Description
You are given a list of numbers A1A2 .. 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 A1A2 .. 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
题目大意：给一个n的序列，若干查询（L,R）。输出F(R)的值。
函数关系：
F(L)=A(L)
F(L+1)=A(L+1)
F(X)=F(X-1)+F(X-2)*A(X) (X-L>=2)
因此每段（L,R）区间
| F(R) | | 1 A(R)|*| 1 A(R-1)| *......* | 1 A(L+2)|*| A(L+1)|
|F(R-1)| = | 1 0 | | 1 0 | ...... | 1 0 | | A(L) |
每次查询(L+2,R)区间的矩阵乘积再稍微处理一下就行了（当R-L>1时）。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 6 #define Mod 1000000007 7 typedef long long LL; 8 const int maxn=100010; 9 LL a[maxn]; 10 11 struct node 12 { 13 LL mat[2][2]; 14 void set(int x)//初始化矩阵 15 { 16 mat[0][0]=1; 17 mat[0][1]=a[x]%Mod; 18 mat[1][0]=1; 19 mat[1][1]=0; 20 21 } 22 }; 23 24 struct IntervalTree 25 { 26 int left,right; 27 node matrix; 28 }f[maxn<<2]; 29 30 31 node mat_mul_mod(node A,node B)//矩阵乘法取模 32 { 33 node ret; 34 ret.mat[0][0]=(A.mat[0][0]*B.mat[0][0]%Mod+A.mat[0][1]*B.mat[1][0]%Mod)%Mod; 35 ret.mat[0][1]=(A.mat[0][0]*B.mat[0][1]%Mod+A.mat[0][1]*B.mat[1][1]%Mod)%Mod; 36 ret.mat[1][0]=(A.mat[1][0]*B.mat[0][0]%Mod+A.mat[1][1]*B.mat[1][0]%Mod)%Mod; 37 ret.mat[1][1]=(A.mat[1][0]*B.mat[0][1]%Mod+A.mat[1][1]*B.mat[1][1]%Mod)%Mod; 38 return ret; 39 } 40 41 void bulid(int left,int right,int i)//建树 42 { 43 int mid; 44 f[i].left=left; 45 f[i].right=right; 46 if(left==right) 47 { 48 f[i].matrix.set(left); 49 return; 50 } 51 mid=(left+right)>>1; 52 bulid(left,mid,i<<1); 53 bulid(mid+1,right,i<<1|1); 54 f[i].matrix=mat_mul_mod(f[i<<1|1].matrix,f[i<<1].matrix); 55 return ; 56 } 57 58 node query(int left,int right,int i)//查询 59 { 60 int mid; 61 if(f[i].left==left && f[i].right==right) return f[i].matrix; 62 mid=(f[i].left+f[i].right)>>1; 63 if(right<=mid) return query(left,right,i<<1); 64 else if(left>mid) return query(left,right,i<<1|1); 65 else return mat_mul_mod(query(mid+1,right,i<<1|1),query(left,mid,i<<1)); 66 } 67 68 int main() 69 { 70 int t,n,m,i,lp,rp; 71 scanf("%d",&t); 72 while(t--) 73 { 74 scanf("%d %d",&n,&m); 75 for(i=1;i<=n;i++) 76 scanf("%lld",&a[i]); 77 bulid(1,n,1); 78 while(m--) 79 { 80 scanf("%d %d",&lp,&rp); 81 if(lp==rp || lp+1==rp) 82 { 83 printf("%lld ",a[rp]%Mod); 84 continue; 85 } 86 node temp=query(lp+2,rp,1); 87 printf("%lld ",(temp.mat[0][0]*a[lp+1]%Mod+temp.mat[0][1]*a[lp]%Mod)%Mod); 88 } 89 } 90 return 0; 91 }