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ZOJ 3772 Calculate the Function(矩陣線段樹）

編輯：關於C++

Description

You are given a list of numbers A₁A₂ .. A_N and M queries. For the i-th query:

The query has two parameters L_i and R_i.The query will define a function F_i(x) on the domain [L_i, R_i] ∈ Z.F_i(L_i) = A_{L_i}F_i(L_i + 1) = A_{(L_i + 1)}for all x >= L_i + 2, F_i(x) = F_i(x - 1) + F_i(x - 2) × A_x

You task is to calculate F_i(R_i) for each query. Because the answer can be very large, you should output the remainder of the answer divided by 1000000007.

Input

There are multiple test cases. The first line of input is an integer T indicates the number of test cases. For each test case:

The first line contains two integers N, M (1 <= N, M <= 100000). The second line contains N integers A₁A₂ .. A_N (1 <= A_i <= 1000000000).

The next M lines, each line is a query with two integer parameters L_i, R_i (1 <= L_i <= R_i <= N).

Output

For each test case, output the remainder of the answer divided by 1000000007.

Sample Input

Sample Output

題意：給你一段序列，每次問你一段區間構造的數列的最後一項。

分析：這題顯然不能暴力，正確做法就是將矩陣套到線段樹裡，每個節點

裡存的都是一個矩陣，然後求區間pushup時候利用矩陣乘法乘一次就行了

注意矩陣相乘的順序。

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int INF=0x3f3f3f3f;
typedef long long LL;
const int mod=1e9+7;
const int maxn=1e5+100;
LL a[maxn];
int t,n,m;
struct Matrix{
    LL mat[2][2];
    Matrix(){};
    Matrix(int x)
    {
        mat[0][0]=mat[1][0]=1;
        mat[0][1]=x;mat[1][1]=0;
    }
}A[maxn<<2];
Matrix mult(Matrix m1,Matrix m2)
{
    Matrix ans;
    for(int i=0;i<2;i++)
    {
        for(int j=0;j<2;j++)
        {
            ans.mat[i][j]=0;
            for(int k=0;k<2;k++)
                ans.mat[i][j]=(ans.mat[i][j]+m1.mat[i][k]*m2.mat[k][j])%mod;
        }
    }
    return ans;
}
void build(int rs,int l,int r)
{
    if(l==r)
    {
        A[rs]=Matrix(a[l]);
        return ;
    }
    int mid=(l+r)>>1;
    build(rs<<1,l,mid);
    build(rs<<1|1,mid+1,r);
    A[rs]=mult(A[rs<<1|1],A[rs<<1]);
}
Matrix query(int x,int y,int l,int r,int rs)
{
    if(l>=x&&r<=y)
        return A[rs];
    int mid=(l+r)>>1;
    if(y<=mid) return query(x,y,l,mid,rs<<1);
    if(x>mid) return query(x,y,mid+1,r,rs<<1|1);
    return mult(query(mid+1,y,mid+1,r,rs<<1|1),query(x,mid,l,mid,rs<<1));
}
int main()
{
    int l,r;
    scanf(%d,&t);
    while(t--)
    {
        Matrix hehe;
        scanf(%d%d,&n,&m);
        REPF(i,1,n)
          scanf(%lld,&a[i]);
        build(1,1,n);
        while(m--)
        {
            scanf(%d%d,&l,&r);
            if(r-l<=1)
                printf(%lld
,a[r]);
            else
            {
                hehe=query(l+2,r,1,n,1);
                LL ans=(hehe.mat[0][0]*a[l+1]%mod+hehe.mat[0][1]*a[l]%mod)%mod;
                printf(%lld
,ans);
            }
        }
    }
    return 0;
}