CodeForces 475D CGCDSSQ

最新推荐文章于 2020-11-30 22:51:05 发布

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最新推荐文章于 2020-11-30 22:51:05 发布

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分类专栏： Codeforces ----栈 ----Gcd 文章标签： codeforces

本文链接：https://blog.youkuaiyun.com/jtjy568805874/article/details/52414176

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----Gcd

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本文介绍了一种针对区间GCD查询的高效算法，通过预处理和数据结构优化实现nlogn的时间复杂度，适用于大量查询场景。文章提供了一个具体的实现案例，并对比了相似题目，如HDU5726。

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Description

Given a sequence of integers a₁, ..., a_n and q queries x₁, ..., x_q on it. For each query x_i you have to count the number of pairs (l, r)such that 1 ≤ l ≤ r ≤ n and gcd(a_l, a_l + 1, ..., a_r) = x_i.

$\text{[math]}$ is a greatest common divisor of v₁, v₂, ..., v_n, that is equal to a largest positive integer that divides all v_i.

Input

The first line of the input contains integer n, (1 ≤ n ≤ 10⁵), denoting the length of the sequence. The next line contains n space separated integers a₁, ..., a_n, (1 ≤ a_i ≤ 10⁹).

The third line of the input contains integer q, (1 ≤ q ≤ 3 × 10⁵), denoting the number of queries. Then follows q lines, each contain an integer x_i, (1 ≤ x_i ≤ 10⁹).

Output

For each query print the result in a separate line.

Sample Input

Input

Output

Input

7
10 20 3 15 1000 60 16
10
1
2
3
4
5
6
10
20
60
1000

Output

14
0
2
2
2
0
2
2
1
1


求所有区间的gcd的值，对于以i为右端点的全部gcd的种类一定是log(a[i])级别的，
往这些区间里再加上a[i+1]，再把相同的区间合并，就可以保证全部的效率是nlogn了。
顺便一提，HDU5726和这题几乎一模一样，那题我用了线段树，时间上要多一个log
我就说怎么大家都会写，原来cf上有过类似的题目啊。
#include<set>
#include<map>
#include<ctime>
#include<cmath>
#include<stack>
#include<queue>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define per(i,j,k) for (int i = j; i >= k; i--)
#define loop(i,j,k) for (int i = j;i != -1; i = k[i])
#define lson x << 1, l, mid
#define rson x << 1 | 1, mid + 1, r
#define fi first
#define se second
#define mp(i,j) make_pair(i,j)
#define pii pair<int,int>
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const double eps = 1e-4;
const int INF = 0x7FFFFFFF;
const int mod = 9973;
const int N = 3e5 + 10;
map<int, LL> M;
int n, x;

int gcd(int x, int y)
{
	return x%y ? gcd(y, x%y) : y;
}

int main()
{
	scanf("%d", &n);
	stack<pii> a, b;
	rep(i, 1, n)
	{
		scanf("%d", &x);  a.push(mp(x, 1));
		while (!a.empty()) b.push(mp(gcd(a.top().fi, x), a.top().se)), a.pop();
		while (!b.empty())
		{
			pii q = b.top();	b.pop();
			M[q.fi] += q.se;
			if (!a.empty() && a.top().fi == q.fi) q.se += a.top().se, a.pop();
			a.push(q);
		}
	}
	scanf("%d", &n);
	while (n--)
	{
		scanf("%d", &x);
		printf("%lld\n", M[x]);
	}
	return 0;
}