Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
uses crt;
var a:array[1..10000]of integer;
i,n,kt,j:integer;
begin
clrscr;
readln(n);
for i:=1 to n do readln(a[i]);
for i:=1 to n do
if a[i]>1 then
begin
kt:=0;
for j:=2 to trunc(sqrt(a[i])) do
if a[i] mod j=0 then kt:=1;
if kt=0 then write(a[i]:4);
end;
readln;
end.
uses crt;
var n,t:integer;
{------------------kiem-tra-so-nguyen-to----------------------------}
function ktnt(x:integer):boolean;
var i:integer;
kt:boolean;
begin
kt:=true;
for i:=2 to trunc(sqrt(x)) do
if x mod i=0 then
begin
kt:=false;
break;
end;
if kt=true then ktnt:=true
else ktnt:=false;
{----------------------chuong-trinh-chinh------------------------}
begin
clrscr;
t:=0;
repeat
write('Nhap n='); readln(n);
if (n mod 2=0) then t:=t+n;
until ktnt(n)=false;
writeln(t);
readln;
end.
uses crt;
var n,i,dem,j,t:integer;
kt:boolean;
begin
clrscr;
readln(n);
t:=0;
for i:=2 to n do
begin
kt:=true;
for j:=2 to i-1 do
if i mod j=0 then kt:=false;
if kt=true then
begin
write(i:4);
t:=t+i;
end;
end;
writeln;
writeln(t);
readln;
end.
program abcdef;
uses Crt;
var
k, lowerLimit, upperLimit, i, j, reversed, temp, remainder: integer;
isPrime, isPalindrome: boolean;
begin
clrscr;
write('Nhap so chu so k (1<=k<=9): ');
readln(k);
lowerLimit := 1;
for i := 1 to k - 1 do
lowerLimit := lowerLimit * 10;
upperLimit := lowerLimit * 10 - 1;
writeln('Cac so nguyen to doi xung co ', k, ' chu so la:');
for i := lowerLimit to upperLimit do
begin
// Kiểm tra số nguyên tố
isPrime := True;
if i < 2 then
isPrime := False
else
for j := 2 to trunc(sqrt(i)) do
if i mod j = 0 then
beginisPrime := False;break;end;
// Kiểm tra số đối xứng
if isPrime then
begin
reversed := 0;
temp := i;
while temp <> 0 dobeginremainder := temp mod 10;
reversed := reversed * 10 + remainder;
temp := temp div 10;
end;
isPalindrome := (i = reversed);
if isPalindrome then
writeln(i);
end;
end;
readln;
end.
uses crt;
var i,n,t,j,kt:integer;
begin
clrscr;
readln(n);
t:=0;
for i:=2 to n do
if n mod i=0 then
begin
kt:=0;
for j:=2 to trunc(sqrt(i)) do
if i mod j=0 then kt:=1;
if kt=0 then t:=t+i;
end;
write(t);
readln;
end.
#include <bits/stdc++.h>
using namespace std;
long long n,i,x,t,dem,j;
int main()
{
cin>>n;
x=n;
t=0;
while (n>0)
{
t=t+n%10;
n=n/10;
}
cout<<t;
dem=0;
for (i=2; i<=x; i++)
{
bool kt=true;
for (j=2; j*j<=i; j++)
if (i%j==0) kt=false;
if (kt==true) dem++;
}
cout<<dem;
return 0;
}
def kiem_tra_nguyen_to(n):
if n < 2:
return False
for i in range(2, int(n ** 0.5) + 1):
if n % i == 0:
return False
return True
def kiem_tra_nguyen_to_cung_nhau(m, n):
if kiem_tra_nguyen_to(m) and kiem_tra_nguyen_to(n):
return True
return False
M = int(input("Nhập số M: "))
N = int(input("Nhập số N: "))
if kiem_tra_nguyen_to_cung_nhau(M, N):
print("Hai số", M, "và", N, "là hai số nguyên tố cùng nhau.")
else:
print("Hai số", M, "và", N, "không phải là hai số nguyên tố cùng nhau.")
câu 1
Program Nguyen_to;
Var n,i:integer;
Function NT(n:integer):Boolean;
Var ok: Boolean;
i: integer;
Begin ok:=true;
for i:=2 to n-1 do if (n mod i)= 0 then ok:=ok and false;
if n < 2 then NT:=false else NT:=ok;
End;
Begin Write('Nhap n: ');
Readln(n); i:=n;
Repeat i:=i+1;
Until NT(i);
Write('So nguyen to nho nhat lon hon ',n, 'la: ',i);
Readln End.
câu 2
uses crt;
const so: set of char=['0','1','2','3','4','5','6','7','8','9'];
var a:array[1..100] of integer;
st,b:string;
c,l,i,n,j:integer;
s, Max: integer;
begin clrscr;
write('Nhap xau:');
readln(st);
l:=length(st);
i:=1;
n:=0;
repeat if (st[i] in so) then begin b:='';
repeat b:=b+st[i];
inc(i);
until (not(st[i] in so)) or (i>l);
inc(n);
val(b,a[n],c);
end;
inc(i);
until i>l;
Max:=a[1];
for i:=2 to n do If Max<A[i] Then Max:=A[i];
Writeln('Phan tu lon nhat cua mang:', Max);
readln;
end.
loi nay la sao a