AtCoder Beginner Contest 162 Participation Report

AtCoder Beginner Contest 162 Participation Report

ABC162A - Lucky 7

Break through in 2 minutes. Just write.

N = int(input())

if str(N).count('7') > 0:
    print('Yes')
else:
    print('No')

ABC162C - Sum of gcd of Tuples (Easy)

Break through in 4 minutes. For some reason I accidentally did the C problem first. It took a long time because the judge system was clogged and I could not use online execution. Well, I just wrote the problem sentence, but it seems to be TLE Because it was, I modified it by a constant number of times.

def main():
    from math import gcd

    K = int(input())

    result = 0
    for a in range(1, K + 1):
        for b in range(1, K + 1):
            t = gcd(a, b)
            for c in range(1, K + 1):
                result += gcd(t, c)
    print(result)


main()

ABC162B - FizzBuzz Sum

Break through in two and a half minutes. Just add except for Fizz, Buzz, FizzBuzz, etc.

N = int(input())

result = 0
for i in range(1, N + 1):
    if i % 3 == 0 or i % 5 == 0:
        continue
    result += i
print(result)

ABC162D - RGB Triplets

It broke through in 24 minutes. TLE1. N ≤ 4000, so I thought it was a double loop. When I did naive, it became triple, but it was not particularly difficult because it can be cumulatively summed.

def main():
    N = int(input())
    S = input()

    rcs = [0] * N
    gcs = [0] * N
    bcs = [0] * N

    for i in range(N):
        if S[i] == 'R':
            rcs[i] = 1
        elif S[i] == 'G':
            gcs[i] = 1
        elif S[i] == 'B':
            bcs[i] = 1

    for i in range(1, N):
        rcs[i] += rcs[i - 1]
        gcs[i] += gcs[i - 1]
        bcs[i] += bcs[i - 1]

    result = 0
    for i in range(N):
        if S[i] == 'R':
            for j in range(i + 1, N):
                if S[j] == 'R':
                    continue
                elif S[j] == 'G':
                    result += bcs[N - 1] - bcs[j]
                    t = 2 * j - i
                    if t < N and S[t] == 'B':
                        result -= 1
                elif S[j] == 'B':
                    result += gcs[N - 1] - gcs[j]
                    t = 2 * j - i
                    if t < N and S[t] == 'G':
                        result -= 1
        elif S[i] == 'G':
            for j in range(i + 1, N):
                if S[j] == 'G':
                    continue
                elif S[j] == 'R':
                    result += bcs[N - 1] - bcs[j]
                    t = 2 * j - i
                    if t < N and S[t] == 'B':
                        result -= 1
                elif S[j] == 'B':
                    result += rcs[N - 1] - rcs[j]
                    t = 2 * j - i
                    if t < N and S[t] == 'R':
                        result -= 1
        elif S[i] == 'B':
            for j in range(i + 1, N):
                if S[j] == 'B':
                    continue
                elif S[j] == 'R':
                    result += gcs[N - 1] - gcs[j]
                    t = 2 * j - i
                    if t < N and S[t] == 'G':
                        result -= 1
                elif S[j] == 'G':
                    result += rcs[N - 1] - rcs[j]
                    t = 2 * j - i
                    if t < N and S[t] == 'R':
                        result -= 1
    print(result)


main()

ABC162E - Sum of gcd of Tuples (Hard)

I couldn't break through. I thought I could solve it. I'm sorry.

Addendum: gcd (f (k, n) with a sequence of length n consisting of integers greater than or equal to 1 and less than or equal to k {A 1 </ sub>, ..., A n </ sub>} If A 1 </ sub>, ..., A n </ sub>) is the number of 1, the answer is ∑ i = 1..K </ sub> f (floor) (K / i), N) * i.

N, K = map(int, input().split())

MOD = 10 ** 9 + 7

cache = {}
def f(k, n):
    if k == 1:
        return 1
    if k in cache:
        return cache[k]
    result = pow(k, n, MOD)
    for i in range(2, k + 1):
        result -= f(k // i, n)
        result %= MOD
    cache[k] = result
    return result


result = 0
for i in range(1, K + 1):
    result += f(K // i, N) * i
    result %= MOD
print(result)

Explanation This is the assumed solution of PDF.

N, K = map(int, input().split())

MOD = 10 ** 9 + 7

c = [0] * (K + 1)
for i in range(K, 0, -1):
    t = pow(K // i, N, MOD)
    for j in range(2, K // i + 1):
        t -= c[i * j]
        t %= MOD
    c[i] = t

result = 0
for i in range(1, K + 1):
    result += c[K // i] * i
    result %= MOD
print(result)

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