Inverse of Euler totient function

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Educational Codeforces Round 173 (Rated for Div. 2)
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aryanc403

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Inverse of Euler totient function

Автор SLIMSHADYNICK, история, 5 лет назад, По-английски

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 лет назад
4

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feodorv

5 лет назад, # |

← 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.

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