10139 - Factovisors

[Sedefcho]

Larry, that is really the most helpful answer I've ever received.
Thank you so much

Well, to be serious, OK, I will think about your hint, as it is
really a sort of hint Although a quite quite small hint.

[Sedefcho]

Larry, thank you for the small hint.
Although I knew it even before I started implementing some code
that we SHOULD NOT pre-generate all the primes up to 2^32 - 1.
It was pretty obvious.

Now I've designed a decent algorithm for solving this problem.
Although not the best one, of course.

So thank you, anyway.

[Wei]

Code: Select all

    Cut after AC


[mf]

Try this test case:

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31 402653184

The answer must be "does not divide," since:

31! = 2^26 3^14 ... 29^1 31^1
402653184 = 2^27 3^1

[Wei]

Well~~Thx~~

I finally got AC~~

[CodeJerk]

Hi
I tried solving Problem: 10139 but got a Time Limit Exceeded error.
What I was doing was, starting from n, i was calculating the gcd(n,m) and dividing m by gcd(n,m).

Code: Select all

 m = m/gcd (m,n) 

then i did the same with n-1, and so on, till m was reduced to 1, or n became 1.

Sometimes, one gets a large prime number as one of the factors, and to handle that i counted the no. of times that the gcd was consecutively 1. In case, this exceeded 50 times, I checked if m was prime, and if it was, exited with a negative answer.

This works fine on almost all inputs, and comes out pretty fast when i try it on my system, but always gives TLE on the programming-challenges website.
the code is here

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#include <iostream.h>
#include <math.h>

long  gcd (long p,long q)
{
 if (q == 0) return p;
 else return (gcd (q, p%q));
}

bool isprime (long n)
{
 
 for (int i = 3; i<=sqrt(n)+1 ; i+=2)
 {
  if (n%i == 0) {return (false);}
 }
 return (true);
}

int main()
{
do{
 long n,m,saven, savem;
 cin >>n>> m;
 saven = n; savem = m;
 bool flag = false;
 int counter= 0;  //prime check
 if (n == 0 && m == 1) {flag = true;goto End;}
 if (n==0 && m>1) {flag = false;goto End;}
 if (m == 0) {flag = false;goto End;}
 
 
 
 // to c if m divides n!62454TK
 
 
 
 while ( n >= 1 && m >= 1)
 {
  if (m <= n) { flag = true; goto End;}
  long g = gcd (m,n);
 
      if (g==1) { counter++; if (counter == 50)
                                   { if (isprime (m)) {flag = false;break;}}
      }
   else counter = 0;
    
   
  m = m/g; n = (n - 1)*(n/g);
  //cout << "n =" << n << " m = " << m << endl; //int x ; cin >> x;
  if (m == 1) { flag = true; break; }
 }
 
 End:
 if (flag == true) cout << savem <<" divides " << saven<<"!";
 else              cout << savem <<" does not divide " << saven<<"!";
 cout << endl;
 } while (true);
 return (0);
}
 
 


Can u plz suggest a better algo to solve this problem. I have tried looking my Number Theory books, but am still in the dark regarding this problem. I would be really grateful if u could help me out.
Thanks,
CJ

[dumb dan]

Code: Select all

do{
    ...
  } while (true);

I think it's the endless loop that gives you TLE.

[CodeJerk]

The question doesnt specify any end condition, like, for e.g., that the last input is 0 0, which is *not* to be processed.
If u got AC could you plz tell me what was the end condition u specified.
Thanks in advance.
CJ

[sumankar]

Read until you hit EOF.Something along the line of...

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 while ( 2 == scanf("%d %d", &a, &b) ) {
 }


Regards,
Suman.

