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That's an interesting problem! It feels like there should be a clever math trick to solve this efficiently, avoiding brute-force looping. I'd probably start by calculating ab, then finding the two multiples of x that bracket it. Then a simple comparison to see which is closer. It reminds me of some of the fun puzzles I enjoy in games. Speaking of which, sometimes when I need a mental break, I mess around with something completely different like Solar Smash and just destroy planets. Great problem though!

import sys import math

def closest_multiple(a, b, x): value = a ** b # this will be a float if b is negative div = value / x

lower = math.floor(div)
upper = lower + 1

lower_mult = lower * x
upper_mult = upper * x

diff_lower = abs(value - lower_mult)
diff_upper = abs(value - upper_mult)

if diff_lower < diff_upper:
    return round(lower_mult)
elif diff_lower > diff_upper:
    return round(upper_mult)
else:
    return round(min(lower_mult, upper_mult))  # tie case

Fast input reading

def main(): input = sys.stdin.read data = input().split()

T = int(data[0])
idx = 1
results = []

for _ in range(T):
    a = int(data[idx])
    b = int(data[idx + 1])
    x = int(data[idx + 2])
    idx += 3

    res = closest_multiple(a, b, x)
    results.append(str(res))

print("\n".join(results))

main()

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