n, m = map(int, input().split())
ans = []
i = 1
while i < m + 2 - i:
    ans.append([i, m + 2 - i])
    i += 1
i = m + 2
while i < 3 * m + 3 - i:
    ans.append([i, 3 * m + 3 - i])
    i += 1
for i in ans:
    print(*i)

Submission status: Accepted

Solution code:
N,M = map(int,input().split())
if N % 4 in [1,2]:
    odd_pair = [N//2//2 , N//2//2+1]
    even_pair = [N-N//2//2-1,N-N//2//2+1]
elif N%4==3:
    odd_pair = [N-N//2//2-1 , N-N//2//2]
    even_pair = [N//2//2,N//2//2+2]
elif N%4==0:
    odd_pair = [N-N//2//2 , N-N//2//2+1]
    even_pair = [N//2//2,N//2//2+2]
cnt = 0
while True:
    print(*odd_pair)
    cnt += 1
    if cnt == M:
        break
    odd_pair[0] -= 1
    odd_pair[1] += 1
    print(*even_pair)
    cnt += 1
    if cnt == M:
        break
    even_pair[0] -= 1
    even_pair[1] += 1

Submission status: Accepted

Solution code:
import math

nums = [int(e) for e in input().split()]
n = nums[0]
m = nums[1]

x = -1
y = -1

ms = []

if n%2 == 0:
    x = 1
    y = n
    count = True

    while x < y:
        if y-x <= n/2 and count:
            y -= 1
            count = False
        ms += [[x,y]]
        x += 1
        y -= 1


else:
    x = 1
    y = n
    while x < y:
        ms += [[x, y]]
        x += 1
        y -= 1


# print(ms)

for i in range(m):
    print(str(ms[i][0]) + " " + str(ms[i][1]))

Submission status: Accepted

Solution code:
N, M = map(int, input().split()) 
ans = []
if M % 2 != 0:
    for i in range((M + 1)//2):
        ans.append((1 + i, M + 1 - i))
        
    for j in range((M - 1)//2):
        ans.append((M + 2 + j, 2*M + 1 - j))
else:
    for i in range(M//2):
        ans.append((1 + i, M + 1 - i))
        ans.append((M + 2 + i, 2*M + 1 - i))
              
for i in range(M):
    print(*ans[i])

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')
    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:
            # 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
    #%%
    N, M = R()
    if N%2 == 0: # even
        j = 2
        for i in range(1, M+1):
            if j == (N+6)//4:
                j+=1
            print(j, N+1-i)
            j += 1
    else: # odd
        for i in range(M):
            print(i+2, N-i)

    #%%
    # 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:
#!/usr/bin/env python3
import sys


def check(N, M, ans):
    p = list(range(1, N+1))
    s = set()
    for _ in range(N):
        for a, b in ans:
            s.add((min(p.index(a), p.index(b)), max(p.index(a), p.index(b))))
        p = [p[-1]] + p[:-1]
    assert len(s) == N * M


def solve(N: int, M: int):
    ans = []
    i, j = 0, N + 1
    diffs = set()
    for _ in range(M):
        i += 1
        j -= 1
        while j - i == i + N - j or j - i in diffs or i + N - j in diffs:
            j -= 1
        diffs.add(j - i)
        diffs.add(i + N - j)
        ans.append((i, j))
    for a, b in ans:
        print(a, b)
    # check(N, M, ans)


# Generated by 1.1.6 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()
    N = int(next(tokens))  # type: int
    M = int(next(tokens))  # type: int
    solve(N, M)

if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
#  =========     /\       /|    |====/|
#      |        /  \       |    |   / |
#      |       /____\      |    |  /  |
#      |      /      \     |    | /   |
#  ========= /        \  =====  |/====|  
#  code

def main():
    n , m = map(int , input().split())
    if n % 2 == 0:
        n -= 1
    k = (n + 1) // 2
    j = 1

    while j < k and m > 0:
        print(j , k)
        j += 1
        k -= 1
        m -= 1

    j = (n + 1) // 2 + 1
    k = n

    while j < k and m > 0:
        if k <= n:
            print(j , k)
            m -= 1
        j += 1
        k -= 1
    
    return

if __name__ == "__main__":
    main()

Submission status: Accepted

Solution code:
n, m = map(int, input().split())
i = 0
c = 0
while (i + 1) + i < (n - i - (i + 1)) and c < m:
    print(i + 1, n - i)
    c += 1
    i += 1
a = n // 2 + n % 2
for i in range(m - c):
    print(a - i - 1, a + i + 1)

Submission status: Accepted

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

    if N % 2:
        left = 1
        right = N - 1
        for _ in range(M):
            print(left, right)
            left += 1
            right -= 1
    else:
        left = 1
        right = N - 1
        flag = False
        for _ in range(M):
            if right - left <= N//2:
                flag = True
            
            if flag:
                print(left, right-1)
            else:
                print(left, right)
            
            left += 1
            right -= 1

        
if __name__ == "__main__":
    main()

Submission status: Accepted

Solution code:
n, m = map(int,input().split())
for i in range(m):
  print(i + 1, 2*m - i + (n%2==0 and 2*(m - i) - 1 >= n//2))

Submission status: Accepted

Solution code:
n,m= map(int, input().split())
ans=[]
if n%2==1:
    for i in range(m):
        ans.append(i+1)
        ans.append(n-1-i)

else:
    if m<=n//4:
        for i in range(m):
            ans.append(n//2+1+i)
            ans.append(n//2-i)

    else:
        for i in range(m-n//4):
            ans.append(2+i)
            ans.append(n-i)
        for i in range(n//4):
            ans.append(n//2+1+i)
            ans.append(n//2-i)


print(*ans)

