Editorial for Four Multiples

Approach

Let $S_x$ be the integers in $[1, n]$ divisible by $x$. The answer is $|S_a \cup S_b \cup S_c \cup S_d|$.

Four-set inclusion-exclusion expands this into the $15$ nonempty subsets of $\{a, b, c, d\}$. For a subset $T$, the integers divisible by every modulus in $T$ are exactly those divisible by $\mathrm{lcm}(T)$, and there are $\lfloor n / \mathrm{lcm}(T) \rfloor$ of them. Add the term if $|T|$ is odd and subtract if $|T|$ is even.

Loop over bitmasks $1..15$ rather than writing out $15$ terms by hand. Hardcoding the terms works for four sets, but it is easy to miss a sign or an lcm, and the mask loop is the same code you would use for more sets.

If $\mathrm{lcm}(T) > n$, that term is $0$. In C++ the raw lcm of four values up to $10^9$ may not fit in a 64-bit integer; stop early when the running lcm already exceeds $n$. Time is $O(1)$.

Solution (Python)

from math import gcd

n, a, b, c, d = map(int, input().split())
mods = [a, b, c, d]

ans = 0
for mask in range(1, 16):
    cur = 1
    bits = 0
    too_big = False
    for i in range(4):
        if mask & (1 << i):
            bits += 1
            g = gcd(cur, mods[i])
            nxt = cur // g * mods[i]
            if nxt > n:
                too_big = True
                break
            cur = nxt
    if too_big:
        continue
    term = n // cur
    if bits % 2:
        ans += term
    else:
        ans -= term

print(ans)