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

 1 //离线处理-树状数组
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cstdio>
 5 #include<cmath>
 6 #include<cstdlib>
 7 #include<algorithm>
 8 #include<queue>
 9 #include<vector>
10 #include<utility>
11 using namespace std;
12 const int maxn=1e5+10;
13 int Arr[maxn];//存数据一开始的值
14 int N,Q;
15 int pos[maxn*10]; //记录gcd的位置
16 vector<pair<int ,int> >querys[maxn];
17 vector<pair<int, int> >gcds[maxn];
18 int ans[maxn];
19 int treeArr[maxn]; //树状数组
20
21 int gcd(int a,int b){ //gcd求最大公约数
22     if(!(a%b))return b;
23     else return gcd(b,a%b);
24 }
25
26 void init(){ //初始化
27     int tmp,ts;
28     for(int i=1;i<=N;i++){
29         tmp=Arr[i];
30         ts=i;                                     //固定右端点，向左遍历
31         for(int j=0;j<gcds[i-1].size();j++){
32             int tmpgcd=gcd(tmp,gcds[i-1][j].first);
33             if(tmpgcd!=tmp){  //如果gcd发生变化
34                 gcds[i].push_back(make_pair(tmp,ts)); //first存gcd，second存gcd的左端点ts
35                 ts=gcds[i-1][j].second; //上一个gcd的右端点就是下一个gcd的左端点
36                 tmp=tmpgcd;
37             }
38         }
39         gcds[i].push_back(make_pair(tmp,ts)); //将与上一个的最左端的gcd存起来
40     }
41     return;
42 }
43
44 int lowbit(int x){  //取最低位的1
45     return x&(-x);
46 }
47
48 void Add(int cur,int num){ //单点更新
49     for(int i=cur;i<=N;i+=lowbit(i))
50         treeArr[i]+=num;   //由叶子节点向上更新树状数组
51     return;
52 }
53
54 int Query(int cur){ //区间查询  从右端点往左加二进制最低位1的
55     int sum=0;
56     for(int i=cur;i>0;i-=lowbit(i))
57         sum+=treeArr[i];
58     return sum;
59 }
60
61 void Solve(){
62     memset(pos,0,sizeof(pos));
63     memset(treeArr,0,sizeof(treeArr));
64     for(int i=1;i<=N;i++){
65         for(int j=0;j<gcds[i].size();j++){
66             if(pos[gcds[i][j].first]){ //如果标记已经存在，就将标记去掉，所以执行单点更新操作
67                 Add(pos[gcds[i][j].first],-1);
68             }
69             Add(gcds[i][j].second,1);//一个新的不同的gcd，从叶子节点开始更新个数+1
70             pos[gcds[i][j].first]=gcds[i][j].second;//记录该gcd的右端点
71         }
72         for(int j=0;j<querys[i].size();j++){
73             ans[querys[i][j].second]=Query(i)-Query(querys[i][j].first-1);//区间右端点-区间左区间
74         }
75     }
76     for(int i=1;i<=Q;i++)
77         printf("%d\n",ans[i]);
78
79     return;
80 }
81 int main(){
82     //reopen("input","r",stdin);
83     while(~scanf("%d%d",&N,&Q)){
84         for(int i=1;i<=N;i++){
85             scanf("%d",&Arr[i]); //将数据存在Arr数组中
86             querys[i].clear(); //清空
87             gcds[i].clear(); //清空
88         }
89         init();
90         int tmp1,tmp2;
91         for(int i=1;i<=Q;i++){
92             scanf("%d%d",&tmp1,&tmp2);
93             querys[tmp2].push_back(make_pair(tmp1,i)); //将tmp1与i成对插入vector的tmp2下标里
94         }
95         Solve();
96     }
97     return 0;
98 }