Submission status: Accepted

Solution code:
from collections import Counter
import sys

sys.setrecursionlimit(10 ** 6)

mod = 1000000007
inf = int(1e18)

dx = [0, 1, 0, -1]
dy = [1, 0, -1, 0]


def inverse(a):
    return pow(a, mod - 2, mod)


def usearch(x, a):
    lft = 0
    rgt = len(a) + 1
    while rgt - lft > 1:
        mid = (rgt + lft) // 2
        if a[mid] <= x:
            lft = mid
        else:
            rgt = mid
    return lft


def main():
    n, m = map(int, input().split())
    ans = []
    for i in range(1, (m+1)//2+1):
        if len(ans) == m:
            break
        ans.append([i, m+2-i])

    for i in range(1, (m+1)//2+1):
        if len(ans) == m:
            break
        ans.append([m+1+i, 2*m+2-i])

    for x in ans:
        print(x[0], x[1])

main()

Submission status: Accepted

Solution code:
from collections import Counter
import sys

sys.setrecursionlimit(10 ** 6)

mod = 1000000007
inf = int(1e18)

dx = [0, 1, 0, -1]
dy = [1, 0, -1, 0]


def inverse(a):
    return pow(a, mod - 2, mod)


def usearch(x, a):
    lft = 0
    rgt = len(a) + 1
    while rgt - lft > 1:
        mid = (rgt + lft) // 2
        if a[mid] <= x:
            lft = mid
        else:
            rgt = mid
    return lft


def main():
    n, m = map(int, input().split())
    ans = []
    for i in range(1, (m+1)//2+1):
        ans.append([i, m+2-i])

    for i in range(1, (m+1)//2+1):
        if len(ans) == m:
            break
        ans.append([m+1+i, 2*m+2-i])

    for x in ans:
        print(x[0], x[1])

main()

Submission status: Accepted

Solution code:
n,m=map(int,input().split())
k=0
if n%2==1:
  for i in range(m):
    print(i+1,end=" ")
    print(n-i)
elif n%4==0:
  for i in range((n+2)//4):
    if k>=m:
      break
    print(i+1,end=" ")
    print(n-i)
    k+=1
  for i in range((n+2)//4,n//2):
    if k>=m:
      break
    print(i+1,end=" ")
    print(n-i-1)
    k+=1
else:
  for i in range((n+2)//4-1):
    if k>=m:
      break
    print(i+1,end=" ")
    print(n-i)
    k+=1
  for i in range((n+2)//4-1,n//2):
    if k>=m:
      break
    print(i+1,end=" ")
    print(n-i-1)
    k+=1

Submission status: Accepted

Solution code:
n,k=map(int,input().split())
ans=[]
a=1
b=n
s=set()
for _ in range(k):
  s.add(abs(a-b))
  s.add(abs(n+a-b)%n)
  ans.append([a,b])
  a+=1
  b-=1
  if abs(a-b) in s or abs(n+a-b)%n in s or abs(a-b)==abs(n+a-b):
    b-=1
for i in range(k):
  print(*ans[i])

Submission status: Accepted

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

if n % 2: # odd
    for i in range(m):
        print(i + 1, n - i)
else: # even
    for i in range(m):
        if 1 + 2 * i >= n - 1 - 2 * i:
            l1 = i
            break
        m -= 1
        print(i + 1, n - i)

    for i in range(m):
        if i + 1 + l1 >= n - i - l1 - 1:
            raise Exception()
        print(i + 1 + l1, n - i - l1 - 1)

Submission status: Accepted

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

M_MAX = (N - 1) // 2

pos = 1
size = M_MAX
i = 0
while i < M and size > 0:
    print(pos, pos + size)
    pos += 1
    i += 1
    size -= 2

pos = 1 + M_MAX + 1
size = M_MAX - 1
while i < M and size > 0:
    print(pos, pos + size)
    pos += 1
    i += 1
    size -= 2

Submission status: Accepted

Solution code:
N, M = map(int, input().split())
matches = []
i = 1
l = N // 2 - 1 if N % 2 == 0 else N // 2
while len(matches) < M:
    if l <= 0:
        i = N // 2 + 1 if N % 2 == 0 else N // 2 + 2
        l = N // 2 - 2 if N % 2 == 0 else N // 2 - 1
        continue
    matches.append((i, i+l))
    i += 1; l -= 2
for a, b in matches: print(a, b)

Submission status: Accepted

Solution code:
n,m=map(int,input().split())
l=m//2
for i in range(l):
    print(1+i,1+2*l-i)
for j in range(m-l):
    print(1+2*l+1+j,2*m+1-j)

Submission status: Accepted

Solution code:
N, M = map(int, input().split())
order = [i for i in range(2, N+1)]
reverse = order[::-1]
d = 0
for i in range(M):
    if N%4 == 0 and reverse[i] - order[i] == N//2:
        d = 1
    if N%2 == 0 and i == (N//2-1) //2:
        d = 1
    print(order[i+d], reverse[i])

Submission status: Accepted

Solution code:
def main():
    d = 0
    for i in range(M):
        if N%4 == 0 and reverse[i] - order[i] == N//2:
            d = 1
        if N%2 == 0 and i == (N//2-1) //2:
            d = 1
        print(order[i+d], reverse[i])

N, M = map(int, input().split())
order = [i for i in range(2, N+1)]
reverse = order[::-1]
main()