codeforces 475D CGCDSSQ

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本文介绍了一个关于整数序列及其查询的问题解决方案，通过使用双 map 数据结构来高效地计算每对区间内所有元素的最大公约数等于特定值的区间数量。

http://www.elijahqi.win/2018/03/02/codeforces-475d-cgcdssq/
Given a sequence of integers a1, …, an and q queries x1, …, xq on it. For each query xi you have to count the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n and gcd(al, al + 1, …, ar) = xi.

is a greatest common divisor of v1, v2, …, vn, that is equal to a largest positive integer that divides all vi.

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, …, an, (1 ≤ ai ≤ 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 xi, (1 ≤ xi ≤ 109).

Output
For each query print the result in a separate line.

Examples
Input

Copy
3
2 6 3
5
1
2
3
4
6
Output
1
2
2
0
1
Input

Copy
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
假设现在以a[i]为左端点向右端移动那么这个区间的所有gcd加起来不超过log(a[i])种因为我gcd如果要改变至少/2

那么我可以从右向左开始模拟开两个map 表示这种gcd的答案是多少然后每次向右端移动的时候相当于都存储着我i+1这个位置的所有gcd的答案然后和我a[i]进行gcd 那么把新获得的值插入新的map中表示我以i为左端点 gcd为x的区间有多少个注意适时累加到答案map中即可

#include<map>
#include<cstdio>
#include<algorithm>
#define N 110000
#define ll long long
using namespace std;
map<int,ll>::iterator it;
map<int,ll> mm[2],ans;
inline char gc(){
    static char now[1<<16],*S,*T;
    if (T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;}
    return *S++;
}
inline int read(){
    int x=0,f=1;char ch=gc();
    while(ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=gc();}
    while(ch<='9'&&ch>='0') x=x*10+ch-'0',ch=gc();
    return x*f;
}
inline int gcd(int x,int y){
    if (!y) return x;else return gcd(y,x%y);
}
int n,m,a[N];
int main(){
    freopen("cf475d.in","r",stdin);
    n=read();
    for (int i=1;i<=n;++i) a[i]=read();int now=1,pre=0;
    for (int i=n;i;--i){
        mm[now].clear();mm[now][a[i]]=1;ans[a[i]]++;
        for (it=mm[pre].begin();it!=mm[pre].end();++it){
            int x=gcd(a[i],(*it).first);mm[now][x]+=(*it).second;ans[x]+=(*it).second;
        }
        now^=1;pre^=1;
    }m=read();
    for (int i=1;i<=m;++i){
        int x=read();printf("%I64d\n",ans[x]);
    }
    return 0;
}