Hdu-5869 Different GCD Subarray Query(区间不同值离线算法)

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 31 3 4 6 93 52 51 5

 

Sample Output

666

 

Source

2016 ACM/ICPC Asia Regional Dalian Online

 

题意：给n个数，m个询问，每次问[l,r]区间内的所有子区间的GCD的不同值有多少个。

分析：区间不同值有一种离线的做法，考虑到以每个数为后缀的所有区间最多有log(n)个不同值，我们可以预处理出以每个数为后缀的不同gcd值，然后按照从左到右的顺序处理每个询问，当区间的右端固定时，我们用树状数组记下每种答案左端最晚出现的位置，这样对于每个询问直接用r结尾的总方案数减去(l-1)结尾的总方案数来得到答案。

#include<iostream>#include<string>#include<algorithm>#include<cstdlib>#include<cstdio>#include<set>#include<map>#include<vector>#include<cstring>#include<stack>#include<cmath>#include<queue>#define INF 0x3f3f3f3f#define N 100005using namespace std; typedef pair<int,int> pii;typedef long long ll;int n,q,ans[N],f[N],a[N],last[N*10];vector<int> G[N];int _gcd(int a,int b){   if(b == 0) return a;   a %= b;   return _gcd(b,a);}int lowbit(int x){return x & (-x);}void Insert(int k,int x){while(k <= n){f[k] += x;k += lowbit(k);}}int Find(int k){int ans = 0;while(k){ans += f[k];k -= lowbit(k);}return ans;}struct thing{int l,r;}ask[N];int main(){while(~scanf("%d%d",&n,&q)){memset(last,0,sizeof(last));memset(f,0,sizeof(f));for(int i = 1;i <= n;i++) {scanf("%d",&a[i]);G[i].clear();}for(int i = 1;i <= q;i++) {scanf("%d%d",&ask[i].l,&ask[i].r);G[ask[i].r].push_back(i);}vector<pii> pre;for(int i = 1;i <= n;i++){vector<pii> tmp;pre.push_back(make_pair(a[i],i));for(pii v : pre){if(tmp.size() && _gcd(v.first,a[i]) == tmp[tmp.size()-1].first) tmp[tmp.size()-1] = make_pair(_gcd(v.first,a[i]),v.second);else tmp.push_back(make_pair(_gcd(v.first,a[i]),v.second));}for(pii v : tmp){if(!last[v.first] || last[v.first] < v.second){if(last[v.first]) Insert(last[v.first],-1);last[v.first] = v.second;Insert(last[v.first],1);}}pre = tmp;for(int v : G[i]) ans[v] = Find(ask[v].r) - Find(ask[v].l-1);}for(int i = 1;i <= q;i++) printf("%d\n",ans[i]);}}