HDU5869-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): 1243    Accepted Submission(s): 467

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 31 3 4 6 93 52 51 5

 

Sample Output

666

 

Source

2016 ACM/ICPC Asia Regional Dalian Online

 

Recommend

wange2014

 

题意：给出n个数，有q个询问，问区间【l，r】的所有子区间有几种公约数
解题思路：固定右端点，预处理 每个点向左延伸区间产生的不同gcd的值，gcd的个数不会超过log（r - l + 1）个，将所有询问离线按右端点从小到大排序，然后树状数组维护，枚举右端点，一个个求出当前右端点固定的时候，枚举所有的gcd值出现的位置，如果这个gcd前面没有出现过，那么它出现的时候就是这个位置并且这个位置 + 1，否则之前出现的位置 - 1，并且这个位置 + 1（因为后来再次出现的位置一定在之前出现的右边），最后区间查询

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int a[100005], n, Q;int c[100005];vector<pair<int, int> >q[100005], b[100005];int vis[1000050];int ans[100005];int gcd(int a, int b) { return b == 0 ? a : gcd(b, a%b); }int lowbit(int x){return x & (-x);}void add(int x, int add){while (x <= n){c[x] += add;x += lowbit(x);}}int getsum(int x){int sum = 0;while (x>0){sum += c[x];x -= lowbit(x);}return sum;}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], pos = i;b[i].push_back(make_pair(x, pos));for (int j = 0; j<b[i - 1].size(); j++){int k = b[i - 1][j].first; pos = b[i - 1][j].second;int g=gcd(k, a[i]);if (g != x){b[i].push_back(make_pair(g, pos));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(vis, 0, sizeof vis);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 (vis[x]) add(vis[x], -1);vis[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] = getsum(i) - getsum(x - 1);}}for (int i = 1; i <= Q; i++) printf("%d\n", ans[i]);}return 0;}