HDU - 1588 矩阵前缀和

题意:给定\(k,b,n,m\),求\(\sum_{i=0}^{n-1}f(g(i))\)

其中\(f(i)=f(i-1)+f(i-2),f(1)=1,f(0)=0\),\(g(i)=k*i+b\)

令矩阵\(A\)为

\[\begin{bmatrix}
1 & 1 \\
1 & 0 \\
\end{bmatrix}
\]

那么

\[\begin{bmatrix}
f(n+1) \\
f(n) \\
\end{bmatrix}=A^n \begin{bmatrix}
1 \\
0 \\
\end{bmatrix}
\]

我们所求的$$S = f(g(1))+f(g(2))+...+f(g(n-1)) $$

\[S=f(b)+f(k+b)+f(k*2+b)+...+f(k*(n-1)+b)
\]

\[S=A^b\begin{bmatrix}1 \\0 \\\end{bmatrix}+A^{k+b}\begin{bmatrix}1 \\0 \\\end{bmatrix}+...+A^{k(n-1)+b}\begin{bmatrix}1 \\0 \\\end{bmatrix}
\]

\[S=A^b[E+(A^k)^1+(A^k)^2...+(A^k)^{n-1}]\begin{bmatrix}1 \\0 \\\end{bmatrix}
\]

中间的前缀和求法可参考我上一篇文章(p讲解都没有):http://www.cnblogs.com/caturra/p/8452828.html

#include<bits/stdc++.h>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define println(a) printf("%lld\n",(ll)a)
using namespace std;
typedef long long ll;
ll read(){
    ll x=0,f=1;register char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
ll k,b,n,m;
struct Matrix{
    ll mt[5][5],r,c;
    void init(int rr,int cc,bool flag=0){
        r=rr;c=cc;
        memset(mt,0,sizeof mt);
        if(flag) rep(i,1,r) mt[i][i]=1;
    }
    Matrix operator * (Matrix rhs){
        Matrix ans; ans.init(r,rhs.c);
        rep(i,1,r){
            rep(j,1,rhs.c){
                int t=max(r,rhs.c);
                rep(k,1,t){
                    ans.mt[i][j]+=(mt[i][k]*rhs.mt[k][j])%m;
                    ans.mt[i][j]=(ans.mt[i][j])%m;
                }
            }
        }
        return ans;
    }
};
Matrix fpw(Matrix A,ll n){
    Matrix ans;ans.init(A.r,A.c,1);
    while(n){
        if(n&1) ans=ans*A;
        n>>=1;
        A=A*A;
    }
    return ans;
}
int bas[3][3]={
	{0,0,0},
	{0,1,1},
	{0,1,0}
};
int bas2[3]={0,1,0};
int main(){
	Matrix A; A.init(2,2);
	rep(i,1,2)rep(j,1,2) A.mt[i][j]=bas[i][j];
	Matrix C; C.init(2,1);
	rep(i,1,2) C.mt[i][1]=bas2[i];
	while(cin>>k>>b>>n>>m){
		Matrix Ak=fpw(A,k);
		Matrix Ab=fpw(A,b);
		Matrix UNIT; UNIT.init(2,2,1);
		Matrix B; B.init(4,4);
		rep(i,1,2)rep(j,1,2) B.mt[i][j]=Ak.mt[i][j];
		rep(i,1,2)rep(j,3,4) B.mt[i][j]=UNIT.mt[i][j-2];
		rep(i,3,4)rep(j,3,4) B.mt[i][j]=UNIT.mt[i-2][j-2];
		Matrix res=fpw(B,n);
		B.init(2,2);
		rep(i,1,2) rep(j,1,2) B.mt[i][j]=res.mt[i][j+2];
		res=Ab*B*C;
		println(res.mt[2][1]);
	}
	return 0;
}