Daily AtCoder # 16 in Python

Introduction

Last time It is C today. Today is the end of the example.

#16

** Thoughts ** ABC049-C I didn't understand at all, so I looked at the explanation. Hmmm, is it not necessary to distinguish er by reversing the character string? After reversing the character string, I just make it an if statement.

s = str(input())
s = ''.join(list(reversed(s)))

t = 0
while t <= len(s):
    if s[t:t+5] == 'maerd':
        t += 5
        continue
    elif s[t:t+7] == 'remaerd':
        t += 7
        continue
    elif s[t:t+5] == 'esare':
        t += 5
        continue
    elif s[t:t+6] == 'resare':
        t += 6
        continue
    elif t == len(s):
        print('YES')
        quit()
    else:
        break
print('NO')

Python3 → 33ms PyPy3 → 189ms Is Python faster than PyPy for slicing?

ABC086-C Required [Manhattan Distance?] When moving from (0,0) to (x, y) (https://ja.wikipedia.org/wiki/%E3%83%9E%E3%83%B3%E3 If% 83% 8F% E3% 83% 83% E3% 82% BF% E3% 83% B3% E8% B7% 9D% E9% 9B% A2) is d, you can arrive in time if t> = d. I will. Also, if (d --t)% 2 == 0, even if you arrive less than t, you can go back and forth between the adjacent square and the target square, so you can arrive at t.

n = int(input())
l = [list(map(int,input().split())) for _ in range(n)]

for i in range(n):
    t = l[i][0]
    x = l[i][1]
    y = l[i][2]
    d = x + y
    if d <= t and (d - t) % 2 == 0:
        continue
    else:
        print('No')
        quit()
print('Yes')

Python3 → 369ms PyPy3 → 585ms

Summary

I've solved all the examples this time, so I'll solve similar problems from the next time! see you