递推

ssl 1532

题目大意

给出数列$a_0,a_1\dots a_n$和$f$的前$n-1$项$f_0,f_1\dots f_{n-1}$
$f_i=a_0\ast f_{i-n}+a_1\ast f_{i-(n-1)}+...+a_{n-1}\ast f_{i-1}+a_n$
现在让你求$f_k$(结果对9973取模）

输入样例

2 10
1 1 0
0 1

输出样例

55

数据范围

$$\begin{gathered} 1⩽n⩽k⩽10^{18} \\ 1⩽a_i,fi⩽10^4 \end{gathered}$$

解题思路

k较大，无法直接递推，我们考虑矩阵乘法
设矩阵
$\left[\begin{array}{ccccc}{f}_{i-n}& f_{i-(n-1)}& \dots & f_{i-1}& 1\end{array}\right]$
$f_{i-n}-f_{i-2}$直接等于下一位
$f_{i-1}$就是前面n项乘对应的a，然后加上最后一个1乘$a_n$
1不变
得出需要乘的矩阵后快速幂即可

代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define wyc 9973
ll n, k;
struct matrix
{ll n, m, a[100][100];matrix operator *(matrix const &b)const{matrix c;c.n = n;c.m = b.m;for (int i = 1; i <= c.n; ++i)for (int j = 1; j <= c.m; ++j)c.a[i][j] = 0;for (int i = 1; i <= c.n; ++i)for (int k = 1; k <= m; ++k)for (int j = 1; j <= c.m; ++j)c.a[i][j] = (c.a[i][j] + a[i][k] * b.a[k][j] % wyc) % wyc;return c;}
}A, B;
void Counting(ll g)
{while(g){if (g&1) A = A * B;B = B * B;g>>=1;}
}
int main()
{scanf("%lld%lld", &n, &k);B.n = B.m = A.m = n + 1;A.n = 1;for (int i = 1; i <= n + 1; ++i){scanf("%lld", &B.a[i][n]);//前n个数乘上对应的aif (i < n) B.a[i + 1][i] = 1;//等于下一个}B.a[n + 1][n + 1] = 1;//最后一个afor (int i = 1; i <= n; ++i)scanf("%lld", &A.a[1][i]);A.a[1][n + 1] = 1;Counting(k);printf("%lld", A.a[1][1]);return 0;
}