树状数组 gcd 查询 Different GCD Subarray Query

Different GCD Subarray Query

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1328    Accepted Submission(s): 504

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

 

 

Recommend

wange2014

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.

树状数组满足这种操作，但是知道这个我也不会啊，不懂后来那种统计操作

#include<bits/stdc++.h>
using namespace std;
const int N=200005;
int c[N],a[N],n,Q,last[N*10],ans[N];
vector < pair<int,int> >q[N],b[N];
void add(int k,int num) {
    while(k<=n) {
        c[k]+=num;
        k+=k&-k;
    }
}
int read(int k) {
    int res=0;
    while(k) {
        res+=c[k];
        k-=k&-k;
    }
    return res;
}
int main() {
    while(~scanf("%d%d",&n,&Q)) {
        for(int i=1; i<=n; i++) {
            scanf("%d",&a[i]);
            q[i].clear();
            b[i].clear();
        }
        for(int i=1; i<=n; i++) {
            int x=a[i],y=i;
            b[i].push_back(make_pair(x,y));
            for(int j=0; j<b[i-1].size(); j++) {
                int t=b[i-1][j].first;
                y=b[i-1][j].second;
                int g=__gcd(t,a[i]);
                if(g!=x) {
                    b[i].push_back(make_pair(g,y));
                    x=g;
                }
            }
        }
        for(int i=1; i<=Q; i++) {
            int L,R;
            scanf("%d%d",&L,&R);
            q[R].push_back(make_pair(L,i));
        }
        memset(c,0,sizeof(c));
        memset(last,0,sizeof(last));
        for(int i=1; i<=n; i++) {
            for(int j=0; j<b[i].size(); j++) {
                int x = b[i][j].first,y= b[i][j].second;
                if(last[x])
                    add(last[x],-1);
                last[x] = y;
                add(y,1);
            }
            for(int j=0; j<q[i].size(); j++) {
                int x=q[i][j].first,y=q[i][j].second;
                ans[y]=read(i)-read(x-1);
            }
        }
        for(int i=1; i<=Q; i++) printf("%d\n",ans[i]);

    }
    return 0;
}