F. Ant colony（Codeforces Round #271)

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.

如果区间[l,r]内所有数的gcd！=该区间内最小的数minv，则对于区间内的每个数都至少有一个数不是其倍数，那么每个数的v值都不可能等于r-l，所以所有的蚂蚁都要被吃掉（统统吃掉），答案是r-l+1。

现在如果gcd=minv，那么对于所有的等于minv的数，所有的数都是他的倍数，对于所有的不等于minv的数，至少有一个不是他的倍数（比如minv），所以答案就是r-l+1-minv的个数。

代码：

//156ms
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn=100000+1000;
const int inf = 0x3f3f3f3f ;
int gcd ( int a , int b ) {
return b ? gcd ( b , a % b ) : a ;
}
struct node
{
int minv;
int num;
int gc;
}tree[maxn<<2];
void build(int rt,int l,int r)
{
if(l==r)
{
scanf("%d",&tree[rt].minv);
tree[rt].num=1;
tree[rt].gc=tree[rt].minv;
return;
}
int mid=(l+r)>>1;
build(rt<<1,l,mid);
build(rt<<1|1,mid+1,r);
if(tree[rt<<1].minv<tree[rt<<1|1].minv)
{
tree[rt].minv=tree[rt<<1].minv;
tree[rt].num=tree[rt<<1].num;
}
else if(tree[rt<<1].minv>tree[rt<<1|1].minv)
{
tree[rt].minv=tree[rt<<1|1].minv;
tree[rt].num=tree[rt<<1|1].num;
}
else
{
tree[rt].minv=tree[rt<<1].minv;
tree[rt].num=tree[rt<<1].num+tree[rt<<1|1].num;
}
tree[rt].gc=gcd(tree[rt<<1].gc,tree[rt<<1|1].gc);
}
int minv,min_num,mgcd;
void query(int L,int R,int rt,int l,int r)
{
if(L<=l&&R>=r)
{
if(minv>tree[rt].minv)
{
minv=tree[rt].minv;
min_num=tree[rt].num;
}
else if(minv==tree[rt].minv)
{
min_num+=tree[rt].num;
}
mgcd=gcd(mgcd,tree[rt].gc);
return;
}
int mid=(l+r)>>1;
if(L<=mid)
{
query(L,R,rt<<1,l,mid);
}
if(R>mid)
{
query(L,R,rt<<1|1,mid+1,r);
}
}
int main()
{
int n;
scanf("%d",&n);
build(1,1,n);
int m,l,r;
scanf("%d",&m);
while(m--)
{
scanf("%d%d",&l,&r);
minv=inf;
min_num=0;
mgcd=0;
query(l,r,1,1,n);
if(minv==mgcd)
{
printf("%d\n",r-l+1-min_num);
}
else
printf("%d\n",r-l+1);
}
return 0;
}