【2018 Multi-University Training Contest 7】Sequence（分块+矩阵快速幂）_let us define a sequence as below f1f2fn=-优快云博客

本文介绍了一种解决特定序列计算问题的高效算法。通过矩阵快速幂和分段计算的方法，实现了对序列Fn的求解，并考虑了大数运算中的模块化处理。适用于序列计算中的边界条件变化和大规模数据处理。

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Problem Description
Let us define a sequence as below

⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ F 1 F 2 F n = = = A B C \cdot F n - 2 + D \cdot F n - 1 + ⌊ P n ⌋

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

Your job is simple, for each task, you should output Fn module $10^9+7.$

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.

$1≤T≤200≤A,B,C,D≤10^91≤P,n≤10^9$

Sample Input
2
3 3 2 1 3 5
3 2 2 2 1 4

Sample Output
36
24

Source
2018 Multi-University Training Contest 7
代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#define ll long long
#define mod 1000000007
using namespace std;

struct M
{
    ll m[3][3];
    M mul(M o)
    {
        M t;
        for(int i=0;i<3;i++)
        for(int j=0;j<3;j++)
        {
            ll temp=0;
            for(int k=0;k<3;k++)
                temp=(temp+m[i][k]*o.m[k][j]%mod)%mod;
            t.m[i][j]=temp;
        }
        return t;
    }
    void zero()
    {
        for(int i=0;i<3;i++)
        for(int j=0;j<3;j++)
        m[i][j]=0;
    }
    void unit()
    {
        zero();
        for(int i=0;i<3;i++)m[i][i]=1;
    }
};
M get(M a,int b)
{
    M ans;
    ans.unit();
    while(b)
    {
        if(b&1)ans=ans.mul(a);
        b>>=1;
        a=a.mul(a);
    }
    return ans;
}
int A,B,C,D,P,n;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        scanf("%d%d%d%d%d%d",&A,&B,&C,&D,&P,&n);
        A%=mod;
        B%=mod;
        if(n==1)cout<<A<<endl;
        else if(n==2)cout<<B<<endl;
        else
        {
            M _a;
            _a.zero();
            _a.m[0][0]=D;_a.m[0][1]=C;_a.m[0][2]=1;
            _a.m[1][0]=1;_a.m[2][2]=1;
            for(ll i=3,last;i<=n;i=last+1)
            {
                ll val=P/i;
                if(val==0)
                {
                    int b=n-i+1;
                    M a=_a;
                    M __a=get(a,b);
                    B=(__a.m[0][0]*B%mod+__a.m[0][1]*A%mod)%mod;
                    //cout<<B<<endl;
                    break;
                }
                last=P/val;
                last=min((int)last,n);
                int b=last-i+1;
                M a=_a;
                M __a=get(a,b);
                ll _B=B;
                B=(__a.m[0][0]*B%mod+__a.m[0][1]*A%mod+__a.m[0][2]*val%mod)%mod;
                __a=get(a,b-1);
                A=(__a.m[0][0]*_B%mod+__a.m[0][1]*A%mod+__a.m[0][2]*(P/(last-1))%mod)%mod;
            }
            printf("%d\n",B);
        }
    }
    return 0;
}