Different GCD Subarray Query
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
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
1 3 4 6 9
3 5
2 5
1 5
Sample Output
6
6
6
6
6
分析：对于每个点来说，向前的gcd很少，可以预处理出gcd出现的最后的左端点；
然后离线查询右端点，每走到一个点，更新gcd出现的左端点，然后树状数组询问答案即可；
代码：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=1e5+;
using namespace std;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}
int n,m,k,t,a[maxn],c[maxn],pr[maxn*],ans[maxn];
vector<pii >b[maxn];
vector<pii >q[maxn];
void add(int x,int y)
{
for(int i=x;i<=n;i+=(i&(-i)))
c[i]+=y;
}
int get(int x)
{
int ret=;
for(int i=x;i;i-=i&(-i))
ret+=c[i];
return ret;
}
void init()
{
for(int i=;i<=n;i++)
{
int x=a[i],y=i;
b[i].pb(mp(x,y));
for(int j=;j<b[i-].size();j++)
{
int now=gcd(b[i-][j].fi,a[i]),pos=b[i-][j].se;
if(now!=x)
{
b[i].pb(mp(now,pos));
x=now;
}
}
}
}
int main()
{
int i,j;
while(~scanf("%d%d",&n,&m))
{
rep(i,,n)scanf("%d",&a[i]),b[i].clear(),q[i].clear();
memset(c,,sizeof c);
memset(pr,,sizeof pr);
init();
rep(i,,m)
{
int x,y;
scanf("%d%d",&x,&y);
q[y].pb(mp(x,i));
}
rep(i,,n)
{
for(j=;j<b[i].size();j++)
{
int x=b[i][j].fi,pos=b[i][j].se;
if(pr[x])add(pr[x],-);
add(pos,);
pr[x]=pos;
}
for(int j=;j<q[i].size();j++)
{
int x=q[i][j].fi,y=q[i][j].se;
ans[y]=get(i)-(x-==?:get(x-));
}
}
rep(i,,m)printf("%d\n",ans[i]);
}
//system("Pause");
return ;
}