F[1] = a, F[2] = b, F[i] = 2 * F[i-2] + F[i-1] + i ^ 4, (i >= 3)
现在要求F[N]
类似于斐波那契数列的递推式子吧， 但是N最大能到int的最大值， 直接循环推解不了
所以就得用矩阵快速幂咯
现在就看转移矩阵长什么样了
Mi表示要求的矩阵 转移矩阵用A表示
A * Mi = Mi+1
矩阵Mi里面至少得有 F[i-1] F[i-2] i ^ 4 Mi+1就相应的有 F[i] F[i-1] (i+1)^4
(i+1)^4 = i^4 + 4 * i ^ 3 + 6 * i ^ 2 + 4 * i + 1
所以Mi中还得有i^3 i^2 i 1
总共就有七个元素
$\begin{pmatrix} 1 & 2 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0& 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 4 & 6 & 4 & 1 \\ 0 & 0 & 0 & 1 & 3 & 3 & 1 \\ 0 & 0 & 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1\end{pmatrix}
\times \begin{pmatrix} f_{i-1} \\ f_{i-2} \\ i^{4} \\ i^{3} \\ i^{2} \\ i \\ 1 \end{pmatrix} = \begin{pmatrix} f_{i} \\ f_{i-1} \\ (i+1)^{4} \\ (i+1)^{3} \\ (i+1)^{2} \\ i+1 \\ 1 \end{pmatrix}$
基本的矩阵运算，就是前面这个相当于系数的矩阵得是 (n-2)次幂 因为f1 f2都求过了
初始的矩阵是
$\begin{pmatrix} f_{2} \\ f_{1} \\ 3^{4} \\ 3^{3} \\ 3^{2} \\ 3 \\ 1 \end{pmatrix}$
也就是
$\begin{pmatrix} b \\ a \\ 81 \\ 27 \\ 9 \\ 3 \\ 1 \end{pmatrix}$
特判一下n == 1 和 2的情况就好啦
代码如下
#include <cstdio>
#define ll long long
#define MOD 2147493647
using namespace std; struct Matrix {
ll m[][];
Matrix() {
for (int i = ; i < ; i++)
for (int j = ; j < ; j++)
m[i][j] = ;
}
}; Matrix mul(Matrix A, Matrix B) {
Matrix temp;
for (int i = ; i < ; i++)
for (int j = ; j < ; j++)
for (int k = ; k < ; k++)
temp.m[i][j] = (temp.m[i][j] % MOD+ A.m[i][k] * B.m[k][j] % MOD) % MOD;
return temp;
} Matrix quick_mod(Matrix A, ll b) {
Matrix ans;
for (int i = ; i < ; i++)
ans.m[i][i] = ;
while (b) {
if (b & ) ans = mul(ans, A);
A = mul(A, A);
b >>= ;
}
return ans;
} int main() {
int T; scanf("%d", &T);
while (T--) {
int n, a, b;
scanf("%d%d%d", &n, &a, &b);
if (n == ) printf("%d\n", a);
else if (n == ) printf("%d\n", b);
else {
Matrix ans;
ans.m[][] = , ans.m[][] = , ans.m[][] = ;
ans.m[][] = ;
ans.m[][] = , ans.m[][] = , ans.m[][] = , ans.m[][] = , ans.m[][] = ;
ans.m[][] = , ans.m[][] = , ans.m[][] = , ans.m[][] = ;
ans.m[][] = , ans.m[][] = , ans.m[][] = ;
ans.m[][] = , ans.m[][] = , ans.m[][] = ;
ans = quick_mod(ans, (ll)n - );
Matrix temp;
temp.m[][] = b, temp.m[][] = a, temp.m[][] = , temp.m[][] = ;
temp.m[][] = , temp.m[][] = , temp.m[][] = ;
ans = mul(ans, temp);
printf("%lld\n", ans.m[][] % MOD);
}
}
return ;
}
其实可以封装一下Matrix 重载一下 * 和 ^ 运算符 这样就很方便也很好看啦