Inverse of Euler totient function

→ Pay attention

Before contest
Refact.ai Match 1 (Codeforces Round 985)
4 days
Register now »

*has extra registration

→ Top rated

#	User	Rating
1	tourist	4009
2	jiangly	3831
3	Radewoosh	3646
4	jqdai0815	3620
4	Benq	3620
6	orzdevinwang	3529
7	ecnerwala	3446
8	Um_nik	3396
9	gamegame	3386
10	ksun48	3373

Countries | Cities | Organizations

View all →

→ Top contributors

#	User	Contrib.
1	cry	164
1	maomao90	164
3	Um_nik	163
4	atcoder_official	160
5	-is-this-fft-	158
6	awoo	157
7	adamant	156
8	TheScrasse	154
8	nor	154
10	Dominater069	153

View all →

→ Find user

→ Recent actions

Detailed →

SLIMSHADYNICK's blog

Inverse of Euler totient function

By SLIMSHADYNICK, history, 5 years ago, In English

There's this problem INVPHI on spoj, I have not been able to solve it. I have been getting NZEC runtime error. Could anyone tell me how to approach as I feel I have exceeded the memory limitations.

SLIMSHADYNICK
5 years ago
4

Comments (4)

Write comment?

feodorv

5 years ago, # |

← Rev. 2 →

Here is the strightforward solution. We can sieve the phi values up to 202918035 (please, see comments to the problem statement).

Code, 9.35sec, 1157M

#include <stdio.h>

#define MAXN 100000000
#define MAXM 202918035

int phi[MAXM+1], minv[MAXN+1];

void gen( void )
{
  int i, j;
  for( i = 1; i <= MAXM; ++i) phi[i] = i;
  for( i = 2; i <= MAXM; ++i)
    if( phi[i] == i )
      for( j = i; j <= MAXM; j += i)
        phi[j] = (phi[j] / i) * (i - 1);
  for( i = 1; i <= MAXM; ++i)
    if( phi[i] <= MAXN && minv[phi[i]] == 0 )
      minv[phi[i]] = i;
}

int main( void )
{
  int t;
  gen();
  scanf( "%d", &t);
  while( t-- > 0 )
  {
    int n;
    scanf( "%d", &n);
    printf( "%d\n", minv[n] ? minv[n] : -1);
  }
  return 0;
}

Also you can note that almost all resulting numbers are odd with three exceptions, so we can sieve only odd values:

Code, 4.65sec, 770M

#include <stdio.h>

#define MAXN 100000000
#define MAXM 202918035

int phi[MAXM/2+1], minv[MAXN+1];

void gen( void )
{
  int i, j;
  for( i = 1; i <= MAXM; i += 2) phi[i>>1] = i;
  for( i = 3; i <= MAXM; i += 2)
    if( phi[i>>1] == i )
      for( j = i; j <= MAXM; j += 2*i)
        phi[j>>1] = (phi[j>>1] / i) * (i - 1);
  for( i = 1; i <= MAXM; i += 2)
    if( phi[i>>1] <= MAXN && minv[phi[i>>1]] == 0 )
      minv[phi[i>>1]] = i;
  minv[16842752] = 33817088;
  minv[33685504] = 67634176;
  minv[67371008] = 135268352;
}

int main( void )
{
  int t;
  gen();
  scanf( "%d", &t);
  while( t-- > 0 )
  {
    int n;
    scanf( "%d", &n);
    printf( "%d\n", minv[n] ? minv[n] : -1);
  }
  return 0;
}

Though I do not know how can we achieve 0.04 sec or how will it run in Java.

→ Reply

SLIMSHADYNICK

5 years ago, # ^ |

Thx for your help,but I did the same thing in java but its not working!

→ Reply

feodorv

5 years ago, # ^ |

Hm, did you read this? Also you can apply segmented phi sieve (as for spoj ETFS problem).

→ Reply

SLIMSHADYNICK

5 years ago, # ^ |

Thanks for your time, I submitted the code in C, it worked fine!!

→ Reply