一个由自然数组成的数列按下式定义：
对于 $i \leq k$ ：$a_i = b_i$
对于 $i > k$ : $a_i = c_1a _ {i-1} + c_2a _ {i-2} + … + c_ka _ {i-k}$
其中 $b_j$ 和 $c_j$ （ $1 \leq j \leq k$）是给定的自然数。写一个程序，给定自然数 $m \leq n$, 计算 $a_m + a _ {m+1} + a _ {m+2} + … + a_n$, 并输出它除以给定自然数 $p$ 的余数的值。
对于 100% 的测试数据：
$1 \leq k \leq 15,1 \leq m \leq n \leq 10^{18},0 \le b_1, b_2,… b_k, c_1, c_2,…, c_k \leq 10^9,1 \leq p \leq 10^8$
链接
Luogu P2461
题解
$$
a_i = \sum _ {j=1}^{k} c_k a _ {j-k}
$$
构造一个矩阵
$$
M_i
=\begin{bmatrix}
S_i \
a_i \
a _ {i-1}\
\vdots\
a _ {i-k+2}\
a _ {i-k+1}\
\end{bmatrix}
$$
转移矩阵为：
$$
Z
=\begin{bmatrix}
1 & c _ {1} & c _ {2} & \cdots & c _ {k-1} & c _ {k} \
0 & c _ {1} & c _ {2} & \cdots & c _ {k-1} & c _ {k} \
0 & 1 & 0 & \cdots & 0 & 0\
\vdots & \vdots & \vdots & \ddots & 0 & 0\
0 & 0 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 0 & 1 & 0 \
\end{bmatrix}
$$
初始的矩阵为
$$
C
=\begin{bmatrix}
S_k\
b_k \
b _ {k-1}\
\vdots\
b _ {2}\
b _ {1} \
\end{bmatrix}
$$
$$
Z \times C = M _ {k+1}\
Z \times M_i = M _ {i+1}\
$$
我们有
$$
Z^{n-k} \times C = M _ {n}
$$
矩阵快速幂即可。
代码
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| #include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#define ll long long
using namespace std;
const int MAXN = 20;
ll N,M,K,P,S;
ll b[MAXN],c[MAXN];
struct Matrix{
ll num[MAXN][MAXN];
Matrix(int op = 0){
memset(num,0,sizeof(num));
if(op){for(int i = 1;i<MAXN;i++) num[i][i] = 1;}
}
ll* operator [] (const int n){
return num[n];
}
};
Matrix mul(Matrix &_x,Matrix &_y){
Matrix ans;
for(int i = 1;i<=S;i++){
for(int j = 1;j<=S;j++){
for(int k = 1;k<=S;k++){
ans[i][j] += _x[i][k] * _y[k][j];
}
ans[i][j] %= P;
}
}
return ans;
}
Matrix pow(Matrix x,ll k){
Matrix ans(1);
for(ll i = k;i;i>>=1,x = mul(x,x)){
if(i & 1) ans = mul(ans,x);
}
return ans;
}
Matrix Z,C;
void init(){
scanf("%lld",&K);
S = K+1;
for(int i = 1;i<=K;i++)
scanf("%lld",&b[i]);
for(int i = 1;i<=K;i++)
scanf("%lld",&c[i]);
scanf("%lld %lld %lld",&M,&N,&P);
for(int i = 1;i<=K;i++)
b[i] %= P,c[i] %= P;
}
ll query(ll x){
ll ans = 0;
if(x <= K){
for(int i = 1;i<=x;i++)
ans += b[i];
}
else{
Matrix a = pow(Z,x-K);
for(int i = 1;i<=S;i++)
ans += C[i][1] * a[1][i];
}
return ans % P;
}
void build(){
ll sum = 0;
for(int i = 1;i<=K;i++){
C[S-i+1][1] = b[i];
sum += b[i];
}
C[1][1] = sum % P;
Z[1][1] = 1;
for(int i = 1;i<=K;i++){
Z[1][i+1] = c[i];
Z[2][i+1] = c[i];
}
for(int i = 2;i<=K;i++){
Z[i+1][i] = 1;
}
}
int main(){
init();
build();
printf("%lld\n",((query(N)-query(M-1))%P+P)%P);
return 0;
}
|
