zoj Calculate the Function

Calculate the Function

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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You are given a list of numbers A₁ A₂ .. 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_Li
• F_i(L_i + 1) = A_{(Li + 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 A₁ A₂ .. 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

1
4 7
1 2 3 4
1 1
1 2
1 3
1 4
2 4
3 4
4 4

Sample Output

1
2
5
13
11
4
4

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Author: CHEN, Weijie
Source: The 14th Zhejiang University Programming Contest

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
using namespace std;

typedef long long LL;

int mod=1000000007;
int ax[100002];
struct node
{
    LL a,b,c,d;
    int l,r;
} f[400008];

void build(int l,int r,int n)
{
    int mid=(l+r)/2;
    f[n].l=l;
    f[n].r=r;
    if(l==r)
    {
        f[n].a=0;
        f[n].b=ax[l];
        f[n].c=1;
        f[n].d=1;
        return;
    }
    build(l,mid,n*2);
    build(mid+1,r,n*2+1);
    f[n].a=((f[n<<1].a*f[(n<<1)+1].a)%mod+f[n<<1].b*f[(n<<1)+1].c)%mod;
    f[n].b=((f[n<<1].a*f[(n<<1)+1].b)%mod+f[n<<1].b*f[(n<<1)+1].d)%mod;
    f[n].c=((f[n<<1].c*f[(n<<1)+1].a)%mod+f[n<<1].d*f[(n<<1)+1].c)%mod;
    f[n].d=((f[n<<1].c*f[(n<<1)+1].b)%mod+f[n<<1].d*f[(n<<1)+1].d)%mod;
}
node serch1(int l,int r,int n)
{
    int mid=(f[n].l+f[n].r)/2;
    node n1,n2,n3;

    if(f[n].l==l && f[n].r==r)
    {
        return f[n];
    }
    if(mid>=r)
        return serch1(l,r,n*2);
    else if(mid<l)
        return serch1(l,r,n*2+1);
    else
    {
        n1=serch1(l,mid,n*2);
        n2=serch1(mid+1,r,n*2+1);
        n3.a=((n1.a*n2.a)%mod+(n1.b*n2.c)%mod)%mod;
        n3.b=((n1.a*n2.b)%mod+(n1.b*n2.d)%mod)%mod;
        n3.c=((n1.c*n2.a)%mod+(n1.d*n2.c)%mod)%mod;
        n3.d=((n1.c*n2.b)%mod+(n1.d*n2.d)%mod)%mod;
    }
    return n3;
}
int main()
{
    int T;
    int i,j,n,m,x,y;
    LL sum1;
    node cur;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(i=1; i<=n; i++)
            scanf("%d",&ax[i]);
        build(1,n,1);
        for(j=1; j<=m; j++)
        {
            scanf("%d%d",&x,&y);
            if(y-x<2)
            {
                printf("%d
",ax[y]);
            }
            else
            {
                cur=serch1(x+2,y,1);
                sum1=((cur.b*ax[x])%mod+(cur.d*ax[x+1])%mod)%mod;
                printf("%lld
",sum1);
            }
        }
    }
    return 0;
}