| Time Limit: 2000MS | Memory Limit: 131072K | |
| Total Submissions: 4727 | Accepted: 1533 |
Description
Ginny’s birthday is coming soon. Harry Potter is preparing a birthday present for his new girlfriend. The present is a magic bracelet which consists of n magic beads. The are m kinds of different magic beads. Each kind of beads has its unique characteristic. Stringing many beads together a beautiful circular magic bracelet will be made. As Harry Potter’s friend Hermione has pointed out, beads of certain pairs of kinds will interact with each other and explode, Harry Potter must be very careful to make sure that beads of these pairs are not stringed next to each other.
There infinite beads of each kind. How many different bracelets can Harry make if repetitions produced by rotation around the center of the bracelet are neglected? Find the answer taken modulo 9973.
Input
The first line of the input contains the number of test cases.
Each test cases starts with a line containing three integers n (1 ≤ n ≤ 109, gcd(n, 9973) = 1), m (1 ≤ m ≤ 10), k (1 ≤ k ≤ m(m − 1) ⁄ 2). The next k lines each contain two integers a and b (1 ≤ a, b ≤ m), indicating beads of kind a cannot be stringed to beads of kind b.
Output
Output the answer of each test case on a separate line.
Sample Input
4 3 2 0 3 2 1 1 2 3 2 2 1 1 1 2 3 2 3 1 1 1 2 2 2
Sample Output
4 2 1 0
Source
#include <iostream> #include <cstdio> #include <cstring> using namespace std; #define ll long long #define N 32000 struct Matric { int size; int a[12][12]; Matric(int s=0) { size=s; memset(a,0,sizeof(a)); } Matric operator * (const Matric &t) { Matric res=Matric(size); for(int i=0;i<size;i++) { for(int k=0;k<size;k++) { if((*this).a[i][k]) for(int j=0;j<size;j++) { res.a[i][j]+=(*this).a[i][k]*t.a[k][j]; if(res.a[i][j]>=9973) res.a[i][j]%=9973; } } } return res; } Matric operator ^ (int n) { Matric ans=Matric(size); for(int i=0;i<size;i++) ans.a[i][i]=1; while(n) { if(n&1) ans=ans*(*this); (*this)=(*this)*(*this); n>>=1; } return ans; } int CalNum(int k) { Matric tmp=(*this); tmp=tmp^k; int ans=0; for(int i=0;i<size;i++) { ans=(ans+tmp.a[i][i])%9973; if(ans>=9973) ans%=9973; } return ans; } }; int tot; int prime[N+10]; bool isprime[N+10]; int phi[N+10]; void prime_pri() { tot=0; phi[1]=1; memset(isprime,true,sizeof(isprime)); isprime[0]=isprime[1]=false; for(int i=2;i<=N;i++) { if(isprime[i]) { prime[tot++]=i; phi[i]=i-1; } for(int j=0;j<tot;j++) { if((ll)i*prime[j]>N) break; isprime[i*prime[j]]=false; if(i%prime[j]==0) { phi[i*prime[j]]=phi[i]*prime[j]; break; } else { phi[i*prime[j]]=phi[i]*(prime[j]-1); } } } } int euler(int n) { if(n<=N) return phi[n]; int ret=n; for(int i=0;(ll)prime[i]*prime[i]<=n;i++) { if(n%prime[i]==0) { ret-=ret/prime[i]; while(n%prime[i]==0) n/=prime[i]; } } if(n>1) ret-=ret/n; return ret; } int exgcd(int a,int b,int& x, int& y) { if(b==0) { x=1; y=0; return a; } int d=exgcd(b,a%b,y,x); y-=a/b*x; return d; } int inv(int a,int MOD) { int x,y; exgcd(a,MOD,x,y); x=(x%MOD+MOD)%MOD; return x; } int main() { prime_pri(); int T; int n,m,k; int MOD=9973; scanf("%d",&T); while(T--) { scanf("%d%d%d",&n,&m,&k); //n个点,m种颜色,k种限制 Matric a=Matric(m); for(int i=0;i<m;i++) for(int j=0;j<m;j++) a.a[i][j]=1; for(int i=0;i<k;i++) { int u,v; scanf("%d%d",&u,&v); a.a[u-1][v-1]=a.a[v-1][u-1]=0; } int ans=0; for(int i=1;i*i<=n;i++) { if(n%i==0) { if(i*i==n) ans+=ans+euler(i)%MOD*a.CalNum(i); else ans+=euler(i)%MOD*a.CalNum(n/i)+euler(n/i)%MOD*a.CalNum(i); if(ans>=9973) ans%=9973; } } ans=ans*inv(n,MOD)%MOD; printf("%d ",ans); } return 0; }