HDU

题目链接

题意：

$
egin{cases}
& F_{1} = A \
& F_{2} = B \
& F_{n} = C·F_{n-2} + D·F_{n-1}+lfloor frac{P}{n} floor
end{cases}
$

给定递推式，求$F_{n}mod{1e9+7}$

思路：

一来就会想到用矩阵快速幂，但是其中有一个下取整就肯定不是裸的矩阵快速幂；

有向下取整我们就可以用分块来做，复杂度为$sqrt{P}log_{2}P$

/*
*  Author: windystreet
*  Date  : 2018-08-14 09:46:10
*  Motto : Think twice, code once.
*/
#include<bits/stdc++.h>

using namespace std;

#define X first
#define Y second
#define eps  1e-5
#define gcd __gcd
#define pb push_back
#define PI acos(-1.0)
#define lowbit(x) (x)&(-x)
#define bug printf("!!!!!
");
#define mem(x,y) memset(x,y,sizeof(x))

typedef long long LL;
typedef long double LD;
typedef pair<int,int> pii;
typedef unsigned long long uLL;

const int maxn = 1e5+7;
const int INF  = 1<<30;
const int mod  = 1e9+7;

struct Matrix
{
    int n,m;
    LL ma[5][5];
    Matrix (int x,int y):n(x),m(y){clear();}
    void set(int n_,int m_){n = n_,m = m_;}
    LL *operator[](int x){return ma[x];}
    Matrix operator*(Matrix x){
        Matrix res(n,x.m);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=x.m;j++)
                for(int k=1;k<=m;k++)
                    (res[i][j]+=ma[i][k]*x[k][j]%mod+mod)%=mod;
        return res;
    }
    Matrix operator ^(int y){
        Matrix x(n,m);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                x[i][j]=ma[i][j];
        Matrix res(x.n,x.n);
        for(int i=1;i<=x.n;i++)
            res[i][i]=1;
        for(;y;y>>=1,x=x*x)
            if(y&1)res=res*x;
        return res;
    }
    void print(){
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                printf("%lld%c",ma[i][j]," 
"[j==m]);
    }
    void clear(){mem(ma,0);}
};
int a ,b,c,d ,p,n;
pii f(int x,int y){
    if(!x)return make_pair(y,n);
    int l = max(y,(p+x+1)/(x+1)),r = min(n,p/x);
    return make_pair(l,r);
}

void solve(){
    scanf("%d%d%d%d%d%d",&a,&b,&c,&d,&p,&n);
    if(n==1)printf("%d
",a);
    if(n==2)printf("%d
",b);
    else{
        Matrix x(1,3),y(3,3);
        x[1][1] = b,x[1][2]=a,x[1][3]=1;
        for(int i=3;i<=n;){
            pii seg = f(p/i,i);
            y[1][1]=d,y[1][2] = 1,y[1][3] = 0;
            y[2][1]=c,y[2][2] = 0,y[2][3] = 0;
            y[3][1]=p/i,y[3][2]=0,y[3][3] = 1;
            x = x*(y^(seg.Y - seg.X  + 1));
            i = seg.Y+1;
        }
        printf("%lld
",x[1][1] );
    }
    
    return;
}

int main()
{
//    freopen("F:\in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
//    ios::sync_with_stdio(false);
    int t = 1;
    scanf("%d",&t);
    while(t--){
    //    printf("Case %d: ",cas++);
        solve();
    }
    return 0;
}