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Câu 1:
program Tinhtong:
var
n, i, sum: integer;
begin
write('Nhap n: ');
readln(n);
sum := 0;
for i := 2 to n do
begin
if i mod 2 = 0 then
begin
sum := sum + i;
end;
end;
writeln('Tong cac so chan tu 2 toi ', n, ' la ', sum);
readln;
end.
Câu 2:
program Tinhtong:
var
n, i, sum: integer;
begin
write('Nhap n: ');
readln(n);
sum := 0;
for i := 1 to n do
begin
if i mod 2 = 1 then
begin
sum := sum + i;
end;
end;
writeln('Tong cac so le tu 1 toi ', n, ' la ', sum);
readln;
end.
1:
uses crt;
var n,i,t:integer;
begin
clrscr;
readln(n);
t:=0;
for i:=1 to n do
t:=t+i*i;
write(t);
readln;
end.
2
program bt2;
var i,n,t:integer;
begin
readln(n);
s:=0;
for i:=1 to n do
if i mod 2 = 1 then s:=s+i;
readln;
end.
Câu 6:
uses crt;
var n,i:integer;
begin
clrscr;
readln(n);
for i:=1 to n do
if n mod i=0 then write(i:4);
readln;
end.
5:
uses crt;
var n,i,dem:integer;
begin
clrscr;
readln(n);
dem:=0;
for i:=0 to n do
if i mod 2=1 then
begin
write(i:4);
dem:=dem+1;
end;
writeln;
writeln(dem);
readln;
end.
Câu 1:
uses crt;
var s,i,n:integer;
begin clrscr;
s:=0;
write('Nhap n: ');readln(n);
for i:=1 to n do
if i mod 2 <> 0 then inc(s,i);
write('Tong cac so le tu 1 den ',n,' la: ',s);
readln
end.
Câu 2:
uses crt;
var s,i,n:integer;
begin clrscr;
s:=0;
write('Nhap n: ');readln(n);
for i:=1 to n do
if i mod 2 = 0 then inc(s,i);
write('Tong cac so le tu 1 den ',n,' la: ',s);
readln
end.
Bài 1:
uses crt;
var i,s:integer;
begin
clrscr;
s:=0;
for i:=10 to 50 do
if i mod 2=0 then s:=s+i;
writeln(s);
readln;
end.
Bài 2:
uses crt;
var a,i,n:integer;
begin
clrscr;
write('Nhap n='); readln(n);
a:=0;
for i:=1 to n do
a:=a+i*(i+2);
writeln(a);
readln;
end.
1.
program tinhtong;
uses crt;
var s,i,n:integer;
begin
s:=0;
for i:=1 to 10 do s:=s+1;
writeln ('tong cua s=',s);
readln
end.
2.
program tinhtong;
uses crt;
var i,L,s:integer;
begin
write('nhap gia tri L='); readln(L);
s:=0;
for i:=1 to L do s:=s+1;
writeln('tong cua s=',s);
readln
end.
uses crt;
var i,n,t:integer;
begin
clrscr;
readln(n);
t:=0;
for i:=1 to n do t:=t+i;
writeln(t);
readln;
end.
Help
#include <bits/stdc++.h>
using namespace std;
double t;
int i,n;
int main()
{
cin>>n;
t=0;
for (i=1; i<=n; i++)
t=t+(1*1.0)/(i*1.0);
cout<<fixed<<setprecision(2)<<t;
return 0;
}