Даны натуральные числа m, n. получить все кратные им числа, меньшие m*n. сделать на паскале )

var m,n,l: integer;

begin

writeln('m,n');

readln(m,n);

if l mod n=0 and l mod m=0 and l< =(m*n) then 

write (l);

readln(l);

end.

 

 

вроде так

import math

import sys

def get_first_max(tree, idx, l, r, L, R):

if r <= L or R <= l:

  return -1

if l >= L and r <= R:

  return tree[idx]

m = (l + r) // 2

return max(get_first_max(tree, idx * 2 + 1, l, m, L, R), get_first_max(tree, idx * 2 + 2, m, r, L, R))

num = input()

k = int(input())

n = len(num)

N = 2**math.ceil(math.log2(n))

M1 = 10 ** 7

M2 = 10 ** 6

tree = [-1] * (2 * N)

for i in range(n):

tree[N - 1 + i] = int(num[i]) * M1  + M2 - i

for i in range(N - 2, -1, -1):

tree[i] = max(tree[2 * i + 1], tree[2 * i + 2])

i = 0

ans = ""

for _ in range(n - k):

maximum = get_first_max(tree, 0, 0, N, i, i + k + 1)

val = maximum // M1

pos = M2 - maximum % M1

ans += str(val)

k -= pos - i

i = pos + 1

if k == 0:

  ans += num[i:]

  break

print(ans)

Объяснение:

import math

import sys

def get_first_max(tree, idx, l, r, L, R):

 if r <= L or R <= l:

   return -1

 if l >= L and r <= R:

   return tree[idx]

 m = (l + r) // 2

 return max(get_first_max(tree, idx * 2 + 1, l, m, L, R), get_first_max(tree, idx * 2 + 2, m, r, L, R))

num = input()

k = int(input())

n = len(num)

N = 2**math.ceil(math.log2(n))

M1 = 10 ** 7

M2 = 10 ** 6

tree = [-1] * (2 * N)

for i in range(n):

 tree[N - 1 + i] = int(num[i]) * M1  + M2 - i

for i in range(N - 2, -1, -1):

 tree[i] = max(tree[2 * i + 1], tree[2 * i + 2])

i = 0

ans = ""

for _ in range(n - k):

 maximum = get_first_max(tree, 0, 0, N, i, i + k + 1)

 val = maximum // M1

 pos = M2 - maximum % M1

 ans += str(val)

 k -= pos - i

 i = pos + 1

 if k == 0:

   ans += num[i:]

   break

print(ans)