Magic Bracelet
Time Limit: 2000MS Memory Limit: 131072K
Total Submissions: 6024 Accepted: 1922
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
题目：POJ2888
题意：把n中珠子串成一个长度为m的环，有k个限制，被限制的两种珠子不能相邻。
思路：裸的polya计数加欧拉函数优化这里就不再赘述如不懂的可以百度或者戳
这里主要是如何求限制条件下的不重复答案。
假设我们循环节的长度为m，那么我们要算长度为m的合法方案数的数。
考虑m<=10，那么我们用一个矩阵记录任意两种珠子之间是否限制。a[i][j]=1表示第i种珠子和第j种珠子之间没有限制。如果将这个矩阵自乘一次，那么a[i][i]则表示∑a[i][j]（1<=j<=m），举个例子，如果m==4.
那么自承一次后，a[4][4]=a[4][1]+a[4][2]+a[4][3]+a[4][4],即长度为2的合法的方案数。按照这个规律，我们只要把这个矩阵n次方，就可以得到长度为n的合法方案的个数。
所以这里配合矩阵快速幂就可以很快的求出我们要的答案了。
Tips：由于矩阵乘法的时候要取多次模，就会TLE，所以取模哪里要优化一下。
AC代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#include<math.h>
#define met(s,k) memset(s,k,sizeof s)
#define scan(a) scanf("%d",&a)
#define scanl(a) scanf("%lld",&a)
#define scann(a,b) scanf("%d%d",&a,&b)
#define scannl(a,b) scanf("%lld%lld",&a,&b)
#define scannn(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define prin(a) printf("%d\n",a)
#define prinl(a) printf("%lld\n",a)
using namespace std;
typedef long long ll;
const int maxn=50000;
const ll mod=9973;
int isprime[maxn],prime[maxn],pcont,n,m,k;
struct Matrix//模仿大牛的炫酷的矩阵快速幂写法
{
ll mat[11][11];
friend Matrix operator* (Matrix x,Matrix y)
{
Matrix c;
met(c.mat,0);
for(int i=0;i<m;i++)
{
for(int j=0;j<m;j++)
{
for(int k=0;k<m;k++)
{
c.mat[i][j]=(c.mat[i][j]+x.mat[i][k]*y.mat[k][j]);
}
c.mat[i][j]%=mod;//因为矩阵的系数不会很大，所以在计算过程中不会溢出，这里把%放在循环外面，才不会TLE
}
}
return c;
}
friend Matrix operator^ (Matrix x,int y)
{
Matrix c;
met(c.mat,0);
for(int i=0;i<m;i++)c.mat[i][i]=1;
while(y)
{
if(y&1)c=c*x;
x=x*x;
y>>=1;
}
return c;
}
}A;
void init()
{
for(int i=2; i<maxn; i++)
{
if(!isprime[i])
{
prime[pcont++]=i;
for(int j=2*i; j<maxn; j+=i)
isprime[j]=1;
}
}
}
int ksm(int x,int y)//最后对答案要取逆元，也要用到快速幂
{
int res=1;
x%=mod;
while(y)
{
if(y&1)res=(res*x)%mod;
x=(x*x)%mod;
y>>=1;
}
return res;
}
int euler(int x)
{
int res=x;
for(int i=0;i<pcont&&prime[i]*prime[i]<=x;i++)
{
if(x%prime[i]==0)
{
res=res/prime[i]*(prime[i]-1);
while(x%prime[i]==0)x/=prime[i];
}
}
if(x!=1)res=res/x*(x-1);
return res%mod;
}
ll cal(int x)
{
ll res=0;
Matrix c=A^x;
for(int i=0;i<m;i++)res=(res+c.mat[i][i])%mod;
return res;
}
ll polya()//裸的polya带欧拉函数优化模板
{
ll ans=0;
for(int i=1; i*i<=n; i++)
{
if(n%i==0)
{
ans=(ans+euler(i)*cal(n/i))%mod;
if(n/i!=i)ans=(ans+euler(n/i)*cal(i))%mod;
}
}
return ans;
}
int main()
{
init();
int t;
scan(t);
while(t--)
{
scannn(n,m,k);
for(int i=0;i<m;i++)
for(int j=0;j<m;j++)A.mat[i][j]=1;
for(int i=0;i<k;i++)
{
int x,y;
scann(x,y);
A.mat[x-1][y-1]=0;
A.mat[y-1][x-1]=0;
}
ll ans=polya();
ans=(ans*ksm(n,mod-2))%mod;//求逆元
prinl(ans);
}
return 0;
}