Question

Cho số tự nhiên n (2 ≤ n ≤ 10000), viết chương trình phân tích số n thành các thừa số nguyên tố.

Ví dụ: n = 12

12 | 2

6 | 2

3 | 3

1 |

Answer 1

uses crt;
var n,i,j,a,b,kt1,kt2,kt,k:longint;
begin
clrscr;
write('nhap n='); readln(n);
if n mod 2=1 then begin
a:=2;
b:=n-a;
kt:=0;
for k:=2 to trunc(sqrt(b)) do
if b mod k=0 then kt:=1;
if kt=0 then writeln(n,'=',a,'+',b)
else write(n,' khong the phan tich thanh tong nguyen to');
end;
if n mod 2=0 then begin
for i:=2 to n div 2 do
begin
a:=i;
b:=n-i;
kt1:=0;
kt2:=0;
for j:=2 to trunc(sqrt(a)) do
if a mod j=0 then kt1:=1;
for j:=2 to trunc(sqrt(b)) do
if b mod j=0 then kt2:=1;
if (kt1=0) and (kt2=0) then writeln(n,'=',a,'+',b);
end;
end;
readln;
end.

Answer 2

uses crt;
var n,kt,i,n1,i1:integer;
begin
clrscr;
write('nhap n='); readln(n);
write(n,'=');
n1:=n;
i1:=2;
repeat
while n1 mod i1<>0 do
i1:=i1+1;
write(i1);
n1:=n1 div i1;
if n1>1 then write('*');
until n1=1;
readln;
end.

Answer 3

uses crt;
var n,kt,i,n1,i1:integer;
begin
clrscr;
write('nhap n='); readln(n);
write(n,'=');
n1:=n;
i1:=2;
repeat
while n1 mod i1<>0 do
i1:=i1+1;
write(i1);
n1:=n1 div i1;
if n1>1 then write('*');
until n1=1;
readln;
end.