ARC110

Kajima Construction Programming Contest 2020 (AtCoder Regular Contest 110) - AtCoder

A

• Multiplying all of them by +1 satisfies the remainder condition, but it is too large, so I found least common denominator.
• Related: Chinese remainder theorem. python

from math import gcd
N = int(input())
x = 1
for i in range(2, N + 1):
    g = gcd(x, i)
    x *= i // g
print(x + 1)

B

• In short, whether this automaton accepts or not.
  • 
• There is a trap when the input is a single character, so it is only the case there. python

def solve(N, T):
    if N == 1:
        if T == "1":
            return 2 * 10 ** 10
        elif T == "0":
            return 10 ** 10

    state = [True, True, True]
    for c in T:
        nextState = [0] * 3
        for i in range(3):
            nextState[i] = state[i - 1] and (c == "011"[i])
        state = nextState
    ret = 0
    for i in range(3):
        if state[i]:
            ret += (3 * 10 ** 10 - ((i - N) % 3 + N)) // 3 + 1
    return ret

ARC110C ARC110D

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