[GDKOI2017Day1T3排队 矩阵快速幂+插头dp

诶总结晚点发

#include 
#include 
#include 
#include 
#define mod 1000000007
#define N 66*2

using namespace std;

int F[10000050],n,p,q,E;

struct Matrix{ 
    int a,d[N][N];
    void print() {
        for (int i=1;i<=a;i++){
            printf("::debug row%d::",i%10);
            for (int j=1;j<=a;j++) printf("%d%c",d[i][j],j==a?'\n':' ');
        }
        for (int i=1;i<=a;i++) printf("-"); puts("");
    }
    void clear() {
        for (int i=1;i<=a;i++)
            for (int j=1;j<=a;j++) d[i][j] = 0;
    }
}id,A,B;

inline void inc(int &x,int y) { x = (x + y) % mod; }
inline int rd() { int r; scanf("%d",&r); return r; }

Matrix operator* (Matrix p1,Matrix p2) {
    Matrix ret = id; ret.a = p1.a;
    for (int i=1;i<=p1.a;i++)
        for (int j=1;j<=p1.a;j++)
            for (int k=1;k<=p1.a;k++) inc( ret.d[i][j] , 1LL * p1.d[i][k] * p2.d[k][j] % mod);
    return ret;
}

inline Matrix qp(Matrix a,int b) {
    Matrix ret = id; ret.a = a.a; 
    for (int i=1;i<=ret.a;i++) ret.d[i][i] = 1;
    while (b) {
        if (b&1) ret = ret * a;
        b >>= 1 , a = a * a;
    }
    return ret;
}

inline void calc() {
    A.a = B.a = 2;
    A.d[1][1] = 3 , A.d[1][2] = 1;
    B.d[1][1] = B.d[1][2] = B.d[2][1] = 1;
    B = qp(B,q-2);
    A = A * B;
    printf("%d\n",A.d[1][1]);
}

inline int momsen(int p1,int p2) {
    int R = p2 & 1;
    for (int _=1;_<=p;_++) {
        int cur = 0;
        if ((p1>>_)&1) cur++;
        if (R > 0) cur++;
        if ((p2>>_)&1) cur++;
        if (_==p) {
            if (p2&1)
                return cur == 1;
            else
                return cur == 2;
        }
        if (cur == 3 || cur == 0) return 0;
        if (cur == 1) R = 1; else R = 0;
    }
    assert(0); return 1;
}


inline Matrix get_Matrix() {
    Matrix ret = id;
    ret.a = E;
    for (int i=0;ifor (int j=0;j1][j+1] = momsen(i,j);
    return ret;
}

inline void solve() {
    E = ( 1<<(p+1) );
    B = get_Matrix();
//  B.print();
    B = qp(B,q);

    int ans = 0;
//  B.print();
    for (int _=1;_<=E;_++) inc(ans,B.d[_][_]);
    printf("%d\n",ans);
}

int main() {
    p = rd() , q = rd();
    if (p > q) swap(p,q);
    if (p == 1) { puts("1"); return 0; }
    if (p == 2) { calc(); return 0; }
    solve();
    return 0;
}