HDU6395矩阵快速幂

题意：

给出A,B,C,D,P,n，求Fn
题解：
把P/n当做一个常数用等值区间，区间变矩阵变，把区间找出来快速幂直接搞定

#include<bits/stdc++.h>
using namespace std;
#define lowerbit(x) ((-x)&(x))
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define MS(x,y) memst(x,y,sizeof(x))
typedef long long ll;
const int maxn=4;
const int mod=1000000007;
struct Matrix{
    ll f[maxn][maxn],n,m;
    Matrix(){memset(f,0,sizeof f);}
    void O(){memset(f,0,sizeof f);}
    void E(){fo(i,1,m)f[i][i]=1;}
};
inline Matrix operator *(Matrix a,Matrix b){
    Matrix c;c.n=a.n;c.m=b.m;
    fo(i,1,c.n)fo(j,1,c.m)fo(k,1,b.n)
    c.f[i][j]=(c.f[i][j]+a.f[i][k]*b.f[k][j]%mod)%mod;
    return c;
}
inline Matrix power(Matrix a,ll p){
    Matrix c;c.n=a.n;c.m=a.m;c.E();
    while(p){
        if(p&1)c=c*a;
        a=a*a;
        p>>=1;
    }
    return c;
}
int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin>>t;
    while(t--){
        int A,B,C,D,P,n;
        cin>>A>>B>>C>>D>>P>>n;
        if(n==1){cout<<A<<endl;continue;}
        else if(n==2){cout<<B<<endl;continue;}
        Matrix K,ans,F;
        K.n=K.m=3;F.n=3;F.m=1;ans.n=ans.m=3;ans.E();
        K.f[1][1]=D;K.f[1][2]=C;
        K.f[2][1]=1;K.f[3][3]=1;
        F.f[1][1]=B;F.f[2][1]=A;F.f[3][1]=1;
        int l=3,r;
        for(;l<=n;l=r+1){
            int ak=P/l;
            if(ak==0)r=n;
            else r=min(n,P/ak);
            K.f[1][3]=ak;
            ans=power(K,r-l+1)*ans;
        }
        ans=ans*F;
        cout<<ans.f[1][1]<<endl;
    }
    return 0;
}