Different GCD Subarray Query

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 681    Accepted Submission(s): 240

Problem Description

This is a simple problem. The teacher gives Bob a list of problems about GCD (Greatest Common Divisor). After studying some of them, Bob thinks that GCD is so interesting. One day, he comes up with a new problem about GCD. Easy as it looks, Bob cannot figure it out himself. Now he turns to you for help, and here is the problem:
  
  Given an array a of N positive integers a1,a2,?aN−1,aN; a subarray of a is defined as a continuous interval between a1 and aN. In other words, ai,ai+1,?,aj−1,aj is a subarray of a, for 1≤i≤j≤N. For a query in the form (L,R), tell the number of different GCDs contributed by all subarrays of the interval [L,R].

Input

There are several tests, process till the end of input.
  
  For each test, the first line consists of two integers N and Q, denoting the length of the array and the number of queries, respectively. N positive integers are listed in the second line, followed by Q lines each containing two integers L,R for a query.

You can assume that 
  
    1≤N,Q≤100000 
    
   1≤ai≤1000000

Output

For each query, output the answer in one line.

Sample Input

5 3

1 3 4 6 9

3 5

2 5

1 5

Sample Output

6

6

6

Source

2016 ACM/ICPC Asia Regional Dalian Online

题意：

题解：

 1 /******************************
 2 code by drizzle
 3 blog: www.cnblogs.com/hsd-/
 4 ^ ^    ^ ^
 5  O      O
 6 ******************************/
 7 #include<bits/stdc++.h>
 8 #include<iostream>
 9 #include<cstring>
10 #include<cstdio>
11 #include<map>
12 #include<algorithm>
13 #include<queue>
14 #define LL __int64
15 #define pii pair<int,int>
16 #define MP make_pair
17 const int N=1000006;
18 using namespace std;
19 int gcd(int a,int b)
20 {
21     return b==0 ? a : gcd(b,a%b);
22 }
23 int n,q,a[N],ans[N];
24 vector<pii> G[N];
25 struct QQ{
26   int l,r,id;
27   bool operator < (const QQ &a) const
28   {
29        return a.r>r;
30   }
31 }Q[N];
32 int C[N],vis[N];
33 void update (int x,int c)
34 {
35     for(int i=x;i<N;i+=i&(-i)) C[i]+=c;
36 }
37 int ask(int x)
38 {
39     int s=0;
40     for(int i=x;i;i-=i&(-i)) s+=C[i];
41     return s;
42 }
43 int main()
44 {
45     while(scanf("%d %d",&n,&q)!=EOF)
46     {
47         for(int i=1;i<=n;i++)
48             scanf("%d",&a[i]);
49         for(int i=0;i<=n;i++)
50             G[i].clear();
51         for(int i=1;i<=n;i++)
52         {
53             int x=a[i];
54             int y=i;
55             for(int j=0;j<G[i-1].size();j++)
56             {
57                 int res=gcd(x,G[i-1][j].first);
58                 if(x!=res)
59                 {
60                     G[i].push_back(MP(x,y));
61                     x=res;
62                     y=G[i-1][j].second;
63                 }
64             }
65             G[i].push_back(MP(x,y));
66         }
67         memset(C,0,sizeof(C));
68         memset(vis,0,sizeof(vis));
69         for(int i=1;i<=q;i++)
70         {
71             scanf("%d %d",&Q[i].l,&Q[i].r);
72             Q[i].id=i;
73         }
74         sort(Q+1,Q+q+1);
75         for(int R=0,i=1;i<=q;i++)
76         {
77             while(R<Q[i].r)
78             {
79                 R++;
80             for(int j=0;j<G[R].size();j++)
81             {
82                 int res=G[R][j].first;
83                 int ids=G[R][j].second;
84                 if(vis[res])
85                     update(vis[res],-1);
86                 vis[res]=ids;
87                 update(vis[res],1);
88             }
89             }
90             ans[Q[i].id]=ask(R)-ask(Q[i].l-1);
91         }
92         for(int i=1;i<=q;i++)
93             cout<<ans[i]<<endl;
94     }
95     return 0;
96 }