Решение олимпиадных задач по информатике: Антон и арбузы

Сегодня решаем олимпиадную задачу по информатике школьного этапа 2022 года. Задача: «Антон и арбузы». 9 — 11 класс.

n, m, d = int(input()), int(input()), int(input())

field = []
for _ in range(n):
    field.append([0] * m)

for _ in range(d):
    xi, yi = int(input()), int(input())
    for x in range(xi):
        for y in range(yi):
            field[x][y] += 1

largest = 0
for row in field:
    for elem in row:
        if elem > largest:
            largest = elem

count = 0
for row in field:
    for elem in row:
        if elem == largest:
            count += 1

print(count, largest)

n, m, d = int(input()), int(input()), int(input())

field = [[0] * m for _ in range(n)]

for _ in range(d):
    xi, yi = int(input()), int(input())
    for x in range(xi):
        for y in range(yi):
            field[x][y] += 1

largest = max([max(row) for row in field])
count = sum([row.count(largest) for row in field])

print(count, largest)

var
  x,y,xi,yi,i,n,m,d,count,largest:integer;
  field:array[,] of integer;
  
begin
  read(n);
  read(m);
  read(d);
  field := new integer[n, m];
  for i := 1 to d do
  begin
    read(xi);
    read(yi);
    for x := 0 to xi - 1 do
      for y := 0 to yi - 1 do
        field[x, y] := field[x, y] + 1
  end;
  
  largest := 0;
  for x := 0 to n - 1 do
    for y := 0 to m - 1 do
      if field[x, y] > largest then
        largest := field[x, y];
      
  count := 0;
  for x := 0 to n - 1 do
    for y := 0 to m - 1 do
      if field[x, y] = largest then
        count := count + 1;
      
  write(count, ' ', largest);  
end.