in ra tất cả các số nguyên tố có 2 chữ số pascal
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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.
#include <bits/stdc++.h>
using namespace std;
long long x;
//chuongtrinhcon
bool ktnt(long long n)
{
for (int i=2; i*i<=n; i++)
if (n%i==0) return(false);
return(true);
}
//chuongtrinhcon
bool ktso(long long n)
{
int a=n%10;
int b=n/10; b=b%10;
int c=n/100; c=c%10;
int d=n/1000; d=d%10;
if (a+b==c+d) return(true);
else return(false);
}
//chuongtrinhchinh
int main()
{
for (x=1000; x<=9999; x++)
if ((ktnt(x)==true) and (ktso(x)==true)) cout<<x<<" ";
return 0;
}
Program HOC24;
var i: 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;
Begin
for i:= 10 to 99 do if nt(i) and i mod 2=0 then write(i,' ');
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 bai_toan;
var
N, i, sum: integer;
begin
write('Nhap so N: ');
readln(N);
write('Cac uoc cua ', N, ' khong ke ', N, ' la: ');
for i := 1 to N - 1 do
if N mod i = 0 then
write(i, ' ');
writeln;
sum := 0;
for i := 1 to N - 1 do
begin
if N mod i = 0 then sum := sum + i;
end;
if sum = N then writeln(N, ' la so hoan hao')
else writeln(N, ' khong phai la so hoan hao');
writeln;
writeln('Tat ca so hoan hao trong pham vi 1 -> ', N, ' la:');
for i := 1 to N do
begin
sum := 0;
for j := 1 to i - 1 do
begin
if i mod j = 0 then sum := sum + j;
end;
if sum = i then writeln(i);
end;
readln;
end.
#include <bits/stdc++.h>
using namespace std;
int diaphuong(int n) {
if (n<2) {
return 0;
}
int squareRoot = (int) sqrt(n);
int i;
for(int i=2; i<=squareRoot; i++) {
if(n%i==0) {
return 0;
}
}
return 1;
}
int main() {
int i, N;
cin >> N;
for(int i=0; i<=N; i++) {
if(diaphuong(i)) {
cout << i << endl;
}
}
}
Chúc bn học tốt!
#include <bits/stdc++.h>
using namespace std;
const long long maxint=1000000;
bool a[maxint];
long long n,i,j;
int main()
{
cin>>n;
for (i=2; i<=n;i++)
a[i]=true;
for (i=2;i<=n; i++)
if (a[i]==true)
{
for (j=i*i;j<=n;j+=i)
a[j]=false;
}
for (i=2; i<=n; i++)
if (a[i]==true) cout<<i<<" ";
return 0;
}
uses crt;
var i,j,kt:integer;
begin
clrscr;
for i:=10 to 99 do
begin
kt:=0;
for j:=2 to trunc(sqrt(i)) do
if i mod j=0 then kt:=1;
if kt=0 then write(i:4);
end;
readln;
end.