题意
Sol
直接矩阵快速幂
推出来的矩阵应该长这样
egin{equation*}
egin{bmatrix}
1&1&1&1&1&1\
1 & 0&0&0&0&0\
0 & 0&1&3&3&1\
0 & 0&0&1&2&1\
0 & 0&0&0&1&1\
0 & 0&0&0&0&1\
end{bmatrix}^{i - 1}*
egin{bmatrix}
F_{1}\
F_0\
1\
1\
1\
1
end{bmatrix}=
egin{bmatrix}
1&1&1&1&1&1\
1 & 0&0&0&0&0\
0 & 0&1&3&3&1\
0 & 0&0&1&2&1\
0 & 0&0&0&1&1\
0 & 0&0&0&0&1\
end{bmatrix}*
egin{bmatrix}
F_{i - 1}\
F_{i - 2}\
i^3\
i^2\
i\
1
end{bmatrix}=
egin{bmatrix}
F_{i}\
F_{i - 1}\
(i + 1)^3\
(i + 1)^2\
i + 1\
1
end{bmatrix}
end{equation*}
#include<cstdio> #include<algorithm> #include<queue> #include<cstring> #define Pair pair<int, int> #define MP(x, y) make_pair(x, y) #define fi first #define se second #define LL long long //#define int long long using namespace std; const int mod = 1e9 + 7; inline LL read() { char c = getchar(); LL x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } int T; LL N; struct Matrix { LL a[10][10], N; Matrix() { N = 6; memset(a, 0, sizeof(a)); } Matrix operator * (const Matrix &rhs) const { Matrix ans; for(int k = 1; k <= N; k++) for(int i = 1; i <= N; i++) for(int j = 1; j <= N; j++) (ans.a[i][j] += (1ll * a[i][k] * rhs.a[k][j]) % mod) %= mod; return ans; } }; Matrix fp(Matrix a, LL p) { Matrix base; // printf("%d", base.a[0][1]); for(int i = 1; i <= 6; i++) base.a[i][i] = 1; while(p) { if(p & 1) base = base * a; a = a * a; p >>= 1; } return base; } const LL GG[10][10] = { {0, 0, 0, 0, 0, 0, 0}, {0, 1, 1, 1, 1, 1, 1}, {0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 3, 3, 1}, {0, 0, 0, 0, 1, 2, 1}, {0, 0, 0, 0, 0, 1, 1}, {0, 0, 0, 0, 0, 0, 1} }; int main() { T = read(); while(T--) { N = read(); if(N == 1) {puts("1"); continue;} if(N == 2) {puts("16"); continue;} Matrix M; memcpy(M.a, GG, sizeof(M.a)); Matrix ans = fp(M, N - 2); LL out = 0; (out += ans.a[1][1] * 16) %= mod; (out += ans.a[1][2] * 1) %= mod; (out += ans.a[1][3] * 27) %= mod; (out += ans.a[1][4] * 9) %= mod; (out += ans.a[1][5] * 3) %= mod; (out += ans.a[1][6]) %= mod; printf("%lld ", out % mod); } return 0; } /* 5 4 1 2 3 100 */