Kertoma

from math import factorial, log10, floor
from time import perf_counter
arr = [int(x) for x in input().split()]
length = sum(arr)

start = perf_counter()
def check_numbers(n):
    cur = [0] * 10
    for i in n:
        cur[int(i)]+=1
    return arr == cur

m = 500
l = 0
r = 1000


while True:
    fac = factorial(m)
    res = int(log10(fac))+1
    if length == res and (res > 6 or check_numbers(str(fac))):
        print(m)
        break
    else:
        if res < length:
            l = m+1
        elif res > length:
            r = m-1
    m = floor((l+r)/2)



print(perf_counter()-start)

Test 1

Group: 1, 2, 3

Verdict: WRONG ANSWER

Test 2

Group: 1, 2, 3

Verdict: TIME LIMIT EXCEEDED

Test 3

Group: 1, 2, 3

Verdict: WRONG ANSWER

Test 4

Group: 1, 2, 3

Verdict: WRONG ANSWER

Test 5

Group: 1, 2, 3

Verdict: WRONG ANSWER

Test 6

Group: 1, 2, 3

Verdict: WRONG ANSWER

Test 7

Group: 2, 3

Verdict: WRONG ANSWER

Test 8

Group: 2, 3

Verdict: WRONG ANSWER

Test 9

Group: 2, 3

Verdict: WRONG ANSWER

Test 10

Group: 2, 3

Verdict: WRONG ANSWER

Test 11

Group: 3

Verdict: TIME LIMIT EXCEEDED

Test 12

Group: 3

Verdict: TIME LIMIT EXCEEDED

Test 13

Group: 3

Verdict: TIME LIMIT EXCEEDED

Test 14

Group: 3

Verdict: TIME LIMIT EXCEEDED

Test 15

Group: 3

Verdict: TIME LIMIT EXCEEDED

Test 16

Group: 3

Verdict: TIME LIMIT EXCEEDED