Codeforces 604D "Discrete mathematical permutation group"

Test instructions

Give you an equation, the equation is too perverted to fight, give you a p a k,p guarantee is a prime number greater than or equal to 3, and K guarantees an integer between 0~p-1. The definition field of the corresponding function is required to be a subset of 0~p-1 in the 0~p-1 domain, so how many ...

Analysis:

"Today, just vaguely listened to the festival set theory, doing to find it seems to be a cyclic group or what = ="

When k=0, it is not difficult to find f (0) = 0, any other.

K=1, f (0) =f (0) mod p, found in addition to any other accidental f (0) is also arbitrary.

When k>=2, find a rule ...

It is not difficult to find out if k=2, then f (2) According to F (1), F (4) According to F (2) ... And so on, until after the MoD operation becomes 1 again ...

So the values of many groups vary according to the value of one of the functions ... So I found a certain rule (which should be proved)

The law is this, found that <k> (the symbol of the interpretation of discrete mathematics = =) is the order of the definition of the domain (this should be deduced from a famous theorem, the cock silk just found the law) ...

OK, finally the problem becomes a regular problem (the dregs pain Pain)

In addition to 0, the definition field can be divided into a W group, each with a <k> order number. The rest of the matter is to arrange a combination, write a quick power solution ...

Reflection:

This problem reflects my analysis of this slag is not comprehensive, although 1A but after the discovery of a place to verify that they have neglected ...

Code:

#include <stdio.h>intMain () {intp,k; intn=1e9+7; scanf ("%d%d",&p,&k); if(k==0)    {        Long LongTmp=p,r=1; P--;  while(p) {if(p&1) {R=r*tmp%N; } tmp=tmp*tmp%N; P>>=1; } printf ("%i64d\n", R); }    Else if(k==1)    {        Long LongTmp=p,r=1;  while(p) {if(p&1) {R=r*tmp%N; } tmp=tmp*tmp%N; P>>=1; } printf ("%i64d\n", R); }    Else    {        Long Longtmp=1; Long Longans;  for(intI=1;; i++) {tmp= (tmp*k)%p; if(tmp==1) {ans=i;  Break; }        }        Long LongR=1; TMP=p; P= (P-1)/ans;  while(p) {if(p&1) {R=r*tmp%N; } tmp=tmp*tmp%N; P>>=1; } printf ("%i64d\n", R); }}

Codeforces 604D "Discrete mathematical permutation group"