[CF Gym 100827F] Knights [2014-2015 ACM-ICPC Pacific Northwest Regional Contest F]

题意

在M*N棋盘上放置任意数量的马，1≤M≤4，1≤N≤109

题解

矩阵快速幂。以两列(M*2)的放置情况作为状态，求出状态间转移的01矩阵，再快速幂求答案。

代码

/****************************************\* Author : ztx* Title  : F - Knights* ALG    :* CMT    :* Time   :\****************************************/#include <cstdio>#define Rep(i,l,r) for(i=(l);i<=(r);i++)#define rep(i,l,r) for(i=(l);i< (r);i++)#define Rev(i,r,l) for(i=(r);i>=(l);i--)#define rev(i,r,l) for(i=(r);i> (l);i--)typedef long long ll ;typedef double lf ;int CH , NEG ;template <typename TP>inline void read(TP& ret) {    ret = NEG = 0 ; while (CH=getchar() , CH<'!') ;    if (CH == '-') NEG = true , CH = getchar() ;    while (ret = ret*10+CH-'0' , CH=getchar() , CH>'!') ;    if (NEG) ret = -ret ;}template <typename TP>inline void readc(TP& ret) {    while (ret=getchar() , ret<'!') ;    while (CH=getchar() , CH>'!') ;}template <typename TP>inline void reads(TP *ret) {    ret[0]=0;while (CH=getchar() , CH<'!') ;    while (ret[++ret[0]]=CH,CH=getchar(),CH>'!') ;    ret[ret[0]+1]=0;}#define  maxs  260LL  /// 2^(2*4)=256#define  mod   1000000009LLstruct Mat {    int n , m ;    int a[maxs][maxs] ;    inline void init(int i,int j) {        n = i , m = j ;        Rep (i,1,n) Rep (j,1,m) a[i][j] = 0 ;    }    Mat operator * (const Mat&B) const {        Mat ret ; ret.init(n,B.m) ; int i , j , k ;        Rep (i,1,n) Rep (j,1,B.m) Rep (k,1,m)            ret.a[i][j] = ((ll)a[i][k]*B.a[k][j]+ret.a[i][j])%mod ;        return ret ;    }    Mat operator ^ (const int&b) const {        /// p > 0        Mat ret=*this , x=*this ;        for (int p=b-1;p>0;p>>=1,x=x*x) if (p&1) ret=ret*x ;        return ret ;    }    inline void print() {        printf("<Mat %d,%d>:\n",n,m) ;        int i , j ;        Rep (i,1,n) {            Rep (j,1,m) {                printf("%d ",a[i][j]) ;            }            puts("") ;        }    }} A , B ;int n , m , sta ;inline void dfs(int a,int b,int c,int dep) {    if (dep > m) {        A.a[a*(1<<m)+b+1][b*(1<<m)+c+1] = 1 ;        // 0->255  --->  1->256        return ;    }    int aa , bb , cc , sta ;    rep (sta,0,8) {        aa = a<<1 , bb = b<<1 , cc = c<<1 ;        if ((sta&1) && ((aa&2)||(bb&4))) continue ;        if ((sta&2) && ((aa&4)||(cc&4))) continue ;        if ((sta&4) && ((bb&4)||(cc&2))) continue ;        if (sta&1) cc ++ ;        if (sta&2) bb ++ ;        if (sta&4) aa ++ ;        dfs(aa,bb,cc,dep+1) ;    }}int main() {int T , i ;ll ans ;//  #define READ    #ifdef  READ        freopen(".in" ,"r",stdin ) ;        freopen(".out","w",stdout) ;    #endif    for (scanf("%d", &T) ; T -- ; ) {        read(m) , read(n) ;        sta = 1 << (m << 1) ;        A.init(sta,sta) , B.init(1,sta) ;        B.a[1][1] = 1 ;        dfs(0,0,0,1) ;        A = A^n ;        B = B*A ;        ans = 0 ;        Rep (i,1,sta) ans += B.a[1][i] , ans %= mod ;        printf("%I64d\n", ans) ;    }    #ifdef  READ        fclose(stdin) ; fclose(stdout) ;    #else        getchar() ; getchar() ;    #endif    return 0 ;}