HDU - 6395 Sequence (+ divisible block matrices fast power)

Define the number of columns:

$\left\{\begin{eqnarray*} F_1 &=& A \\ F_2 &=& B \\ F_n &=& C\cdot{}F_{n-2}+D\cdot{}F_{n-1}+\left\lfloor\frac{P}{n}\right\rfloor \end{eqnarray*}\right.$

The requirements of a number of n-th column.

Obviously divisible blocking problem, the $ \ left \ lfloor \ frac {P} {n} \ right \ rfloor $ n are the same groups into a matrix to fast power. Complexity $ O (3 ^ 3 \ sqrt nlogn) $

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int N=1e5+10,mod=1e9+7;
 5 int n,A,B,C,D,P;
 6 struct Mat {
 7     int a[3][3],n;
 8     Mat() {memset(a,0,sizeof a),n=3;}
 9     int* operator[](int x) {return a[x];}
10     Mat operator+(Mat b) {
11         Mat c;
12         for(int i=0; i<n; ++i)
13             for(int j=0; j<n; ++j)
14                 c[i][j]=(a[i][j]+b[i][j])%mod;
15         return c;
16     }
17     Mat operator*(Mat b) {
18         Mat c;
19         for(int i=0; i<n; ++i)
20             for(int j=0; j<n; ++j)
21                 for(int k=0; k<n; ++k)
22                     c[i][j]=(c[i][j]+(ll)a[i][k]*b[k][j])%mod;
23         return c;
24     }
25 };
26 Mat Pow(Mat x,int p) {
27     Mat ret;
28     for(int i=0; i<ret.n; ++i)ret[i][i]=1;
29     for(; p; p>>=1,x=x*x)if(p&1)ret=ret*x;
30     return ret;
31 }
32 int solve() {
33     if(n==1)return A;
34     Mat t;
35     for(int i=0; i<t.n; ++i)t[i][i]=1;
36     for(int l=3,r,k; l<=n; l=r+1) {
37         k=P/l,r=k&&P/k<n?P/k:n;
38         Mat x;
39         x[0][0]=D;
40         x[0][1]=C;
41         x[0][2]=k;
42         x[1][0]=1;
43         x[2][2]=1;
44         x=Pow(x,r-l+1);
45         t=x*t;
46     }
47     return ((ll)t[0][0]*B%mod+(ll)t[0][1]*A%mod+t[0][2])%mod;
48 }
49 int main() {
50     int T;
51     for(scanf("%d",&T); T--;) {
52         scanf("%d%d%d%d%d%d",&A,&B,&C,&D,&P,&n);
53         printf("%d\n",solve());
54     }
55     return 0;
56 }