Pascal finds the divisor and divisor of a number

program divided;var n,i,s:integer;begin s:=0;counters, which are used to count about a number of numbers; readln(n);Enter the original number; for i:

1 to n do if n mod i=0 then begin s:=s+1; write(i:5); end;Let i start looking from 1 to n, and once you find that it is divisible by n, it is the divisor of n, and the counter is further one; writeln; write(s);Output the approximate number in the next row (i.e. the final result of the counter); readln;Wait for the user to enter a carriage return to exit the program.

end.

Up to 2 (n-1)-1 , when n is larger there will be a lot of repeating divisors (e.g. *3 this is repeated).

That is, it needs to be deduplicated!

Use this number firstRepresentation of the product of equal prime numbers, for example.

Then add one to the exponents of each prime number and multiply them to be the approximate number of this number, such as (3+1)*(1+1)=4*2=8, which means that 24 has 8 divisors.

varn,i,ans:integer;

beginreadln(n);ans:=0;

Read in, and set the counter to 0

fori:=1ton

doifnmod

i=0then

inc(ans);

Cycle from 1 to n.

Find a respectful one, and add a number of his factors to the counter, and I will accompany you.

writeln(ans);

The most stupid output result.

end.

function add(f:longint):longint;

varii,ss:longint;

beginfor ii:=1 to n do

if f mod ii=0 then ss:=ss+ii;

add:=ss;

end;This is the sum of the non-repeating factor.

Example: add(18)=39

Note: The upstairs program is the same as this one, but I am better at it, and the time and space complexity is higher than mine).

dividing line**********==function add(f:longint):longint;

varii,ss:longint;

beginfor ii:=2 to trunc(sqrt(f)) dowhile f mod ii=0 do beginss:=ss+ii;

f:=f div ii;

end;add:=ss;

end;This is the sum of the prime factors after breaking them down.

Example: add(18)=8

Dividing line**********== Note: I didn't compile the above program, so it said...

Find the greatest common divisor of two integers (tossing and dividing).

The procedure is as follows: varm, n : integer;

function fac(i, j:integer):integer;

beginif j=0 then begin fac:=i;exit;end;

fac:=fac(j ,i mod j);

end;begin

readln(n, m);

writeln(fac(n,m));

end.or var

a,b:integer;

procedure qw(a,b:integer);

varv:integer;

beginif a mod b=0 then writeln(b)else if b mod a=0 then writeln(a)else

beginv:=a;

a:=a mod b;

b:=b mod v;

qw(a,b);

end;end;

beginreadln(a,b);

qw(a,b);

readln;

end.

Put the 1 in line 3...n read 1...10 (for a fixed value, all variables under var will think your array is quite large, more than the computer can handle).

Toss and turn divides, the big one is divided by the small one.

I would like to add that in order to get a full score, you need to use high-precision calculations.