[naguib]

#include<cstdio>
#include<string>
#include<cmath>
using namespace std;
//FILE *in=fopen("fact.in","r");
bool iscom[46502];
int prime[5000];
#ifdef __GNUC__
#define longlong long long
#else
#define longlong __int64
#endif
int howfac(longlong* n,int f)
{
//return how many factor F is in n
longlong ret=0;
while(((*n)%f)==0)
{
(*n)/=f;
ret++;
}
return ret;
}
int main()
{
longlong i,j,k,n,m,c=0,f,x,y,m2,div;
memset(iscom,0,sizeof(iscom));
for(i=2;i<=46350;i++)
{
if(!iscom[i])
{
prime[c]=i;
c++;
for(j=(i+i);j<=46500;j+=i)
iscom[j]=1;
}
}
//while(EOF!=fscanf(in,"%lld %lld",&n,&m))
while(EOF!=scanf("%lld %lld",&n,&m))
{
m2=m;
if(n>=m)
div=1;
else if(m==0)
div=0;
else if(n==0)
div=1;
else
{
f=1;
for(i=0;i<c && prime[i]*prime[i]<=m;i++) //check if M is prime
if(m%prime[i]==0){f=0;break;}

if(f)
div=0;
else
for(i=0;i<c;i++) //check if all prime factors in M is in N! with enough quantity
{
k=howfac(&m,prime[i]);
if(k==0)continue;
x=0;
for(j=prime[i];j<=n;j+=prime[i])
{
y=j;
x+=howfac(&y,prime[i]);
if(x>=k)
break;
}
if(x<k)
{div=0;break;}
if(m==1)
{div=1; break;}
}
}
if(!div)
printf("%lld does not divide %lld!\n",m2,n);
else
printf("%lld divides %lld!\n",m2,n);
}
return 0;
}

If any body can find a mistake or a bug plz tell me I hope that the comments will clarify my algorithm

[naguib]

I just rewrote the code and I got Accepted thx anyway

[binusian0600618124]

Never mind it. Found the bug.
Thanx anyway to those who tried to help me.

[Sumon]

Thanks a lot.There was a silly mistake in my code and the test case 94764610 284293833 helped me finding that.

[ashipm01]

You can get the prime factors of the m and n (from two to n since it is n!) numbers and keep dividing out the prime factors from m until no more prime factors remain in m (return true) or until you reach n (return false).

I would not do it with BigInteger though because it seems the judge does not accept the BigInteger class. I got a compile error before on the Square root problem when I tried to use BigInteger, so I had to write my own BigInteger class.

[newton]

i am in great problem!
please say me why RE [signal 11]?

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#include <stdio.h>
#include <string.h>

void str_mul(char str1[],int n );
long is_divisible(char str[],long);
char *x;
main()
{
   char str[2600];
   char res[1001][2600];
   long num,fact,temp;

   x = str;
   strcpy(res[0],"1");
   fact = 1;
   strcpy(str,"1");
   while(fact!=1001)
   {
      x = str;
      str_mul(str,fact);
      strcpy(res[fact],str);
      fact = fact + 1;
   }
   while(scanf("%ld %ld",&fact,&num)==2)
   {
      temp = fact;
      if(!is_divisible(res[fact],num))
         printf("%ld divides %ld!\n",num,temp);
      else
         printf("%ld does not devide %ld!\n",num,temp);
   }
   return 0;
}


void str_rev(char *str)
{
   long len,i;
   char *p,*q,temp;
   len = strlen(str);
   p = str;
   q = &str[len-1];
   for(i=0;i<len/2;i++)
   {
      temp = *p;
      *p = *q;
      *q = temp;
      p++;
      q--;
   }
}


void str_mul(char str1[], int n)
{
   char *p;
   char res[10000],tem[10000];
   int len1,carry,i,j,temp;
   len1 = strlen(str1);
   str_rev(str1);
   carry=0;
   memset(res,'0',sizeof(res));
   p = res;
   for(i=0;i<len1;i++)
   {
      carry = 0;
      p = &res[i];
      int t = n;
      while(t)
      {
         int a = str1[i]-48;
         int b = t%10;
         temp = *p-48 + carry+ (a*b);
         *p = (*p-48 + carry + (a*b))%10 + 48;
         carry=(temp/10);

         p++;
         t/=10;
      }
      if(carry)
         *p = carry + 48;
   }
   if(carry)
      p++;
   *p = '\0';
   for(i=strlen(res)-1;i>=0;i--)
   {
      *x = res[i];
      x++;
   }
   *x = '\0';
}

long is_divisible(char p[], long num)
{
   long mod=0;
   long len,i;
   len = strlen(p);
   for(i=0;i<=len-1;i++)
      mod=((p[i]-48)+mod*10)%num;
   return (mod%num);
}