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].
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.
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
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 ;
}