2016 ACM/ICPC Asia Regional Dalian Online 1002/HDU 5869

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：

 /******************************
 code by drizzle
 blog: www.cnblogs.com/hsd-/
 ^ ^    ^ ^
  O      O
 ******************************/
 #include<bits/stdc++.h>
 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<map>
 #include<algorithm>
 #include<queue>
 #define LL __int64
 #define pii pair<int,int>
 #define MP make_pair
 const int N=;
 using namespace std;
 int gcd(int a,int b)
 {
     return b== ? a : gcd(b,a%b);
 }
 int n,q,a[N],ans[N];
 vector<pii> G[N];
 struct QQ{
   int l,r,id;
   bool operator < (const QQ &a) const
   {
        return a.r>r;
   }
 }Q[N];
 int C[N],vis[N];
 void update (int x,int c)
 {
     for(int i=x;i<N;i+=i&(-i)) C[i]+=c;
 }
 int ask(int x)
 {
     int s=;
     for(int i=x;i;i-=i&(-i)) s+=C[i];
     return s;
 }
 int main()
 {
     while(scanf("%d %d",&n,&q)!=EOF)
     {
         for(int i=;i<=n;i++)
             scanf("%d",&a[i]);
         for(int i=;i<=n;i++)
             G[i].clear();
         for(int i=;i<=n;i++)
         {
             int x=a[i];
             int y=i;
             for(int j=;j<G[i-].size();j++)
             {
                 int res=gcd(x,G[i-][j].first);
                 if(x!=res)
                 {
                     G[i].push_back(MP(x,y));
                     x=res;
                     y=G[i-][j].second;
                 }
             }
             G[i].push_back(MP(x,y));
         }
         memset(C,,sizeof(C));
         memset(vis,,sizeof(vis));
         for(int i=;i<=q;i++)
         {
             scanf("%d %d",&Q[i].l,&Q[i].r);
             Q[i].id=i;
         }
         sort(Q+,Q+q+);
         for(int R=,i=;i<=q;i++)
         {
             while(R<Q[i].r)
             {
                 R++;
             for(int j=;j<G[R].size();j++)
             {
                 int res=G[R][j].first;
                 int ids=G[R][j].second;
                 if(vis[res])
                     update(vis[res],-);
                 vis[res]=ids;
                 update(vis[res],);
             }
             }
             ans[Q[i].id]=ask(R)-ask(Q[i].l-);
         }
         for(int i=;i<=q;i++)
             cout<<ans[i]<<endl;
     }
     return ;
 }