ответ:
uses graphabc;
var xx,yy,u: integer;
procedure growl(x,y,r: integer);
var i: byte;
begin
for i: =1 to r do
begin
circle(x,y,i);
sleep(5);
end;
end;
procedure growr(x,y: integer);
var i: byte;
begin
for i: =1 to 20 do
begin
ellipse(x-5,y-(i),x-(i*3),y+(i));
ellipse(x+5,y-(i),x+(i*3),y+(i));
ellipse(x-(i),y-5,x+(i),y-(i*3));
ellipse(x-(i),y+5,x+(i),y+(i*3));
sleep(3);
end;
end;
procedure fall;
var r,prer,i: byte;
x,y: integer;
begin
for i: =1 to 10 do
begin
x: =random(160)-80+windowwidth div 2;
prer: =0;
r: =random(10)+10;
for y: =80+windowheight div 2 to windowheight+30 do
begin
setpencolor(clwhite);
circle(x,y-1,prer);
setpencolor(clblack);
circle(x,y,r);
prer: =r;
sleep(3);
end;
end;
end;
procedure growd (x,y: integer);
var i: byte;
begin
for i: =20 downto 5 do
begin
clearwindow;
setbrushcolor(clgreen);
ellipse(x-5,y-(20),x-(60),y+(20));
ellipse(x+5,y-(20),x+(60),y+(20));
ellipse(x-(20),y-5,x+(20),y-(60));
ellipse(x-(20),y+5,x+(20),y+(60));
setbrushcolor(rgb(255,255-((20-i)*10),255-((20-i)*;
circle(x-i,y-i,i);
circle(x+i,y-i,i);
circle(x+i,y+i,i);
circle(x-i,y+i,i);
sleep(5);
end;
end;
var j: integer;
begin
repeat
setbrushcolor(clgreen);
growr(windowwidth div 2,windowheight div 2);
setbrushcolor(clwhite);
for j: =1 to 360 do
if j mod 10=0 then
begin
xx: =round(windowwidth div 2+20*cos(pi*(u+(j*10))/180));
yy: =round(windowheight div 2+20*sin(pi*(u+(j*10))/180));
growl(xx,-j) div 10));
end;
sleep(100);
fall;
growd (windowwidth div 2,windowheight div 2);
sleep(100);
until false;
end.
объяснение:
№3
#include <iostream>
using namespace std;
int main() {
float y, x = 0.2;
if (x < 1)
y = abs(x);
else if (x > 4)
y = 1 + sqrt(x);
else if (0 <= x <= 4)
y = 3.5 + 2 * x;
//наш X войдет в 2 промежутка, выведем первое выполненное условие
cout << y;
return 0;
}
№4
#include <iostream>
#include <cmath>
using namespace std;
int main() {
int x = 3;
float y;
if (x < -2)
y = pow((x+2),2);
else if (-2 <= x <= 0)
y = x+2;
else if (x>= 0)
y = pow((x+2),3);
cout << y;
return 0;
}
