Sequence——HDU6395

Let us define a sequence as below F1=A F2=B Fn=C⋅Fn−2+D⋅Fn−1+⌊P/n⌋   Your job is simple, for each task, you should output Fn module 109+7 .   Input The first line has only one integer T , indicates the number of tasks. Then, for the next T lines, each line consists of 6 integers, A , B , C , D , P , n . 1≤T≤200≤A,B,C,D≤1091≤P,n≤109   Sample Input 2 3 3 2 1 3 5 3 2 2 2 1 4   Sample Output 36 24

#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;

typedef long long ll;
const int Mod=1e9+7;
int A,B,C,D,n,P;

struct Mat
{
    int t[3][3];
    
    Mat(){memset(t,0,sizeof t);}
}I;

Mat operator * (Mat a,Mat b){
    Mat c;
    for(int i=0;i<3;i++)
        for(int j=0;j<3;j++){
            ll t=0;
            for(int k=0;k<3;k++)
                t+=(ll)a.t[i][k]*b.t[k][j];
            c.t[i][j]=t%Mod;
        }
    return c;
}

Mat Pow(Mat a,int b){
    Mat c=I;
    while(b){
        if(b&1)
            c=c*a;
        a=a*a;b>>=1;
    }
    return c;
}

int main()
{
    I.t[1][1]=I.t[0][0]=I.t[2][2]=1;
    int t;cin>>t;
    int ok;
    while(t--){
	    scanf("%d%d%d%d%d%d",&A,&B,&C,&D,&P,&n);
	    if(n==1){
	        cout<<A<<endl;
	        continue;
	    }
	    ok=0;
	    Mat f;
	    f.t[0][0]=D;
	    f.t[0][1]=C;
	    f.t[1][0]=1;
	    f.t[2][2]=1;
	    for(int i=3;i<=n;){
	        if(P/i==0){
	            Mat w=f;
	            w=Pow(w,n-i+1);
	            cout<<(w.t[0][0]*(ll)B+w.t[0][1]*(ll)A+w.t[0][2])%Mod<<endl;ok=1;break;
	        }
	        int j=min(n,P/(P/i));
	        Mat w=f;w.t[0][2]=P/i;
	        w=Pow(w,j-i+1);
	        ll a=(w.t[1][0]*(ll)B+w.t[1][1]*(ll)A+w.t[1][2])%Mod,b=(w.t[0][0]*(ll)B+w.t[0][1]*(ll)A+w.t[0][2])%Mod;
	        A=a;B=b;
	        i=j+1;
	    }
	    if(ok==0)
	    	cout<<B<<endl;
	} 
    return 0;
}