Daily AtCoder # 23 in Python

Introduction

Last time Today is also C

#23 ABC154-C 41diff If len (set (A)) = len (A), all the elements are different, so just set

n = int(input())
a = list(map(int,input().split()))

a_s = set(a)
if len(a) == len(a_s):
    print('YES')
else:
    print('NO')

ABC153-C 36diff Sort H. Since the number of attacks will decrease if you use the special move in descending order of physical strength, add from H [k:]

n, k = map(int,input().split())
h = list(map(int,input().split()))
h.sort(reverse=True)
ans = 0
for i in range(k,n):
    ans += h[i]
print(ans)

ABC152-C 119diff If you try all the (i, j) pairs as it is, it will be TLE, so update the minimum value. Then you can determine if $ P_j \ leq P_i $ is satisfied.

n = int(input())
p = list(map(int,input().split()))

ans = 0
for i in range(n):
    if i == 0:
        ans += 1
        m = p[0]
        continue
    if p[i] <= m:
        ans += 1
        m = p[i]

print(ans)

ABC151-C 239diff Manage the correct questions with bool. WA is just added when AC.

n, m = map(int,input().split())
ps = list(list(map(str,input().split())) for _ in range(m))

check = [True] * (n+1)
wa = [0] * (n+1)
wa_ans = 0
for i in range(m):
    if ps[i][1] == 'AC' and check[int(ps[i][0])]:
        check[int(ps[i][0])] = False
        wa_ans += wa[int(ps[i][0])]
    if ps[i][1] == 'WA':
        wa[int(ps[i][0])] += 1
ans = 0
for i in check:
    if not i:
        ans += 1

print(ans,wa_ans)

ABC150-C 94diff itertools god. Since it is $ N \ le 8 $, all combinations are listed.

import itertools
n = int(input())
p = tuple(((map(int,input().split()))))
q = tuple(((map(int,input().split()))))

num = []
for i in range(1,n+1):
    num.append(i)
ans = []
for i in itertools.permutations(num,n):
    ans.append(i)

print(abs(ans.index(p)-ans.index(q)))

Summary

I can still solve it. see you