HDU 5869 Different GCD Subarray Query 树状数组+离线

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

**题意：**给你n个数Q个查询，每次查询询问[l，r]内不同gcd的个数
**思路：**Q很大，按照一般思路使用线段树在线操作似乎不可行，所以考虑使用离线操作。但重点在于如何预处理出GCD。
在这里我们固定右端点，枚举上一个端点所存的所有不同GCD值，求GCD，并记录不同的，延伸右端点时，再重复操作即可。
在树状数组更新时，当出现一个重复的GCD值，始终把标记维护到最靠右的，并把之前出现的相同GCD的位置所在的标记消掉。这样就能使数量数组拥有前缀特性，可以使用sum[r]-sum[l]了
容器套容器很好用阿，弱要加快学了。

/** @Date    : 2016-11-13-16.10

  * @Author  : Lweleth (SoungEarlf@gmail.com)

  * @Link    : https://github.com/

  * @Version :

  */

#include <stdio.h>

#include <iostream>

#include <string.h>

#include <algorithm>

#include <utility>

#include <vector>

#include <map>

#include <set>

#include <string>

#include <stack>

#include <queue>

#define pii pair<int , int>

#define FF first

#define SS second

#define MP(x,y) make_pair((x), (y))

#define LL long long

#define MMF(x) memset((x),0,sizeof(x))

#define MMI(x) memset((x), INF, sizeof(x))

using namespace std;



const int INF = 0x3f3f3f3f;

const int N = 1e5+2000;



int n,q;

int c[N];

int a[N];

int ans[N];

vector< pair<int,int> >gp[N];

int vis[1000010];



struct yuu

{

    int l, r;

    int m;

}Q[N];



int cmp(yuu a, yuu b)

{

    return a.r < b.r;

}

int gcd(int a, int b)

{

    return b?gcd(b, a % b):a;

}



void add(int x, int y)

{

    while(x < N)

    {

        c[x] += y;

        x += x & (-x);

    }

}



int sum(int x)

{

    int ans = 0;

    while(x)

    {

        ans += c[x];

        x -= x & (-x);

    }

    return ans;

}



void init()

{



    MMF(c);

    MMF(vis);

    MMF(ans);

    for(int i = 1; i <= n; i++)

        {

            int x = a[i];

            int p = i;

            for(int j = 0; j < gp[i-1].size(); j++)

            {

                int g = gcd(gp[i-1][j].FF, x);

                if(x != g)

                {

                    gp[i].push_back(MP(x, p));

                    x = g;

                    p = gp[i-1][j].SS;

                }

            }

            gp[i].push_back(MP(x, p));

        }

}

int main()

{

    while(~scanf("%d%d", &n, &q))

    {



        MMF(a);

        for(int i = 1; i <= n; i++ )

        {

            scanf("%d", a + i);

            gp[i].clear();

        }

        init();

        for(int i = 0; i < q; i++)

        {

            scanf("%d%d", &Q[i].l, &Q[i].r);

            Q[i].m = i;

        }



        sort(Q, Q + q, cmp);

        //////

        int pos = 0;

        for(int i = 1; i <= n; i++)

        {

            for(int j = 0; j < gp[i].size(); j++)//

            {

                if(vis[gp[i][j].FF])

                    add(vis[gp[i][j].FF], -1);

                vis[gp[i][j].FF] = gp[i][j].SS;

                add(gp[i][j].SS, 1);

            }

            while(Q[pos].r == i)

            {

                ans[Q[pos].m] = sum(Q[pos].r) - sum(Q[pos].l-1);

                pos++;

            }

        }

        for(int i = 0; i < q; i++)

            printf("%d\n", ans[i]);



    }

    return 0;

}