poj-2888 Magic Bracelet

Test instructions

Give the necklace of n beads and the M-type beads;

There is a K-pair relationship between the beads, these beads can not be adjacent;

Cannot be turned into the same necklace by rotation as essentially different;

To find the essence of the number of different necklaces, the answer to 9973 modulo;

N<=10^9,GCD (n,9973) =1,m<=10;

Exercises

This is obviously an issue of permutation counting;

Burnside lemma or choose POI?

The Burnside lemma, because the poi theorem on the color limit to be very broad!

First consider a displacement posture, rotate x beads;

The necklace is then divided into segments of the GCD (x,n) size, and the required number of invariant permutations requires that these segments be identical;

The problem is asking how many kinds of pendulum the beads can have in the chain of GCD (x,n) +1 length;

Here can be recursive to obtain, matrix optimization;

The specific matrix is related to the relationship between beads, not detailed;

At this point, if we enumerate 1-n's posture must be timed out, then enumeration n of the approximate;

For each approximate p of N, consider several numbers with N's gcd as p;

The number here is not greater than N, so here the number/p must be less than n/p;

And GCD is p so it must be coprime with n/p, then it is Euler function;

The implementation is not too complicated, but the details are very much?

Anyway, I did not add the opposite side WA a lot of a lot of hair (laughter);

Orz wzq god Ben;

Time complexity O (√N*M^3LOGN);

Code:

#include <stdio.h> #include <string.h> #include <algorithm> #define N 10#define mod 9973using namespace Std;struct matrix{int a[n][n];int val (int m) {int ret=0;for (int i=0;i<m;i++) Ret+=a[i][i];return ret%mod;} Friend Matrix operator * (Matrix X,matrix y) {matrix ret;memset (&ret,0,sizeof (Matrix)); for (int. i=0;i<n;i++) for ( int j=0;j<n;ret.a[i][j]%=mod,j++) for (int k=0;k<n;k++) ret.a[i][j]+=x.a[i][k]*y.a[k][j];return ret;}} Temp,t,in;matrix pow (Matrix X,int y) {matrix ret=in;while (y) {if (y&1) ret=ret*x;x=x*x;y>>=1;} return ret;} int pow (int x,int y) {int ret=1;while (y) {if (y&1) ret=ret*x%mod;x=x*x%mod;y>>=1;} return ret;} int phi (int x) {int ret=x;for (int i=2;i*i<=x;i++) {if (x%i==0) {ret/=i,ret*=i-1;while (x%i==0) x/=i;}} if (x!=1) ret/=x,ret*=x-1;return ret;} int main () {int c,tc,n,m,i,j,k,x,y,ans;scanf ("%d", &AMP;TC), for (i=0;i<n;i++) in.a[i][i]=1;for (c=1;c<=tc;c++) { scanf ("%d%d%d", &n,&m,&k); memset (&t,0,sizeof (T)); for (i=0;i<m;i++) for (j=0;j<m; j + +) T.a[i][j]=1;for (i=1;i<=k;i++) {scanf ("%d%d", &x,&y); x--, y--; t.a[x][y]=t.a[y][x]=0;} for (i=1,ans=0;i*i<=n;i++) {if (n%i==0) {Temp=pow (t,n/i); Ans+=phi (i)%mod*temp.val (m)%mod;ans%=mod;if (i*i==n) Continue;temp=pow (t,i); Ans+=phi (n/i)%mod*temp.val (M)%mod;ans%=mod;}} printf ("%d\n", Ans*pow (n%mod,mod-2)%mod);} return 0;}

poj-2888 Magic Bracelet