Daily AtCoder # 18 in Python

Introduction

Last time It's a similar subject today. Mathematical wings second day

#18 ** Thoughts ** ABC095-A Just count and then * 100

s = str(input())
print(700+s.count('o')*100)

ABC085-A Since the 4th part of s is the common part with the output, it is held as a slice. 2019 is linked to it.

s = str(input())
print('2018'+s[4:])

ABC069-B Just subtract the first and last two letters from len (s) and solve

s = str(input())
print(s[0]+str(len(s)-2)+s[-1])

ABC082-B 1WA. I'm not good at sorting strings I forgot to sort the string, so I converted it to a number with ord and sorted it. To maximize the difference in dictionary order, s is a → z order and t is z → a. All you have to do is compare.

s = str(input())
t = str(input())

s = list(map(ord,s))
t = list(map(ord,t))
s.sort()
t.sort(reverse=True)
if s < t:
    print('Yes')
else:
    print('No')

Summary

I was inspired by the wings of mathematics, so I will do my best for quantum computers. see you

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