what is the fastest way to find the divisors of a number?

In the task of problem #1118 the triviality(1) was not defined.
test #4 has 1 as i.
after putting triviality(1) be equal to 0, i've gotten AC.
So, in my opinion, this moment should be fixed in the task.

P.S.
sorry for my bad English

A tried a lot but wa#3 always shown!
I find the largest prime in the range,
If there isnt so I just use brute force.
I know that triviality(1)=0 ... but Why WA???

const
a:array[1..168] of longint=(
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,
73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167
,173,
179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271
,277,281,
283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397
,401,409,
419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521
,523,541,
547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647
,653,659,
661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787
,797,809,
811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929
,937,941,947,953,967,971,977,983,991,997);

var
  m,n,minn,i,k:longint;
  y,min:real;

function prime(x:longint):boolean;
var
  j:longint;
begin
  for j:=1 to 168 do
  begin
    if a[j]>trunc(sqrt(x)) then
    begin
      prime:=true;
      exit;
    end;
    if x mod a[j]=0 then
    begin
      prime:=false;
      exit;
    end;
  end;
end;

function count(x:longint):longint;
var
  s,i:longint;
begin
  s:=1;
  for i:=2 to x-1 do
  if x mod i=0 then s:=s+i;
  count:=s;
end;

begin
  readln(n,m); if (n=1) or (m=1) then begin writeln(1); halt; end;
  if n=m then begin writeln(n); halt; end;

  for i:=m downto n do
  if prime(i) then begin writeln(i); halt; end;

  min:=2147483647;
  for i:=n to m do
  begin
    y:=count(i)/i;
    if y<min then begin min:=y; minn:=i; end;
  end;
  writeln(minn);
end.

I get too many WA with this problem (1118). In test number 3
I just use a simple algorithm, if lower number is 1 then print 1
else go in a loop from j to l, I mean from the max number until the min. number and start to calculate the nontrivial number of every number until I find a prime, if I find a prime then I print the least nontrivial number until this number (including the first prime that I found)
here is my code:

#include <stdio.h>
#include <math.h>
double s,res=2147483647;
int r=1,rs;

calc(int n,int p){
    int i,k=1;
    p++;
    for(i=0;i<p;i++) k*=n;
    k=(k-1)/(n-1);
    r*=k;
}

fact(int n){
    int p=0,i=3,l=sqrt(n)+1;
    double t=n;
    while(n%2==0){
         n/=2;
         p++;
    }
    if(p>0){
        calc(2,p);
        p=0;
    }
  while(i<l){
     while(n%i==0){
         n/=i;
         p++;
     }
     if(p>0){
         calc(i,p);
         p=0;
     }
     i+=2;
  }
    if(n>1) calc(n,1);
    r-=t;
    s=r/t;
    if(s<res){
        rs=t;
        res=s;
    }
  r=1;
}

is_prime(n){
    int i=3,l=sqrt(n)+1;
    if(n%2==0) return 0;
    for(i=3;i<l;i+=2)
     if(n%i==0) return 0;
     return 1;
}

int main(){
    int l,i,j;
    scanf("%d %d",&l,&j);
    if(l==1) printf("1\n");
    else{
     for(i=j;i>=l;i--) {
       fact(i);
       if(is_prime(i)) break;
   }
  printf("%d",rs,res);
  }
  return 0;
}

if someone can help me I would appreciate very much!!
thanks

#include<iostream.h>
#include<math.h>
#include<limits.h>
main()
{
    long D,U,i,sum,crno,minN,j;
    double min,ratio,nouse;
    int flag=0;
    cin>>D>>U;
    min=1000000;
    minN=D;
    for(long t=1000000;t>=2;t--)
    {
        U=t;
    if(D==1)
        cout<<'0';
    else
    {
        for(i=U;i>=D;i--)
        {
            sum=1;

            for(j=2;j<=sqrt((double)i);j++)
            {
                if(i%j==0)
                {
                    sum+=j+i/j;
                }
            }
            if(i%(j-1)==0)
                sum-=j;

            ratio=double(sum)/(double)i;
            if(ratio<=min)
            {

                min=ratio;
                minN=i;
                if(sum==1)
                {
                    for(long k=U;k>=D;k--)
                    {

                        if(modf(sqrt((double)k),&nouse)==0)
                        {
                            ratio=(1+sqrt((double)k))/(double)k;
                            if(ratio>min)
                                break;
                            else
                            {

                                min=k;
                                minN=k;
                                break;
                            }
                        }
                    }
                    break;
                }

            }
        }
        cout<<minN;
    }
    }
}

How could somebody get AC in 0.01 sec !!!
and uses mem lower than 50K
plz. show your algorithm

This is my program.
I will really appreciate your help.

program nontrivial;
var
  m,n,i,j,k,s,mi:longint;
  min:real;
begin
  readln(m,n);
  min:=1e+10; mi:=0;
  for k:=n downto m do
  begin
    s:=0;
    for j:=1 to k-1 do
      if k mod j=0 then inc(s,j);
    if s=1 then begin writeln(k); halt end;
    if s/k<min then begin min:=s/k; mi:=k end;
  end;
  writeln(mi);
end.

var sol,gam,i,j,k,m,n:longint;
    min,min1:real;

procedure readdata;
  begin
    readln (n,m);
  end;

procedure solve;
  begin
    min:=n;
    if n=1 then
     begin
       writeln (1);
       halt;
     end;
    min1:=m;
    for i:=n to m do
      begin
        k:=i div 2;
        for j:=2 to k do
          if i mod j = 0 then
            gam:=gam+j;
        min:=gam/i;
        if min1>min then
          begin
            sol:=i;
            min1:=min;
          end;
        gam:=1;
      end;
  end;

procedure print;
  begin
    writeln (sol);
  end;

begin
  readdata;
  solve;
  print;
end.

