AtCoder Panasonic Programming Contest 2020 Participation Report

AtCoder Panasonic Programming Contest 2020 Participation Report

panasonic2020A - Kth Term

Break through in two and a half minutes. Well, just write.

K = int(input())

t = [1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51]
print(t[K - 1])

panasonic2020B - Bishop

Break through in about 6 minutes. 1WA. I completely forgot the case where H and W are 1.

H, W = map(int, input().split())

if H == 1 or W == 1:
    print(1)
elif W % 2 == 0:
    print(H * W // 2)
else:
    if H % 2 == 0:
        print(H * W // 2)
    else:
        print((W + 1) // 2 + (H - 1) * W // 2)

panasonic2020C - Sqrt Inequality

Lost. Even though I knew that I had to calculate with an integer, I couldn't drop it into an integer formula. If I was told that I should square it twice, I would understand it immediately. For some reason, I don't know during the contest. orz. I hate math problems.

a, b, c = map(int, input().split())

if c - a - b > 0 and (c - a - b) * (c - a - b) > 4 * a * b:
    print('Yes')
else:
    print('No')

panasonic2020D - String Equivalence

Break through in 32 and a half minutes. 1WA. No matter how many times I read it, I was in trouble because the definition did not come to my mind. So, I completely misunderstood the definition and ate WA, and then until about N = 4 I finally understood by writing everything by hand AC. In short, the answer is that I added the character string from a to the next one in the largest dictionary order of the characters that have appeared so far.

N = int(input())

q = ['a']
for i in range(N - 1):
    nq = []
    for s in q:
        stop = ord(max(s)) + 2
        for i in range(ord('a'), stop):
            nq.append(s + chr(i))
    q = nq
for s in q:
    print(s)

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