AtCoder Sumitomo Mitsui Trust Bank Programming Contest 2019 Participation Report

AtCoder Sumitomo Mitsui Trust Bank Programming Contest 2019 Participation Report

A - November 30

Break through in 2 minutes. Just write.

M1, D1 = map(int, input().split())
M2, D2 = map(int, input().split())

if D2 == 1:
    print(1)
else:
    print(0)

B - Tax Rate

It broke through in four and a half minutes. I didn't know what to do if it didn't become an integer, and I issued it properly, but it was AC.

N = int(input())

X = N / 1.08
if int(X) * 108 // 100 == N:
    print(int(X))
elif int(X + 1) * 108 // 100 == N:
    print(int(X + 1))
else:
    print(':(')

C - 100 to 105

Break through in 7 and a half minutes. WA1. I thought it was more difficult than the recent ABC C problem. It wasn't because it was difficult, it was just Chombo.

X = int(input())

dp = [0] * (X + 1 + 105)
dp[0] = 1

for i in range(X):
    if dp[i] == 1:
        for j in range(100, 106):
            dp[i + j] = 1
print(dp[X])

D - Lucky PIN

It breaks through in 22 minutes and a half. Since there are 3 characters, there are only 1000 patterns at the maximum from 000 to 999, so I can afford to try all patterns, but if N ≤ 30000, it seems to be TLE, so if 4 or more of the same characters are consecutive, it will be 3 I compressed and searched. Since only 3 characters are used anyway, a sequence of 4 or more characters is meaningless.

N = int(input())
S = input()

t = [S[0]]
p = S[0]
r = 1
for i in range(1, N):
    if S[i] == p:
        r += 1
        if r < 4:
            t.append(S[i])
    else:
        r = 1
        t.append(S[i])
T = ''.join(t)

result = 0
for i in range(10):
    a = T.find(str(i))
    if a == -1:
        continue
    for j in range(10):
        b = T.find(str(j), a + 1)
        if b == -1:
            continue
        for k in range(10):
            if T.find(str(k), b + 1) != -1:
                result += 1
print(result)

Addendum: I didn't need to compress it (^^; I thought there was an extreme test case where there was a continuation after a series of about 30,000 characters. Otherwise, it's not a D problem difficulty.

N = int(input())
S = input()

result = 0
for i in range(10):
    a = S.find(str(i))
    if a == -1:
        continue
    for j in range(10):
        b = S.find(str(j), a + 1)
        if b == -1:
            continue
        for k in range(10):
            if S.find(str(k), b + 1) != -1:
                result += 1
print(result)

E - Colorful Hats 2

Break through in 28 minutes. WA1. I didn't want to trust that A1 was 0. Just keep applying the possible patterns from the beginning. E Easy for problems.

N = int(input())
A = list(map(int, input().split()))

result = 1
t = [0, 0, 0]
for i in range(N):
    a = A[i]
    f = -1
    k = 0
    for j in range(3):
        if t[j] == a:
            k += 1
            if f == -1:
                t[j] += 1
                f = j
    result = (result * k) % 1000000007
print(result)

F - Interval Running

I couldn't break through. But it's very easy as an F problem. I was disappointed that I missed it even though I seemed to be able to complete it for the first time.

from sys import exit

T1, T2 = map(int, input().split())
A1, A2 = map(int, input().split())
B1, B2 = map(int, input().split())

if A1 > B1:
    A1, B1 = B1, A1
    A2, B2 = B2, A2

if A2 < B2:
    print(0)
    exit()

if A1 * T1 + A2 * T2 < B1 * T1 + B2 * T2:
    print(0)
    exit()

if A1 * T1 + A2 * T2 == B1 * T1 + B2 * T2:
    print('infinity')
    exit()

a = B1 * T1 - A1 * T1
b = (A1 * T1 + A2 * T2) - (B1 * T1 + B2 * T2)
t = a // b

if a % b == 0:
    print(t * 2)
else:
    print(t * 2 + 1)

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