This is a review article for beginners of competition professionals.
The solution I write here is written while looking at the commentary and other people's submissions. It may not be what you actually submitted.
The question is to answer if there are two equal numbers out of the three given numbers.
You can simply compare them. This problem can be rephrased as being asked "Is there two types of elements?", So let's investigate that. It seems convenient to use a set type object to remove duplicate elements in an array.
A = list(map(int, input().split()))
if len(set(A)) == 2:
print('Yes')
else: print('No')
The question is to find out if the elements of a given array meet the condition that "if even, it is divisible by 3 or 5".
Examine all the elements and reject if they are even but not divisible by either.
N = int(input())
A = list(map(int, input().split()))
for a in A:
if a%2 == 0 and not (a%3 == 0 or a%5 == 0):
print('DENIED')
break
else:
print('APPROVED')
Since some character strings are given, it is a problem to output the character string with the highest number of occurrences (if the number is the same, all in dictionary order).
Store strings in an array. Aggregate using collections.Counter. Take the maximum value with max (), and put only the character string with the maximum number of occurrences into one array. I rearranged this with sorted () and output it.
import collections
N = int(input())
S = [input() for _ in range(N)]
count = collections.Counter(S)
maxV = max(count.values())
c = [k for k, v in count.items() if v == maxV]
print('\n'.join(sorted(c)))
When all the products of pairs made from a given array are arranged in order, the question is what is the K-th smallest value.
I do not know. I gave up. Refer to other answers.
The following code is written to make it easier for you to understand, referring to most of the other answers. I have explained as much as possible in the comments, but honestly there are many parts that I do not understand well. I'm not sure if the comment is correct (especially the part with?).
import numpy as np
N, K = map(int, input().split())
A = np.array(list(map(int, input().split())))
A = np.sort(A)
G = A[A > 0]
Z = A[A == 0]
L = A[A < 0]
l, r = 10**18, -10**18
while l-r > 1:
#Check "the number of pairs whose product is m or less".
m = (l+r) // 2
# A[A > 0]Number of items that meet the conditions?
Pk = np.searchsorted(A, m//G, side="right").sum()
# A[A < 0]Number of items that meet the conditions?
Nk = (N - np.searchsorted(A, (-m-1)//(-L), side="right")).sum()
# A[0]Number of items that meet the conditions?
Zk = 0
if m >= 0:
Zk += len(Z) * N
#Since the product of the same elements cannot be selected, reduce it if the conditions are met.
duplicate = np.count_nonzero(A*A <= m)
#Match the number of elements that meet the conditions
k = Pk + Nk + Zk - duplicate
#All elements are doubled. Remove duplicate elements
k //= 2
#If the number of elements k that satisfy the condition is K or more, m is lowered, and if it is less than K, m is raised.
if k >= K:
l = m
else:
r = m
#If l and r match, m is uniquely determined
print(l)
I passed by this.
I don't understand the following part at all. I was wondering why I could find the number that meets the conditions, but I gave up.
# A[A > 0]Number of items that meet the conditions?
Pk = np.searchsorted(A, m//G, side="right").sum()
# A[A < 0]Number of items that meet the conditions?
Nk = (N - np.searchsorted(A, (-m-1)//(-L), side="right")).sum()
It is a problem to think about how to pay the money so that the "total of the number of sheets to be put out and the number of changes" is the smallest. However, the cash in this world is only in the $ 10 ^ n $ unit.
Let's count from the bottom. If the number of sheets to be paid is 5 or less, it is more efficient to pay as it is, and if it is 6 or more, it is more efficient to move up to the next digit and receive the change.
Bad guy.py
N = list(map(int, input()))
N = N[::-1] + [0]
count = 0
for i, n in enumerate(N):
if n <= 5:
count += n
elif n > 5:
count += 10 - n
N[i+1] += 1
print(count)
This is WA. With the above code, there are some omissions. For example, if you want to pay $ 95 $ yen, it is more efficient to pay $ 100 $ yen and receive the change (6 cards in total). At this rate, I paid $ 105 $ for a total of 7 cards.
The condition branches further only when 5 is issued. If the next digit is 5 or more, you can reduce the change by one by moving up.
The one who passed.py
N = list(map(int, input()))
N = N[::-1] + [0]
count = 0
for i, n in enumerate(N):
if n < 5:
count += n
elif n > 5:
count += 10 - n
N[i+1] += 1
elif n == 5:
if N[i+1] >= 5:
N[i+1] += 1
count += 5
print(count)
Looking at the explanation, the solution was different.
We will check the number of sheets in order from the top. You can calculate the carry-up by asking for the amount of money "when you pay exactly" and "when you pay one more".
I will write according to the explanation.py
N = list(map(int, input()))
m = 0 #Minimum number of sheets
m_ = 1 #Secure the number of sheets at the time of carry-up, always thinking that n is one more
for n in N:
m, m_ = min(m + n, m_ + 10-n), min(m + (n+1), m_ + 10-(n+1))
print(m)
How simple is it to write?
That's all for this article.
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