# Код на ruby 2.2.3p173 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)
1) var n:integer; begin readln(n); if(n<0)then write(0) else write(n+4); end.
2) var a, b: integer; begin readln(a, b); if(a = b) then write(1) else write(0); end.
3) var a, b, c, s: integer; begin s := 0; readln(a, b, c); if(a >= 0) then s := s + 1; if(b >= 0) then s := s + 1; if(c >= 0) then s := s + 1; write(s); end.
4) var a, b, c, s: integer; begin s := 0; readln(a, b, c); if(a > 0) then s := s + a; if(b > 0) then s := s + b; if(c > 0) then s := s + c; write(s); end. 5) var a, b, c: integer; begin readln(a, b, c); if (a < b) then swap(a, b); if (a < c) then swap(a, c); if (b < c) then swap(b, c); write(a, ' ', b, ' ', c); end.
6) var n, k, s, i: integer; begin s := 0; readln(n, k); for i := 1 to n.ToString.Length do if(strtoint(n.ToString[i]) > k) then s := s + strtoint(n.ToString[i]); write(s); end.
7) var n, s, i: integer; begin s := 0; readln(n); for i := 1 to n do s := s + i; write(s); end.
9) var n, i: integer; begin readln(n); for i := n downto 1 do write(i); end.
12 var n, i: integer; s: real; begin s := 0; readln(n); for i := n downto 1 do s:=s+(0.1*i); write(s); end.
13) var n, i: integer; begin readln(n); for i := 1 to n do write(sqr(i), ' '); end.
14) var n, i, c: integer; begin i := 1; c := 0; readln(n); while (c <> n) do begin if(i mod 3 = 0) then begin write(i, ' ');c := c + 1; end; i := i + 1; end; end.
15) var n, i, a: integer; c: real; begin i := 1; c := 0; readln(n); for i := 1 to n do begin read(a); c := c + a; end; write(c / n) end.
16) var n, i, a: integer; begin i := 1; while true do begin read(a); if(a = 0) then break; n := n + 1; end; write(n) 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