ABC177

I was aware of the symptoms of being sick before it started, but I'm a wreck. 3 completions in and the rating has dropped for the first time.

B - Substring

• When comparing a 1000 letter S to a 500 letter T, you need to compare 500 x 501, which is about 250,000 letters, but you can do it because that's about as high as you can go. python

def solve(S, T):
    buf = []
    for i in range(len(S) - len(T) + 1):
        diff = 0
        for j in range(len(T)):
            if S[i + j] != T[j]:
                diff += 1
        buf.append(diff)
    return min(buf)

• Forgot +1 in len(S) - len(T) + 1 run-time error
  • When the lengths are equal, it stops looping and the target of min becomes empty.

ABC177C

D - Friends

• 
• I saw the problem statement and thought UnionFind was asking for the number of friend groups.
  • Should have asked for the size of the largest group instead of asking for the number of groups
  • We will break up that group one by one.
• I had implemented it in UnionFind, but the answer doesn't match due to the above misunderstanding.
• So I tried to confirm it by drawing Sample 1 in the diagram.
• I saw the input for sample 1 and mistook 5 3 for "5 and 3 are friends" (really 3 relationships between 5 people).
  • 
• Diagrams actually drawn during the contest
  • 
  • Reinterpret the question so that the answer is 3 and assume "I see, friendships are not symmetrical.
  • I thought this was a trap to make me think it was UnionFind, but actually it's not! Or so I thought.
  • If the friendship is one-way, the question becomes one of finding the length of the longest path.
  • Even with this wrong policy, samples 1 and 2 will be correct.
    • I kept thinking the policy was right, and sample 3 was stuck because it didn't fit.

python

def main():
    global parent, rank
    # parse input
    N, M = map(int, input().split())
    parent = [-1] * N
    rank = [0] * N
    for q in range(M):
        a, b = map(int, input().split())
        unite(a - 1, b - 1)

    xs = list(find_root(x) for x in range(N))
    # debug(": xs", xs)
    # print(len(set(xs)))  # WRONG!
    from collections import Counter
    print(max(Counter(xs).values()))

• I crossed out the one wrong line for the number of groups and added the two lines for the size of the largest group, and I got AC.
  • Implementation of unite, find_root is here: https://github.com/nishio/atcoder/blob/master/yosupo/unionfind.py

E - Coprime - Eratosthenes' sieve to determine the prime number from the smallest, while dividing N numbers by that prime number. - No need to retain the result of prime factorization, since it is obtained by paying attention to each prime factor for the determination of the problem - When there are d numbers out of N that have some prime factor p, - Cannot be PAIRWISE if d is greater than or equal to 2 - If d is N, it is fixed to NOT COPRIME. - Unfortunately, TLE - Eleven of the ones in between were the right answers. python

def solve(N, AS):
    is_pairwise = True
    maxAS = max(AS)
    sieved = [False] * (maxAS + 10)
    for p in range(2, maxAS + 1):
        if sieved[p]:
            continue
        # p is prime
        x = p + p
        while x <= maxAS:
            sieved[x] = True
            x += p
        sum_div = 0
        for i, a in enumerate(AS):
            dividable = 0
            while a % p == 0:
                a //= p
                dividable = 1
            AS[i] = a
            sum_div += dividable
        if sum_div >= 2:
            is_pairwise = False
        if sum_div == N:
            return "not coprime"

    if is_pairwise:
        return "pairwise coprime"
    return "setwise coprime"

• Fast prime factorization of multiple numbers seems to finish Eratosthenes first.
  • It seems to build a tree of divisors by writing the divisor prime first, rather than simply bool
  • AC easily with Fast Factorization.
    • I didn't want to keep the result of prime factor decomposition, so I wrote it in the above way, but I had no problem with the policy of plugging it into the list in a messy way.

python

def solve(N, AS):
    maxAS = max(AS)
    eratree = [0] * (maxAS + 10)
    for p in range(2, maxAS + 1):
        if eratree[p]:
            continue
        # p is prime
        eratree[p] = p
        x = p * p
        while x <= maxAS:
            if not eratree[x]:
                eratree[x] = p
            x += p

    count = defaultdict(int)
    for a in AS:
        factors = []
        while a > 1:
            d = eratree[a]
            factors.append(d)
            a //= d
        for f in set(factors):
            count[f] += 1

    if any(x == N for x in count.values()):
        return "not coprime"
    if any(x >= 2 for x in count.values()):
        return "setwise coprime"
    return "pairwise coprime"

Counted and compared the number of divisions.

• 100000 100001 100003, I found that my original version had 28792 times while the fast prime factorization had 13 times, so they are not comparable at all! Computational Considerations
• Let N be the number of numbers and M the largest number (this time both 10^6).
• The number of prime numbers is $M / \log(M)$ about
• No, you can't find a prime number and then divide it because $NM/\log(M)$.
  • I was under the impression that the prime numbers were much smaller, $\log(M)$ about
• Fast prime factorization finds prime factors without trial division.
  • This can really be done $\log(M)$ times
  • The most common case is at 2^x, but at most 20.
  • I'm not sure why it wasn't a problem to shove them into a messy list, or why it was only that many pieces in the first place.
    • In this case, I only wanted to know whether it existed or not, so I uniq'd it with set, but if I wanted to know the number of pieces, it would be better to just put it into Counter.

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