Lala Land and Apple Trees

Source

It's easy to see that we're only going to be able to get to l trees on each side were l is the minimum number of trees on both sides. If the lengths are different, we can get one more (if there are 4 trees on the left and 3 on the right, start by the left side to get all 4 trees)

n = int(input())
low = []
high = []
for i in range(n):
    x, apples = map(int, input().split())
    if x < 0:
        low.append((x,apples))
    else:
        high.append((x, apples))
low.sort(reverse = True)
high.sort()

possible = min(len(low), len(high)) + (1 if len(low) != len(high) else 0)
ans = sum([x[1] for x in low[:possible]]) + sum([x[1] for x in high[:possible]])

print(ans)

Currency System in Geraldion

Source

Is there a banknote with the number 1? then all sums are possible. There isn't? ok, 1 is the smallest sum.

input()
a = list(map(int, input().split()))
print(-1 if min(a) == 1 else 1)

Simple Game

Source

We only need to beat Misha by one. Choose the largest available segment and choose the number in front of him. Look out for the corner case.

n,m = map(int, input().split())
print( 1 if n == 1 else m + 1 if n - m > m - 1 else m - 1 )

Elections

Source Just do what the statement says.

n, c = map(int, input().split())
score = [0] * n
for _ in range(c):
    city = list(map(int, input().split()))
    score[city.index(max(city))] += 1
print(score.index(max(score)) + 1)

Arrays

Source

Choose the k smallest numbers from the first array and the m largest numbers from the second one. A single comparison can tell us the answer.

input() # Useless first line

n,m = map(int, input().split())
a1 = list(map(int, input().split()))[:n]
a2 = list(map(int, input().split()))[-m:]

print("YES" if a1[-1] < a2[0] else "NO")

Order Book

Source This problem is similar to the previous one. Aggregate the buy and sell orders and sort in descending order. Take the s best ones (smallest sell orders and largest buy orders).

n,s = map(int, input().split())

sell = {}
buy = {}

for i in range(n):
    type, price, vol = input().split()
    price = int(price)
    vol = int(vol)
    if type == 'S':
        sell[price] = sell.get(price, 0) + vol
    else:
        buy[price]  = buy.get(price, 0) + vol
for order in sorted(sell.items(), reverse = True )[-s:]:
    print("S", order[0], order[1])
for order in sorted(buy.items(), reverse = True )[:s]:
    print("B", order[0], order[1])

Bear and Elections

Source

Steal a vote from the rival the with most votes until Limak has a sure win. A linear search for the best rival on each step or sorting the rivals' vote array is enough to solve the problem. A heap is also an option.

import heapq

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

limak = votes[0]
votes = [-x for x in votes[1:]]
heapq.heapify(votes)

ans = 0
while limak <= -votes[0]:
    vote = -heapq.heappop(votes)
    vote -= 1
    limak += 1
    ans += 1
    heapq.heappush(votes, -vote)
print(ans)

Multiplication Table

Source

$x$ can appear once on each row $i$ where the modulo between $x$ and $i$ is 0. Additionally, x will be present only if $x / i \leq n$ (we would count numbers outside of the table if we ignore this check).

n,x = map(int, input().split())
ans = 0
for i in range(1, n + 1):
    if x % i == 0 and x / i <= n:
        ans+=1
print(ans)

Conclusion

I've only used python for competitive programming and some scripting during my thesis project and I certainly do love how I don't need 20 ~ 30 lines of code only for reading and parsing input ♥.