zoj Calculate the Function

Calculate the Function
Time Limit: 2 Seconds      Memory Limit: 65536 KB

You are given a list of numbers A1 A2 .. AN and M queries. For the i-th query:

 

  • The query has two parameters Li and Ri.
  • The query will define a function Fi(x) on the domain [Li, Ri] ∈ Z.
  • Fi(Li) = ALi
  • Fi(Li + 1) = A(Li + 1)
  • for all x >= Li + 2, Fi(x) = Fi(x - 1) + Fi(x - 2) × Ax

 

You task is to calculate Fi(Ri) 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 .. AN (1 <= Ai <= 1000000000).

The next M lines, each line is a query with two integer parameters Li, Ri (1 <= Li <= Ri <= 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

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

 

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<cstring>
 4 #include<cstdlib>
 5 using namespace std;
 6 
 7 typedef long long LL;
 8 
 9 int mod=1000000007;
10 int ax[100002];
11 struct node
12 {
13     LL a,b,c,d;
14     int l,r;
15 } f[400008];
16 
17 void build(int l,int r,int n)
18 {
19     int mid=(l+r)/2;
20     f[n].l=l;
21     f[n].r=r;
22     if(l==r)
23     {
24         f[n].a=0;
25         f[n].b=ax[l];
26         f[n].c=1;
27         f[n].d=1;
28         return;
29     }
30     build(l,mid,n*2);
31     build(mid+1,r,n*2+1);
32     f[n].a=((f[n<<1].a*f[(n<<1)+1].a)%mod+f[n<<1].b*f[(n<<1)+1].c)%mod;
33     f[n].b=((f[n<<1].a*f[(n<<1)+1].b)%mod+f[n<<1].b*f[(n<<1)+1].d)%mod;
34     f[n].c=((f[n<<1].c*f[(n<<1)+1].a)%mod+f[n<<1].d*f[(n<<1)+1].c)%mod;
35     f[n].d=((f[n<<1].c*f[(n<<1)+1].b)%mod+f[n<<1].d*f[(n<<1)+1].d)%mod;
36 }
37 node serch1(int l,int r,int n)
38 {
39     int mid=(f[n].l+f[n].r)/2;
40     node n1,n2,n3;
41 
42     if(f[n].l==l && f[n].r==r)
43     {
44         return f[n];
45     }
46     if(mid>=r)
47         return serch1(l,r,n*2);
48     else if(mid<l)
49         return serch1(l,r,n*2+1);
50     else
51     {
52         n1=serch1(l,mid,n*2);
53         n2=serch1(mid+1,r,n*2+1);
54         n3.a=((n1.a*n2.a)%mod+(n1.b*n2.c)%mod)%mod;
55         n3.b=((n1.a*n2.b)%mod+(n1.b*n2.d)%mod)%mod;
56         n3.c=((n1.c*n2.a)%mod+(n1.d*n2.c)%mod)%mod;
57         n3.d=((n1.c*n2.b)%mod+(n1.d*n2.d)%mod)%mod;
58     }
59     return n3;
60 }
61 int main()
62 {
63     int T;
64     int i,j,n,m,x,y;
65     LL sum1;
66     node cur;
67     scanf("%d",&T);
68     while(T--)
69     {
70         scanf("%d%d",&n,&m);
71         for(i=1; i<=n; i++)
72             scanf("%d",&ax[i]);
73         build(1,n,1);
74         for(j=1; j<=m; j++)
75         {
76             scanf("%d%d",&x,&y);
77             if(y-x<2)
78             {
79                 printf("%d\n",ax[y]);
80             }
81             else
82             {
83                 cur=serch1(x+2,y,1);
84                 sum1=((cur.b*ax[x])%mod+(cur.d*ax[x+1])%mod)%mod;
85                 printf("%lld\n",sum1);
86             }
87         }
88     }
89     return 0;
90 }

 

posted @ 2014-04-08 08:46  芷水  阅读(264)  评论(0)  编辑  收藏  举报