【Codeforces Round 271 (Div 2)F】【贪心线段树】Ant colony 区间段内是其他所有数因子的数的个数

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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 s_i.

In order to make his dinner more interesting, Mole organizes a version of «Hunger Games» for the ants. He chooses two numbers l andr (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 s_i divides s_j (also, ant j gets one battle point only if s_j divides s_i).

After all fights have been finished, Mole makes the ranking. An ant i, with v_i battle points obtained, is going to be freed only if v_i = 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 [l_i, r_i] 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, ..., s_n (1 ≤ s_i ≤ 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 l_i and r_i (1 ≤ l_i ≤ r_i ≤ 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 [l_i, r_i].

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.

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1,class T2>inline void gmax(T1 &a,T2 b){if(b>a)a=b;}
template <class T1,class T2>inline void gmin(T1 &a,T2 b){if(b<a)a=b;}
const int N=0,M=0,Z=1e9+7,ms63=1061109567;
struct A
{int l,r;int gcd,minv,num;
}a[1<<18];
int n,m;
int l,r;
int GCD,MINV,NUM;
int gcd(int x,int y)
{return y==0?x:gcd(y,x%y);
}
void pushup(int o)
{a[o].gcd=gcd(a[ls].gcd,a[rs].gcd);a[o].minv=min(a[ls].minv,a[rs].minv);a[o].num=0;if(a[o].minv==a[ls].minv)a[o].num+=a[ls].num;if(a[o].minv==a[rs].minv)a[o].num+=a[rs].num;
}
void build(int o,int l,int r)
{a[o].l=l;a[o].r=r;if(a[o].l==a[o].r){scanf("%d",&a[o].gcd);a[o].minv=a[o].gcd;a[o].num=1;return;}int m=(l+r)>>1;build(ls,l,m);build(rs,m+1,r);pushup(o);
}
void get(int o,int l,int r)
{if(a[o].l==l&&a[o].r==r){if(GCD==-1)GCD=a[o].gcd;else GCD=gcd(GCD,a[o].gcd);if(a[o].minv<MINV){MINV=a[o].minv;NUM=a[o].num;}else if(a[o].minv==MINV){NUM+=a[o].num;}return;}int m=(a[o].l+a[o].r)>>1;if(r<=m)get(ls,l,r);else if(l>m)get(rs,l,r);else{get(ls,l,m);get(rs,m+1,r);}
}
int main()
{while(~scanf("%d",&n)){build(1,1,n);scanf("%d",&m);for(int i=1;i<=m;++i){scanf("%d%d",&l,&r);MINV=1e9;NUM=0;GCD=-1;get(1,l,r);if(GCD%MINV==0)printf("%d\n",r-l+1-NUM);else printf("%d\n",r-l+1);}}return 0;
}
/*
【trick&&吐槽】【题意】
给你一个长度为n（1e5）的数列。
每个数的数值都在[1,1e9]范围。
然后有m（1e5）个询问。
对于每个询问，问你区间为[l,r]内，有多少个数，是这个区间内其他所有数的因子【类型】
贪心 线段树【分析】
呀嘿嘿，这题被我秒掉了。很显然，有一个贪心结论，如果一个数，是区间内所有数的因子的话。
那么必要条件是，
1，这个数是这区间段最小的数
2，这个数是这区间段内所有数的因子，也就是说，这个数是这区间段所有数gcd的因子。于是，我们只要求出区间最小数和（同时也记录个数），区间gcd，然后就可以AC这道题啦。【时间复杂度&&优化】
O（mlogn）*/

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