HDU 5869 Different GCD Subarray Query rmq+离线+数状数组

题目链接：

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

Different GCD Subarray Query

Time Limit: 6000/3000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
#### 问题描述
> 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].

输入

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

输出

For each query, output the answer in one line.

样例输入

5 3

1 3 4 6 9

3 5

2 5

1 5

样例输出

6

6

6

题意

给你n个数排成一行，每次询问求一段区间内的所有子串的不同的gcd的种数有多少。

题解

首先，要处理出所有的子区间是比较困难的，而且事实上很多子区间的gcd都是重复出现的，我们把区间右端点固定，那么随着左区间从右往左移，区间gcd的值成倍递减，所以我们取到的不同的gcd的值只有logAi种，处理出来之后，我们就可以用离线的方式，用线段树来求区间不同的值有几个，每个事件(处理出来的sigma(logA[i])个区间gcd)以左端点为准插入树状数组中维护一下（离线查询的线段树/树状数组可以参考[这个])。

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
//#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=1e5+10;
const int maxa=1e6+10;

int gcd(int a,int b) {
    return b==0?a:gcd(b,a%b);
}

struct Node {
    int l,r,v;
    Node(int l,int r,int v):l(l),r(r),v(v) {}
};

bool cmp1(const Node& t1,const Node& t2) {
    return t1.r<t2.r;
}

int n,m;
int arr[maxn],ans[maxn];
int dp[maxn][20];
void rmq_init() {
    for(int i=1; i<=n; i++) dp[i][0]=arr[i];
    for(int j=1; (1<<j)<=n; j++) {
        for(int i=1; i+(1<<j)-1<=n; i++) {
            dp[i][j]=gcd(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
        }
    }
}

int query(int l,int r) {
    int k=0;
    while(l+(1<<k)-1<=r) k++;
    k--;
    return gcd(dp[l][k],dp[r-(1<<k)+1][k]);
}

int sumv[maxn],mp[maxa];

void add(int x,int v) {
    while(x<maxn) {
        sumv[x]+=v;
        x+=(-x)&x;
    }
}

int sum(int x) {
    int ret=0;
    while(x>0) {
        ret+=sumv[x];
        x-=x&(-x);
    }
    return ret;
}

void init(){
    clr(mp,0);
    clr(sumv,0);
}

int main() {
    while(scf("%d%d",&n,&m)==2&&n) {
        init();
        for(int i=1; i<=n; i++) scanf("%d",&arr[i]);
        rmq_init();

        //v表示区间的gcd.
        //处理出代表性的事件
        vector<Node> events;
        for(int i=1; i<=n; i++) {
            int r=i;
            while(r>0) {
                int l=0,v=query(r,i);
                events.pb(Node(r,i,v));
                while(l+1<r) {
                    int mid=l+(r-l)/2;
                    int tmp=query(mid,i);
                    if(tmp==v) r=mid;
                    else l=mid;
                }
                r--;
            }
        }

        //v表示查询的id
        vector<Node> que;

        for(int i=0; i<m; i++) {
            int l,r;
            scf("%d%d",&l,&r);
            que.pb(Node(l,r,i));
        }

        sort(all(que),cmp1);
        sort(all(events),cmp1);

        //离线处理
        int j=0;
        rep(i,0,que.sz()) {
            while(j<events.sz()&&events[j].r<=que[i].r){
                Node& e=events[j];
                if(mp[e.v]){
                    add(mp[e.v],-1);
                }
                add(e.l,1);
                mp[e.v]=e.l;
                j++;
            }
            ans[que[i].v]=sum(que[i].r)-sum(que[i].l-1);
        }

        rep(i,0,m) prf("%d\n",ans[i]);
    }

    return 0;
}

//end-----------------------------------------------------------------------