题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5869

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Different GCD Subarray Query

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1172 Accepted Submission(s): 444

Problem Description
This is a simple problem. The teacher gives Bob a list of problems about GCD (Greatest Common Divisor). After studying some of them,Bob thinks that GCD is so interesting. One day,he comes up with a new problem about GCD. Easy as it looks,Bob cannot figure it out himself. Now he turns to you for help,and here is the problem:

Given an array a of N positive integers a1,a2,⋯aN−1,aN; a subarray of a is defined as a continuous interval between a1 and aN. In other words,ai,ai+1,⋯,aj−1,aj is a subarray of a,for 1≤i≤j≤N. For a query in the form (L,R),tell the number of different GCDs contributed by all subarrays of the interval [L,R].

Input
There are several tests,process till the end of input.

For each test,the first line consists of two integers N and Q,denoting the length of the array and the number of queries,respectively. N positive integers are listed in the second line,followed by Q lines each containing two integers L,R for a query.

You can assume that

1≤N,Q≤100000

1≤ai≤1000000

Output
For each query,output the answer in one line.

Sample Input
5 3
1 3 4 6 9
3 5
2 5
1 5

Sample Output
6
6
6

Source
2016 ACM/ICPC Asia Regional Dalian Online

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题目大意:
给长度为N的序列,M次询问，每次询问【L,R】之间所有子区间的不同GCD有多少个、

解题思路:
固定左端点，然后预处理出结果，用树状数组或者线段树均可。

然后vis【1e6】居然RE。。。。最后改了map才过、

时间复杂度$O(N*logA*log_2n)$

附本题代码
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#include <bits/stdc++.h>

using namespace std;

#define INF (~(1<<31))
#define INFLL (~(1ll<<63))
#define pb push_back
#define mp make_pair
#define abs(a) ((a)>0?(a):-(a))
#define lalal puts("*******");
#define s1(x) scanf("%d",&x)
#define Rep(a,b,c) for(int a=(b);a<=(c);a++)
#define Per(a,c) for(int a=(b);a>=(c);a--)
#define no puts("NO")

typedef long long int LL ;
typedef unsigned long long int uLL ;

const int    N   = 100000+7;
const int    MOD = 1e9+7;
const double eps = 1e-6;
const double PI  = acos(-1.0);

inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void fre(){
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
}
inline int gcd(int a,int b){return (b==0)?a:gcd(b,a%b);}
/***********************************************************************/
int sum[N];
#define lowbit(x) (x&-x)
void update(int index,int val){
    for(int i=index;i<N;i+=lowbit(i)) sum[i]+=val;
}
int getSum(int index){
    int ans = 0;
    for(int i=index;i;i-=lowbit(i)) ans += sum[i];
    return ans;
}
vector<pair<int,int> >E[N];
int a[N],ans[N];
struct node {
    int l,r,id;
    bool operator <(const node &p) const{
        //if(r==p.r) return l<p.l;
        return r<p.r;
    }
}q[N];
map<int,int>vis;
int main(){
    int n,m;
    while(~scanf("%d %d",&n,&m)){
        memset(sum,0,sizeof(sum));
        for(int i=1;i<=n;++i) scanf("%d",&a[i]);
        for(int i=1;i<=m;++i) scanf("%d %d",&q[i].l,&q[i].r),q[i].id=i,ans[i]=0;
        sort(q+1,q+m+1);

        for(int i=0;i<=n;++i)E[i].clear();
        for(int i=1;i<=n;++i){
            int x=a[i];
            int y=i;
            for(int j=0;j<E[i-1].size();++j){
                int res=gcd(x,E[i-1][j].first);
                if(x!=res){
                    E[i].pb(mp(x,y));
                    x=res;
                    y=E[i-1][j].second;
                }
            }
            E[i].pb(mp(x,y));
        }

        vis.clear();
        for(int R=0,i=1;i<=m;++i){
            while(R < q[i].r){
                R++;
                for(int j=0;j<E[R].size();++j) {
                    int res=E[R][j].first;
                    int ids=E[R][j].second;
                    if(vis[res]) update(vis[res],-1);
                    vis[res] = ids;
                    update(vis[res],1);
                }
            }

            ans[q[i].id] = getSum(R) - getSum(q[i].l-1);
        }

        for(int i=1;i<=m;++i)      printf("%d\n",ans[i]);
    }
    return 0;
}