ZOJ 3772 Calculate the Function 线段树+矩阵

最新推荐文章于 2014-10-01 22:57:29 发布

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原文链接：http://www.cnblogs.com/xiong-/p/3752965.html

本文介绍了一种计算特定序列函数的方法，并通过矩阵乘法优化了查询效率。

Calculate the Function

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

Submit Status

Appoint description: System Crawler (2014-04-09)

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

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\n",a[rp]%Mod);
84                 continue;
85             }
86             node temp=query(lp+2,rp,1);
87             printf("%lld\n",(temp.mat[0][0]*a[lp+1]%Mod+temp.mat[0][1]*a[lp]%Mod)%Mod);
88         }
89     }
90     return 0;
91 }