题目描述：
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.
思路：
题目的意思是说，给一个数列，看里面有多少个数，这样的数可以被数列中的其他所有数整除。显然这个数就是数列的gcd啦！为什么呢？首先gcd可以满足条件，然后如果不是gcd，那就是gcd的因数，可是数列中的数如果有一个是gcd的因数那它小于等于gcd，而它又不可能比gcd小，只能相等。（为什么，它要是比gcd小，那它才会是gcd）。我怎么会有这么奇怪的想法怀疑它不是gcd (ｷ｀ﾟДﾟ´)!!
又因为是区间查询问题，整一个线段树来维护区间gcd，和等于gcd的数目。注意的是build函数里这么pushup，还有查询函数怎么统计结果。
pushup就是对一个节点求左右两个节点的gcd，如果左边的节点的gcd与这个gcd相等，统计数目加左边的相等数目，如果右边的等就再加右边的数目，不等就是零。
query函数求答案的时候要看一下当前区间答案来自哪里，是左区间，还是右区间，还是两边都有？分别处理一下就好。
这道题竟然连懒标记都没用，就是静态查询√
代码：
1 #include <iostream> 2 #define max_n 100005 3 using namespace std; 4 int n; 5 int t; 6 struct node 7 { 8 int num; 9 int gcd; 10 int id; 11 }tree[max_n<<2]; 12 int a[max_n]; 13 14 int GCD(int a,int b) 15 { 16 if(a<b) swap(a,b); 17 int r = a%b; 18 if(r==0) 19 { 20 return b; 21 } 22 return GCD(b,r); 23 } 24 void build(int id,int l,int r) 25 { 26 if(l==r) 27 { 28 tree[id].gcd = a[l]; 29 tree[id].num = 1; 30 return; 31 } 32 int mid = (l+r)>>1; 33 build(id<<1,l,mid); 34 build(id<<1|1,mid+1,r); 35 int gcd = GCD(tree[id<<1].gcd,tree[id<<1|1].gcd); 36 tree[id].num = 0; 37 tree[id].gcd = gcd; 38 if(tree[id<<1].gcd==gcd) 39 { 40 tree[id].num += tree[id<<1].num; 41 } 42 if(tree[id<<1|1].gcd==gcd) 43 { 44 tree[id].num += tree[id<<1|1].num; 45 } 46 } 47 pair<int,int> query(int id,int L,int R,int l,int r) 48 { 49 //cout << "l " << l << " r " << r << endl; 50 if(L<=l&&r<=R) 51 { 52 int gcd = tree[id].gcd; 53 int num = tree[id].num; 54 //cout << "gcd " << gcd << "num " << num << endl; 55 return pair<int,int>(gcd,num); 56 } 57 int mid = (l+r)>>1; 58 int ans = 0; 59 pair<int,int> res1,res2; 60 if(L<=mid){ res1 = query(id<<1,L,R,l,mid); } 61 if(mid<R) {res2 = query(id<<1|1,L,R,mid+1,r);} 62 int gcd; 63 if(L<=mid) 64 { 65 if(mid<R) 66 { 67 gcd = GCD(res1.first,res2.first); 68 //cout << "gcd " << gcd << endl; 69 if(res1.first==gcd) 70 ans+=res1.second; 71 if(res2.first==gcd) 72 ans+=res2.second; 73 } 74 else 75 { 76 gcd = res1.first; 77 ans += res1.second; 78 } 79 } 80 else 81 { 82 gcd = res2.first; 83 ans += res2.second; 84 } 85 //cout << "gcd " << gcd << " ans " << ans << endl; 86 return pair<int,int>(gcd,ans); 87 } 88 int main() 89 { 90 //cout << GCD(1,3) << endl; 91 cin >> n; 92 for(int i = 1;i<=n;i++) 93 { 94 cin >> a[i]; 95 } 96 build(1,1,n); 97 cin >> t; 98 for(int q = 0;q<t;q++) 99 { 100 int L,R; 101 cin >> L >> R; 102 cout << R-L+1-query(1,L,R,1,n).second << endl;; 103 } 104 return 0; 105 }