POJ

题目：有n个珠子的环，有m种颜色，旋转视为相同。但是有一些颜色限制，规定某两种种颜色不能相邻。

思路：运用矩阵，a->b->c->a，表示一个长度为3的一种方案，而且最终是回到起点的。

这样颜色限制解决了，然后用Burnside引理去做。

注意：要尽量少取模，不然会超时。

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<ctime>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}// 0x3f3f3f3f3f3f3f3fconst int MOD=9973;const int maxn=4e4+50;bool isprime[maxn];int prime[maxn],tol;void make_prime(int n){    for(int i=0;i<=n;i++)        isprime[i]=true;    isprime[0]=isprime[1]=false;    tol=0;    for(int i=2;i<=n;i++){        if(isprime[i])            prime[tol++]=i;        for(int j=0;j<tol;j++){            if(i*prime[j]<=n)                isprime[i*prime[j]]=false;            else                break;            if(i%prime[j]==0)                break;        }    }}const int MAXN = 13;int sz;struct Matrix {int a[MAXN][MAXN];void toZero() {for(int i = 0; i < sz; ++i)for(int j = 0; j < sz; ++j)a[i][j] = 0;}void toOne() {for(int i = 0; i < sz; ++i) {for(int j = 0; j < sz; ++j)a[i][j] = (int)(i == j);}}Matrix operator + (const Matrix &B) const {Matrix A(*this);int i, j;for(i = 0; i < sz; ++i) {for(j = 0; j < sz; ++j)//A.a[i][j] += B.a[i][j];A.a[i][j] = (A.a[i][j] + B.a[i][j]) % MOD;} return A;}Matrix operator * (const Matrix &B) const {Matrix A(*this), res;res.toZero();int i, j, k;for(i = 0; i < sz; ++i) {for(j = 0; j < sz; ++j) {if(A.a[i][j] == 0) continue;for(k = 0; k < sz; ++k){//res.a[i][k] += A.a[i][j] * B.a[j][k];res.a[i][k] = (res.a[i][k] + (A.a[i][j] * B.a[j][k]));res.a[i][k]%=MOD;}}} return res;}Matrix operator ^ (int p) const {Matrix res, X(*this);res.toOne();for(; p; p >>= 1) {if(p & 1) res = res * X;X = X * X;} return res;}void Disp() {printf("---------------------\n");for(int i = 0; i < sz; ++i) {for(int j = 0; j < sz; ++j)printf("%4d", a[i][j]);printf("\n");}printf("---------------------\n");}}A;int phi(int n){    int res=n;    for(int i=0;i<tol&&prime[i]*prime[i]<=n;i++)        if(n%prime[i]==0){            res=res-res/prime[i];            while(n%prime[i]==0){                n/=prime[i];            }        }    if(n>1)        res=res-res/n;    return res%MOD;}int solve(int k){    Matrix B=A^k;    int sum=0;    for(int i=0;i<sz;i++)        sum=(sum+B.a[i][i])%MOD;    return sum;}int pow_mod(int a,int b){    int res=1;    a=a%MOD;    while(b){        if(b&1) res=res*a%MOD;        a=a*a%MOD;        b/=2;    }    return res;}int main(){    int T,n,m,k,u,v;    make_prime(maxn-50);    scanf("%d",&T);    while(T--){        scanf("%d%d%d",&n,&m,&k);        sz=m;        for(int i=0;i<sz;i++)            for(int j=0;j<sz;j++)                A.a[i][j]=1;        while(k--){            scanf("%d%d",&u,&v);u--;v--;            A.a[u][v]=A.a[v][u]=0;        }        int ans=0,i;        for(i=1;i*i<n;i++){            if(n%i)                continue;            ans=(ans+phi(i)*solve(n/i)%MOD)%MOD;            ans=(ans+phi(n/i)*solve(i)%MOD)%MOD;        }        if(i*i==n)            ans=(ans+phi(i)*solve(i)%MOD)%MOD;        ans=ans*pow_mod(n,MOD-2)%MOD;        printf("%d\n",ans);    }    return 0;}