program ;
var
i, a: longint;
x,
u, j,
n, m : longint;
function fdiv (a, left : longint) : longint;
var
d,
i : longint;
begin
d : = 1;
for i : = left to trunc (sqrt (a)) do
if (a mod i = 0)
then
begin
d : = i;
break;
end;
if (d < > 1)
then
fdiv : = d
else
fdiv : = a;
end;
function g (n, u : longint) : longint;
var
i : longint;
begin
i : = 0;
while (n mod u = 0)
do
begin
inc (i);
n : = n div u;
end;
g : = i;
end;
begin
assign (input, 'input.txt');
assign (output, 'output.txt');
reset (input);
read (a);
close (input);
m : = 0;
n : = 1;
i : = fdiv (a, 2);
u : = 2;
while (a > 1)
do
begin
n : = n * i;
j : = 0;
while (a mod i = 0)
do
begin
inc (j);
a : = a div i;
end;
if (j > m)
then
begin
m : = j;
u : = i;
end;
i : = fdiv (a, i + 2);
end;
x : = 1;
while (g (n * x, u) * n * x < m)
do
inc (x);
n : = n * x;
rewrite (output);
write (n);
close (output);
end.
def zadanie(jump1, jump2)
min = jump1
for i in 1..1000
next if (i*jump1)%jump2 == 0 or (i*jump2)%jump1 == 0
if min > (i*jump1)%jump2
min = (i*jump1)%jump2
p [min, "#{i} * #{jump1} - #{i*jump1/jump2} * #{jump2}"]
end
if min > (i*jump2)%jump1
min = (i*jump2)%jump1
p [min, "#{i} * #{jump2} - #{i*jump2/jump1} * #{jump1}"]
end
end
return min
end
# # Примеры применения
p zadanie(33, 55)
Вывод
[22, "1 * 55 - 1 * 33"]
[11, "2 * 33 - 1 * 55"]
11
Т.е. минимум при 2 * 33 - 1 * 55 = 11