Today's problems are very simple but allowed me to remember some things I had not used in a while; I even had a hard time reimplementing a solution to the Kefa and Park problem using C++.

As a bonus, I found out that XCode is a pretty decent C++ IDE for competing, so I guess I'll be using it more from now on. If you find yourself wanting to use it but can't get <bits/stdc++.h> to work, it is because XCode isn't using GCC anymore, so you must add the header file by yourself to XCode's header directory route. See more here.

Apple and Orange

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Check whether the falling location of each apple/orange is between the bounds of the house.

s, t = list(map(int, input().split()))
a, b = list(map(int, input().split()))
m, n = list(map(int, input().split()))
apples = [x + a for x in list(map(int, input().split()))]
oranges = [x + b for x in list(map(int, input().split()))]

print(sum(s <= i <= t for i in apples))
print(sum(s <= i <= t for i in oranges))

Day of the Programmer

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The only special case happens when the given year is $1918$ , compute it and the rest is a piece of cake. The answer is September 12 on leap years and 13 in non-leap ones no matter the type of calendar used.

def leap(y):
    if y <= 1917:
        return y % 4 == 0
    else:
        return y % 400 == 0 or (y % 4 == 0 and y % 100 != 0)


y = int(input())
print('26.09.1918' if y == 1918 else '{}.09.{}'.format(12 if leap(y) else 13, y))

Divisible Sum Pairs

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The simple $O(n^2)$ is enough to solve the problem due to its constraints. If you want, spice the solution up using overly complicated functions just as I did.


from itertools import combinations

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

print(len([(x, y) for x, y in combinations(enumerate(a), 2) if x[0] < y[0] and (x[1] + y[1]) % k == 0]))

Grading Students

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We can find $n$ 's closest multiple of five using the modulo operator ( $n + 5 - (n % 5)$ ). To only round the grade up when the difference is less than 3 we check that the modulo is greater or equal to 3.

#include<bits/stdc++.h>

using namespace std;

int main(){
	std::ios::sync_with_stdio(false);
	int n, g;
	cin >> n;
	while(n--){
		cin >> g;
		cout << (g < 38 ? g : g + (g % 5 >= 3 ? 5 - g % 5 : 0)) << endl;
	}
	return 0;
}

Migratory Birds

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Count the number of times each bird has been seen and return the id of the first bird with the maximum amount of appearances. C++'s max_element does just this, so my solution is based on it.

#include<bits/stdc++.h>

using namespace std;

int a[5], n;
int main(){
    std::ios::sync_with_stdio(false);
    cin >> n;
    for(int i = 0, x; i < n; i++) cin >>x, a[x - 1]++;
    cout << (max_element(a, a + 5) - a) + 1 << endl;
    return 0;
}

Some Last Words

Don't fear using pointer arithmetic.