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=1000006;
using namespace std;
int gcd(int a,int b)
{
    return b==0 ? 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=0;
    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=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(int i=0;i<=n;i++)
            G[i].clear();
        for(int i=1;i<=n;i++)
        {
            int x=a[i];
            int y=i;
            for(int j=0;j<G[i-1].size();j++)
            {
                int res=gcd(x,G[i-1][j].first);
                if(x!=res)
                {
                    G[i].push_back(MP(x,y));
                    x=res;
                    y=G[i-1][j].second;
                }
            }
            G[i].push_back(MP(x,y));
        }
        memset(C,0,sizeof(C));
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=q;i++)
        {
            scanf("%d %d",&Q[i].l,&Q[i].r);
            Q[i].id=i;
        }
        sort(Q+1,Q+q+1);
        for(int R=0,i=1;i<=q;i++)
        {
            while(R<Q[i].r)
            {
                R++;
            for(int j=0;j<G[R].size();j++)
            {
                int res=G[R][j].first;
                int ids=G[R][j].second;
                if(vis[res])
                    update(vis[res],-1);
                vis[res]=ids;
                update(vis[res],1);
            }
            }
            ans[Q[i].id]=ask(R)-ask(Q[i].l-1);
        }
        for(int i=1;i<=q;i++)
            cout<<ans[i]<<endl;
    }
    return 0;
}