HDU 6395 Sequence（矩阵快速幂）

Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 1407    Accepted Submission(s): 512   Problem Description Let us define a sequence as below   ⎧⎩⎨⎪⎪⎪⎪⎪⎪F1F2Fn===ABC⋅Fn−2+D⋅Fn−1+⌊Pn⌋   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   Source 2018 Multi-University Training Contest 7   Recommend chendu   |   We have carefully selected several similar problems for you:  6396 6395 6394 6393 6392

根据递推式求第n项，线性递推考虑矩阵快速幂，因为递推式中有向下取整，所以可以把n项拆成一段段的，每一段中的常数项相同，就可以分段矩阵快速幂，二分一直有些迷，膜了大佬之后才明白了。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
ll A,B,C,D,P,N,T,f1,f2;
struct matrix
{
    ll a[3][3];
    matrix()
    {
        memset(a,0,sizeof(a));
    }
    matrix operator*(matrix &p)
    {
        matrix ans;
        for(int i=0;i<3;i++)
        {
            for(int j=0;j<3;j++)
            {
                for(int k=0;k<3;k++)
                    ans.a[i][j]=(ans.a[i][j]+a[i][k]*p.a[k][j]);
                ans.a[i][j]%=mod;
            }
        }
        return ans;
    }
}U,X;
matrix quick_pow(matrix A,int b)
{
    matrix ans=U;
    while(b)
    {
        //printf("111\n");
        if(b&1) ans=ans*A;
        A=A*A;
        b>>=1;
    }
    return ans;
}
ll srch(ll i)
{
    ll key=P/i;
    ll l=i,r=N;
    while(l<r)
    {
        int mid=r-(r-l)/2;
        if(P/mid==key) l=mid;
        else if(P/mid<key)
        {
            r=mid-1;
        }
        else l=mid+1;
    }
    return l;
}
int main()
{
    U.a[0][0]=U.a[1][1]=U.a[2][2]=1;
    scanf("%lld",&T);
    while(T--)
    {
        scanf("%lld%lld%lld%lld%lld%lld",&A,&B,&C,&D,&P,&N);
        if(N==1)
        {
            printf("%lld\n",A);
            continue;
        }
        if(N==2)
        {
            printf("%lld\n",B);
            continue;
        }
        memset(X.a,0,sizeof(X.a));
        X.a[0][0]=D;
        X.a[0][1]=1;
        X.a[1][0]=C;
        X.a[2][2]=1;
        f1=B%mod;
        f2=A%mod;
        ll i=3;
        while(i<=N)
        {
            //printf("i=%lld\n",i);
            ll j=srch(i);
            X.a[2][0]=P/i;
            matrix ans=quick_pow(X,j-i+1);
            ll f11=(f1*ans.a[0][0]%mod+f2*ans.a[1][0]%mod+ans.a[2][0])%mod;
            ll f22=(f1*ans.a[0][1]+f2*ans.a[1][1]+ans.a[2][1])%mod;
            f1=f11;
            f2=f22;
            i=j+1;
            //printf("i=%lld %lld %lld\n",i,j,N);
        }
        printf("%lld\n",f1);
    }
    return 0;
}