ACdream 1093 女神的正多面体 矩阵快速幂

#include <stdio.h>
#include <string.h>
#define REP(i, n) for(int i = 0; i < n; ++i)
typedef long long ll;
const int mod = 1000000007;
const int O = 8;
int N, K;
typedef struct M{
    ll mat[O][O];
    M operator * (const M &t) const{//重载矩阵乘法
        M tmp;
        memset(tmp.mat, 0, sizeof(tmp.mat));
        REP(i, N) REP(j, N) REP(k, N) tmp.mat[i][j] = (tmp.mat[i][j] + mat[i][k] * t.mat[k][j]) % mod;
        return tmp;
    }
    
    //不使用函数的原因是为了让快速幂看起来更加自然
    
    M operator + (const M &t) const{//重载矩阵加法
        M tmp;
        REP(i, N) REP(j, N) tmp.mat[i][j] = (mat[i][j] + t.mat[i][j]) % mod;
        return tmp;
    }
}M;
M A[3] = {
    {{//正四边形
    {0, 1, 1, 1},
    {1, 0, 1, 1},
    {1, 1, 0, 1},
    {1, 1, 1, 0},
    }},
    {{//正六边形
    {0, 1, 0, 1, 1, 0, 0, 0},
    {1, 0, 1, 0, 0, 1, 0, 0},
    {0, 1, 0, 1, 0 ,0, 1, 0},
    {1, 0, 1, 0, 0, 0, 0, 1},
    {1, 0, 0, 0, 0, 1, 0, 1},
    {0, 1, 0, 0, 1, 0, 1, 0},
    {0, 0, 1, 0, 0 ,1, 0, 1},
    {0, 0, 0, 1, 1, 0, 1, 0},
    }},
    {{//正八边形
    {0, 1, 1, 1, 1, 0},
    {1, 0, 1, 0, 1, 1},
    {1, 1, 0, 1, 0, 1},
    {1, 0, 1, 0, 1, 1},
    {1, 1, 0, 1, 0, 1},
    {0, 1, 1, 1, 1, 0},
    }},
}, E;
M pow(ll k){//矩阵快速幂求A^k.
    M res = E, tmp = A[K];
    for(; k; k >>= 1){
        if(k & 1) res = res * tmp;
        tmp = tmp * tmp;
    }
    return res;
}
M _pow(ll k){//递归快速幂求A^1 + A^2 + A^3 + ...... + A^k.
    if(k == 1) return A[K];
    M res = _pow(k >> 1) * (E + pow(k >> 1));
    if(k & 1) res = res + pow(k);
    return res;
}
void work(){
    int t, n, s, e;
    ll k;
    REP(i, O) REP(j, O) E.mat[i][j] = (i == j);//初始化单位矩阵
    scanf("%d", &t);
    while(t--){
        scanf("%d%lld%d%d", &n, &k, &s, &e);
        N = (n == 4 ? 4 : (n == 6 ? 8 : 6));//选择矩阵需要的范围
        K = (n == 4 ? 0 : (n == 6 ? 1 : 2));//选择正多边形的类型
        printf("%lld\n", _pow(k).mat[s - 1][e - 1]);
    }
}
int main(){
    work();
    return 0;
}