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const fi='dulieu.txt';
fo='ketqua.txt';
var f1,f2:text;
a,b:array[1..100]of integer;
n,i:integer;
begin
assign(f1,fi); reset(f1);
assign(f2,fo); rewrite(f2);
n:=0;
while not eof(f1) do
begin
n:=n+1;
readln(f1,a[n],b[n]);
end;
for i:=1 to n do
begin
if (a[i]<b[i]) then writeln(f2,a[i])
else writeln(f2,b[i]);
end;
close(f1);
close(f2);
end.
const fi='hcn.inp';
fo='hcn.out';
var f1,f2:text;
a,b:array[1..100]of integer;
i,j,n:integer;
begin
assign(f1,fi); reset(f1);
assign(f2,fo); rewrite(f2);
n:=0;
while not eof(f1) do
begin
inc(n);
readln(f1,a[n],b[n]);
end;
for i:=1 to n do
writeln(f2,2*(a[i]+b[i]),' ',a[i]*b[i]);
close(f1);
close(f2);
end.
const fi='hcn.inp';
fo='hcn.out';
var f1,f2:text;
a,b:array[1..100]of integer;
i,j,n:integer;
begin
assign(f1,fi); reset(f1);
assign(f2,fo); rewrite(f2);
n:=0;
while not eof(f1) do
begin
inc(n);
readln(f1,a[n],b[n]);
end;
for i:=1 to n do
writeln(f2,2*(a[i]+b[i]),' ',a[i]*b[i]);
close(f1);
close(f2);
end.
const fi='tong.inp';
fo='tong.out';
var f1,f2:text;
a:array[1..100]of integer;
n,i,t:integer;
begin
assign(f1,fi); reset(f1);
assign(f2,fo); rewrite(f2);
readln(f1,n);
for i:=1 to n do
read(f1,a[i]);
t:=0;
for i:=1 to n do
t:=t+a[i];
writeln(f2,t);
close(f1);
close(f2);
end.
#include <bits/stdc++.h>
using namespace std;
long long a,b;
double tb;
int main()
{
freopen("dulieu.inp","r",stdin);
freopen("ketqua.out","w",stdout);
cin>>a>>b;
cout<<a<<" "<<b;
cout<<fixed<<setprecision(2)<<(a*1.0+b*1.0)/(2*1.0);
return 0;
}
Program HOC24;
var i,n: integer;
a: array[1..1000] of integer;
t: longint;
f1,f2: text;
const fi='DATA1.TXT';
fo='KQ1.TXT';
begin
assign(f1,fi);
assign(f2,fo);
reset(f1);
rewrite(f2);
readln(f1,n);
for i:=1 to n do read(f1,a[i]);
t:=0;
for i:=1 to n do if a[i] mod 2=0 then t:=t+a[i];
writeln(f2,t);
for i:=1 to n do if a[i] mod 5=0 then write(f2,a[i],' ');
close(f1); close(f2);
end.
uses crt;
const fi='input.txt';
fo='output.txt';
var f1,f2:text;
a,b:integer;
begin
assign(f1,fi); reset(f1);
assign(f2,fo); rewrite(f2);
readln(f1,a,b);
if (a=0) and (b=0) then writeln(f2,'Phuong trinh co vo so nghiem');
if (a<>0) then writeln(f2,-b/a:4:2);
if (a=0) and (b<>0) then writeln(f2,'Phuong trinh vo nghiem');
close(f1);
close(f2);
end.
4P+5O2-to>2P2O5
0,2---0,25-----0,1 mol
nP=6,2\31=0,2 mol
=>VO2=0,25.22,4=5,6l
=>mP2O5=0,1.142=14,2g
=>Vkk=5,6.5=28l