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

Sample Output

Source

2016 ACM/ICPC Asia Regional Dalian Online

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题意：输入N和Q，表示有N个数的一个序列，Q次询问，每次输入 l 和 r 表示一个区间，求这个区间不同的最大公倍数的个数（由这个区间的子区间得到）；

思路：对数列进行GCD离散处理（~我也是才知道还有这样的离散~） 

for(int i=1;i<=N;i++){int tot=a[i],pos=i;for(int j=0;j<v[i-1].size();j++){int  r=__gcd(a[i],v[i-1][j].first);if(tot!=r){v[i].push_back(make_pair(tot,pos));tot=r;  pos=v[i-1][j].second;}}v[i].push_back(make_pair(tot,pos));}

然后对Q次询问离线处理，先输入Q次询问的区间，然后按右端点从小到大排序，i从1~N循环，当i==node[len].r  则 ans[node[len].id]=Sum(i)-Sum(node[len].l-1) ;

可以方便快速的用树状数组处理；

代码如下：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <map>
#include <vector>
using namespace std;
int a[100005];
int c[1000005];
int vis[1000005];
int sum[100005];
struct Node
{int l,r;int id;
}node[100005];
bool cmp(const Node s1,const Node s2)
{return s1.r<s2.r;
}
vector<pair<int,int> > v[100005];int __gcd(int x,int y)
{int r=x%y;x=y;y=r;if(r==0) return x;return __gcd(x,y);
}
int Lowbit(int t)
{return t&(t^(t-1));
}
int Sum(int x)
{int sum = 0;while(x > 0){sum += c[x];x -= Lowbit(x);}return sum;
}
void add(int li,int t)
{while(li<=1000005){c[li]+=t;li=li+Lowbit(li);}
}
int main()
{int N,Q;while(scanf("%d%d",&N,&Q)!=EOF){for(int i=1;i<=N;i++) scanf("%d",&a[i]);for(int i=1;i<=N;i++){int tot=a[i],pos=i;for(int j=0;j<v[i-1].size();j++){int  r=__gcd(a[i],v[i-1][j].first);if(tot!=r){v[i].push_back(make_pair(tot,pos));tot=r;  pos=v[i-1][j].second;}}v[i].push_back(make_pair(tot,pos));}for(int i=0;i<Q;i++)scanf("%d%d",&node[i].l,&node[i].r),node[i].id=i;sort(node,node+Q,cmp);memset(c,0,sizeof(c));memset(vis,0,sizeof(vis));int len=0;for(int i=1;i<=N;i++){for(int j=0;j<v[i].size();j++){int s1=v[i][j].first;int s2=v[i][j].second;if(vis[s1]){add(vis[s1],-1);}vis[s1]=s2;add(s2,1);}while(node[len].r==i){sum[node[len].id]=Sum(i)-Sum(node[len].l-1);len++;}}for(int i=0;i<Q;i++)printf("%d\n",sum[i]);for(int i=0;i<=N;i++)v[i].clear();}return 0;
}