{$N+}
var
  t1,t2:extended;
  num,i,j,k,divisor:longint;
  s:string;
function triv(n:longint):extended;
var i,j, counter:longint;
begin
  j:=round(sqrt(n)); counter:=0;
  divisor:=0;
  for i:=1 to j do
    if n mod i=0 then begin
      inc(counter,i);
      inc(divisor);
    end;
  if counter>1 then
    for i:=j+1 to n div 2 do
      if n mod i =0 then
      begin
        inc(counter,i);
        inc(divisor);
      end;
  triv:=counter/n;
end;
procedure die(s:string);
begin
  writeln(s); halt(0);
end;

begin
  writeln('Input I and J');
  read(i,j);
  if i<=0 then die('wrong input');
  if j>=1000000 then die('wrong input');
  if i=j then begin
    str(i,s);
    die(s);
  end;
  if i=1 then begin
    str(1,s);
    die(s);
  end;
  t1:=maxlongint;
  for k:=j downto i do
  begin
    t2:=triv(k);
    if t2<t1 then begin
      t1:=t2;
      num:=k;
      if divisor=1 then begin str(k,s); die(s) end;
    end;
  end;
  writeln(num);
end.

const
ss:array[1..169] of longint=(
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,
73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167
,173,
179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,27
1,277,281,
283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,39
7,401,409,
419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,52
1,523,541,
547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,64
7,653,659,
661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,78
7,797,809,
811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,92
9,937,941,
947,953,967,971,977,983,991,997,0);
var
  a:array[1..169] of longint;
  i,j,k,x,y:longint;
function judge(m:longint):boolean;
var
  i,j,k,n,q:longint;
begin
  n:=m;
  fillchar(a,sizeof(a),0);
  judge:=false;
  i:=1;
  while (n>1)and(i<169) do begin
    while n mod ss[i]=0 do begin
      inc(a[i]);
      n:=n div ss[i];
    end;
    inc(i);
  end;
  if n>1 then begin ss[169]:=n;a[169]:=1;end;
  k:=1;
  for i:=1 to 169 do if a[i]>0 then begin
    q:=1;
    for j:=1 to a[i] do q:=q*ss[i]+1;
    if m*2<k*q then exit;
    k:=k*q;
  end;
  judge:=true;
end;

begin
  readln(x,y);
  for i:=x to y do
    if judge(i) then begin writeln(i);break;end;
end.

var
  z,y,x,a,b,tt,i,j,k,p,q,r:longint;
  ss:array[1..10000] of longint;
  s:string;
begin
  tt:=1;ss[1]:=2;
  for i:=3 to trunc(sqrt(1000000)) do begin
    for j:=1 to tt do if (i mod ss[j]=0)or(ss[j]>trunc(sqrt(i))) then
break;
    if i mod ss[j]<>0 then begin
      inc(tt);
      ss[tt]:=i;
    end;
  end;
  readln(a,b);
  for y:=a to b do begin
    q:=1;
    x:=y;
    r:=1;
    k:=1;
    while (x<>1 )and(k<>tt) do begin
      for j:=k to tt do
        if x mod ss[j]=0 then break;
      k:=j;
      if x mod ss[j]=0 then begin
        z:=1;
        p:=1;
        while x mod ss[j]=0 do begin
          z:=z*ss[j];
          p:=z+p;
          q:=q*ss[j];
          x:=x div ss[j];
        end;
        r:=r*p;
        if (r>=2*q) then break;
      end;
    end;
    if x<>1 then begin r:=r*(x+1);q:=y;end;
    if r<2*q then begin

      writeln(y);
      halt;
    end;
  end;
end.

type
  c=record
    d:longint;
    h:real;
end;

var sol,gam,i,j,k,l,m,n,b:longint;
    a:array [1..10000] of c;
    min:real;

procedure readdata;
  begin
    readln (n,m);
  end;

procedure check;
  begin
    if n=1 then
      begin
        writeln (0);
        halt;
      end;
  end;

procedure solve;
  begin
    for i:=n to m do
      begin
        k:=i div 2;
        for j:=1 to k do
          if i mod j = 0 then
            gam:=gam+j;
        inc (b);
        a[b].d:=i;
        a[b].h:=gam/i;
        gam:=0;
        if i>=m then
          break;
      end;
    min:=a[1].h;
    for i:=1 to b do
      if min>a[i].h then
        begin
          min:=a[i].h;
          sol:=a[i].d;
        end;
  end;

procedure print;
  begin
    writeln (sol);
  end;

begin
  readdata;
  check;
  solve;
  print;
end.

what exactly means this problem (Nonetrivial numbers1118):
  1.to find the smallest trivial number
 or
  2.to find the number in the input range that have the smallest
    trivialaty?
if it is 1 then what is trivial number???