【NOIP2015模拟10.28A组】递推
思路一
对于 \(30%\) 的数据，由于 \(n\) 和 \(x_i\) 都比较小，所以依题暴力枚举每个整点的坐标算贡献即可
思路二
对于额外 \(20%\) 的数据，发现 \(n=1\) 且有数列 \(F\) 为斐波那契数列，于是就变成求 \(\sum_{i=0}^{x_0 - 1}Fib_i\)
于是我们可以矩阵优化求和
思路三
既然提到矩阵，我们不妨顺着这个思路来想
如果只有一维，我们很容易用矩阵加速递推切掉它
那么考虑高维
发现唯一剩下的问题是如何计算括号中的矩阵之和
因为他们出现了等比
于是考虑暂且抛开单位矩阵 \(I\)
记 F = \(A + A^2 + A^3 + A^4 + ... + A^n\)
\(mid = \lfloor \frac{n}{2} \rfloor\)
\[F(n) =
\left \{
\begin{aligned}
(A + A^2 + A^3 + A^4 + ... + A^{mid})(A^{mid} + I) & & (\texttt{n is even}) \\
(A + A^2 + A^3 + A^4 + ... + A^{mid})(A^{mid} + I) + A^n & & (\texttt{n is odd})
\end{aligned}
\right.
\]
由于题目比较恶心，即使 \(A^n\) 用矩阵快速幂算也会 \(T\) 掉
因为快速幂和分治过程性质一样
所以我们考虑在分治的过程中算出 \(A^n\)
详见代码
\(Code\)
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
typedef long long LL;
const int N = 15 , M = 45;
const int P = 1e9 + 9;
int n , m , c[M] , f[M];
LL ans;
struct matrix{
int m[M][M];
}A , Q , I , Now , Sum , res;
inline matrix Mul(matrix a , matrix b) //矩阵乘法
{
memset(res.m , 0 , sizeof res.m);
for(register int i = 1; i <= m; i++)
for(register int j = 1; j <= m; j++)
for(register int k = 1; k <= m; k++)
res.m[i][j] = (res.m[i][j] + 1LL * a.m[i][k] * b.m[k][j] % P) % P;
return res;
}
inline matrix Plus(matrix a , matrix b) //矩阵加法
{
memset(res.m , 0 , sizeof res.m);
for(register int i = 1; i <= m; i++)
for(register int j = 1; j <= m; j++)
res.m[i][j] = (a.m[i][j] + b.m[i][j]) % P;
return res;
}
inline matrix divide(int x)
{
if (x == 1) return Q = A;
int mid = x >> 1;
matrix tmp = divide(mid); //分而治之
matrix temp;
temp = Mul(tmp , Q);
tmp = Plus(tmp , temp);
Q = Mul(Q , Q); //平方算A^{2*mid}
if (!(x & 1)) return tmp;
Q = Mul(Q , A); //奇数时再乘个A，和快速幂同理
tmp = Plus(tmp , Q);
return tmp;
}
int main()
{
freopen("recursion.in" , "r" , stdin);
freopen("recursion.out" , "w" , stdout);
scanf("%d%d" , &n , &m);
for(register int i = 1; i <= m; i++) scanf("%d" , &c[i]);
for(register int i = 0; i < m; i++) scanf("%d" , &f[i]);
for(register int i = 1; i <= m; i++) A.m[i + 1][i] = 1 , I.m[i][i] = Sum.m[i][i] = 1;
for(register int i = 1; i <= m; i++) A.m[i][m] = c[m - i + 1];
int x;
for(register int i = 1; i <= n; i++)
{
scanf("%d" , &x);
Q = I;
Now = divide(x - 1);
for(register int j = 1; j <= m; j++) Now.m[j][j] = (Now.m[j][j] + 1) % P; //加上单位矩阵I
Sum = Mul(Now , Sum); //先算括号中的
}
for(register int i = 1; i <= m; i++) //把f乘上来
ans = (ans + 1LL * f[i - 1] * Sum.m[i][1] % P) % P;
printf("%lld" , ans);
}