#include <bits/stdc++.h>

using namespace std;

long long n,i,dem,t;

int main()

{

cin>>n;

dem=0;

t=0;

for (i=1; i<=n; i++)

if (i%2==0) 

{

t=t+i;

dem++;

}

cout<<t<<" "<<dem;

return 0;

}

const fi='songuyen.txt'

fo='kluan.txt'

var f1,f2:text;

n:integer;

begin

assign(f1,fi); reset(f1);

assign(f2,fo); rewrite(f2);

readln(f1,n);

if n mod 2=0 then writeln(f2,'La so chan')

else writeln(f2,'La so le');

close(f1);

close(f2);

end.

uses crt;

var n,i,o,d:integer;

function ktnt(n:integer): integer;

var i,d:integer;

begin

d:=0;

for i:=1 to sqrt(n) do

if (n mod i=0) then d:=d+1;

if d=2 then ktnt=0

else ktnt=1;

end;

begin

readln(n);

writeln(' so nguyen to be hon hoac bang n la'); {a}

for i:=1 to n do

if ktnt(i)=0 then writeln(i);

writeln('so nguyen to nho nhat khong be hon n');

o:=n;

while o>0 do

begin

if ktnt(o)=0 then

begin

write(o);

break;

end;

o:=o+1;

end;

writeln('cặp số nguyên tố là hai số nguyên lẻ liên tiếp nhỏ hơn hoặc bằng n');

o:=0;

o:=1;

d:=0;

for i:=o+2 to n do

begin

if ktnt(i)=0 then

begin

d:=d+1;

write(i,' ');

if d<2 then continue;

end;

d:=0;

writeln;

end;

readln;

end.

uses crt;

const fi='songuyen.inp'

var f1:text;

a:array[1..100]of integer;

i,n,t:integer;

begin

clrscr;

assign(f1,fi); reset(f1);

n:=0;

while not eoln(f1) do 

  begin

n:=n+1;

read(f1,a[n]);

end;

t:=0;

for i:=1 to n do 

 if a[i] mod 2=0 then t:=t+a[i];

writeln(t);

readln;

end.

ai giup e voi dc k ah:( đang thi ạ

alo mng ơi=(((

const fi='songuyen.txt'

fo='soduong.out'

var f1,f2:text;

a:array[1..100]of integer;

i,n:integer;

begin

assign(f1,fi); reset(f1);

assign(f2,fo); rewrite(f2);

n:=0;

while not eof(f1) do 

  begin

inc(n);

read(f1,a[n]);

end;

for i:=1 to n do 

  if a[i]>0 then write(f2,a[i]:4);

close(f1);

close(f2);

end.

Program HOC24;

const fi='SONGUYEN.TXT';

fo='KETQUA.TXT';

var n,i,d:integer;

a: array[1..32000] of integer;

function nt(x: integer): boolean;

var j: integer;

begin

nt:=true;

if (x=2) or (x=3) then exit;

nt:=false;

if (x=1) or (x mod 2=0) or (x mod 3=0) then exit;

j:=5;

while j<=trunc(sqrt(x)) do

begin

if (x mod j=0) or (x mod (j+2)=0) then exit;

j:=j+6;

end;

nt:=true;

end;

procedure ip;

begin

assign(f,fi);

reset(f);

readln(f,n);

for i:=1 to n do read(f,a[i]);

close(f);

end;

procedure out;

begin

assign(f,fo);

rewrite(f);

d:=0;

for i:=1 to n do

if nt(a[i]) then inc(d);

write(f,d);

close(f);

end;

begin

ip;

out;

end.

#include <bits/stdc++.h>

using namespace std;

long long m,n,t,i;

int main()

{

freopen("dulieu.inp","r",stdin);

freopen("ketqua.out","w",stdout);

cin>>m>>n;

t=0;

for (i=m; i<=n; i++)

if (i%2!=0) t+=i;

cout<<t;

return 0;

}

1.

Program bai1;

uses crt;

var n,i: integer;

begin

clrscr;

write('nhap so n ='); readln(n);

for i := 1 to n do

if n mod i = 0 then

write(i :2 , ' , ');

readln;

end.

2.

program bai2;

uses crt;

var i,n: integer;

begin

clrscr;

write('nhap so n ='); readln(n);

for i: 1 to n do

if i mod 2 = 0 then

write(i: 3 ,',');

readln;

end.