Codeforces Round #271 (Div. 2) F题 Ant colony（线段树求区间gcd）

F. Ant colony

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mole is hungry again. He found one ant colony, consisting of n ants, ordered in a row. Each ant i (1 ≤ i ≤ n) has a strength si.

In order to make his dinner more interesting, Mole organizes a version of «Hunger Games» for the ants. He chooses two numbers l and r(1 ≤ l ≤ r ≤ n) and each pair of ants with indices between l and r (inclusively) will fight. When two ants i and j fight, ant i gets one battle point only if si divides sj (also, ant j gets one battle point only if sj divides si).

After all fights have been finished, Mole makes the ranking. An ant i, with vi battle points obtained, is going to be freed only if vi = r - l, or in other words only if it took a point in every fight it participated. After that, Mole eats the rest of the ants. Note that there can be many ants freed or even none.

In order to choose the best sequence, Mole gives you t segments [li, ri] and asks for each of them how many ants is he going to eat if those ants fight.

Input

The first line contains one integer n (1 ≤ n ≤ 105), the size of the ant colony.

The second line contains n integers s1, s2, ..., sn (1 ≤ si ≤ 109), the strengths of the ants.

The third line contains one integer t (1 ≤ t ≤ 105), the number of test cases.

Each of the next t lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n), describing one query.

Output

Print to the standard output t lines. The i-th line contains number of ants that Mole eats from the segment [li, ri].

Sample test(s)

input

5
1 3 2 4 2
4
1 5
2 5
3 5
4 5

output

4
4
1
1

Note

In the first test battle points for each ant are v = [4, 0, 2, 0, 2], so ant number 1 is freed. Mole eats the ants 2, 3, 4, 5.

In the second test case battle points are v = [0, 2, 0, 2], so no ant is freed and all of them are eaten by Mole.

In the third test case battle points are v = [2, 0, 2], so ants number 3 and 5 are freed. Mole eats only the ant 4.

In the fourth test case battle points are v = [0, 1], so ant number 5 is freed. Mole eats the ant 4.

题意：已给数列，给定一个区间，求区间内的某个数x可以整除区间内所有值，求此x的个数

显然个数就是区间内gcd（al，al+1，al+2，...，ar-1，ar）的个数

求任何区间的gcd，可以用线段树解决

求任何区间内某个数的个数可以：先用pair记录值和index，再排序用二分确定上下界，单次复杂度是Ologn

#include
#include
#include
#include
using namespace std;

#define root 1,n,1
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

const int N = 1e5+100;
typedef pairpii;
int d[N<<2];
int a[N];
pii b[N];
int gcd(int a,int b)
{
    if(b==0) return a;
    else return gcd(b,a%b);
}
void pushup(int rt)
{
    d[rt]=gcd(d[rt<<1],d[rt<<1|1]);
}
void build(int l,int r,int rt)
{
    if(l==r) {d[rt]=a[l];return;}
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    pushup(rt);
}
int query(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R) return d[rt];
    int m=(l+r)>>1;
    int x,y;
    x=y=0;
    if(L<=m) x=query(L,R,lson);
    if(R>m) y=query(L,R,rson);
    return gcd(x,y);
}

int main()
{
    int n,t;
    cin>>n;
    for(int i=1;i<=n;i++) cin>>a[i],b[i].first=a[i],b[i].second=i;
    build(root);
    sort(b+1,b+1+n);
    cin>>t;
    for(int i=1;i<=t;i++)
    {
        int l,r;
        scanf("%d%d",&l,&r);
        int gd=query(l,r,root);
        int fr=lower_bound(b+1,b+n+1,pii(gd,l))-(b+1);
        int ba=lower_bound(b+1,b+1+n,pii(gd,r+1))-(b+1);
        printf("%d\n",(r-l+1)-(ba-fr));
    }

}