Cheesy 7 | CodeChef

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who can make (m+1) first will win the game

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3

now (m+1) is possible only if the remainder is in the range (1 to m)

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4

n=6 , m=4 now 6%5=1 so A will select 1 and make the number 6-1=5 so A wins

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so divide the number with n%(m+1) so if the remainder comes in the range 1 to m then the particular guy can select that number and make (m+1)

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I am so much glad that you understand the problem and able to solve it. :blush: :stuck_out_tongue::stuck_out_tongue:

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Refer these notes for this problem

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@cubefreak777 Wow, that is impressive. btw how did you find it?

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By searching online

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@cubefreak777 Can you tell me what is wrong in my code?

import random
t = int(input())
for i in range(t):
    a, b = map(int, input().split())
    n = []
    k = 0
    for _ in range(b):
        n.append(k+1)
        k = k + 1
    m = 0
    while True:
        ra = 0 + random.choice(n)
        if ra == a:
            break
            ai = m
            if ai % 2 == 0:
                print('A')
            else:
                print('B')
        else:
            if ra == 2:
                print('A')
            elif ra == 1:
                print('B')
            else:
                pass
            break
        m = m + 1

Please help.

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I got a TLE

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Constraints are large and problem expects an O(1) solution and your code doesn’t looks like a constant time solution to me

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