HDU 5869 Different GCD Subarray Query 离线+树状数组

Different GCD Subarray Query

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：

　　长度n的序列, m个询问区间[L, R], 问区间内的所有子段的不同GCD值有多少种.

题解：

　　固定右端点

　　预处理出 每个点向左延伸 的 不同gcd值

　　这样的 值不会超过log a 个

　　然后问题就变成了 问你一段区间内不同 gcd 值有多少，值是很少的 （询问一个区间有多少颜色的题型）

　　树状数组维护就可以了

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
using namespace std;

#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair

typedef long long LL;
const long long INF = 1e18;
const double Pi = acos(-1.0);
const int N = 1e6+, M = 1e2+, mod = 1e9+, inf = 2e9;

int n,q,a[N],ans[N];
int gcd(int a, int b) { return b ==  ? a : gcd(b, a%b);}
vector<pii> G[N];
struct QQ{
        int l,r,id;
        bool operator < (const QQ &a) const {
             return a.r > r;
        }
}Q[N];
int C[N],vis[N];
void update(int x,int c) {
        for(int i =x; i < N; i+=i&(-i)) C[i] += c;
}
int ask(int x) {
        int s = ;
        for(int i = x; i; i-= i & (-i))s += C[i];
        return s;
}
int main() {
        while(scanf("%d%d",&n,&q)!=EOF) {
            for(int i = ; i <= n; ++i) scanf("%d",&a[i]);
            for(int i = ; i <= n; ++i) G[i].clear();
            for(int i = ; i <= n; ++i) {
                    int x = a[i];
                    int y = i;
                    for(int j = ; j < G[i-].size(); ++j) {
                        int res = gcd(x,G[i-][j].first);
                        if(x != res) {
                                G[i].push_back(MP(x,y));
                                x = res;
                                y = G[i-][j].second;
                        }
                    }
                    G[i].push_back(MP(x,y));
            }
            memset(C,,sizeof(C));
            memset(vis,,sizeof(vis));
            for(int i = ; i <= q; ++i) {scanf("%d%d",&Q[i].l,&Q[i].r),Q[i].id = i;}
            sort(Q+,Q+q+);
            for(int R = , i = ; i <= q; ++i) {
                    while(R < Q[i].r) {
                        R++;
                        for(int j = ; j < G[R].size(); ++j) {
                            int res = G[R][j].first;
                            int ids = G[R][j].second;
                            if(vis[res]) update(vis[res],-);
                            vis[res] = ids;
                            update(vis[res],);
                        }
                    }
                    ans[Q[i].id] = ask(R) - ask(Q[i].l-);
            }
            for(int i = ; i <= q; ++i) cout<<ans[i]<<endl;
        }
        return ;
}