2016 ICPC 大连网络赛 HDU 5869 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): 714    Accepted Submission(s): 261


[align=left]Problem Description[/align]
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].

  
 

[align=left]Input[/align]
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
 

[align=left]Output[/align]
For each query, output the answer in one line.
 

[align=left]Sample Input[/align]

5 3
1 3 4 6 9
3 5
2 5
1 5

 

[align=left]Sample Output[/align]

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。

【解题方法】fzu月赛原题。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loga 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）。这种类型就随便处理了，bit线段树都可以。

【AC 代码】

//
//Created by just_sort 2016/9/13 10:57
//Copyright (c) 2016 just_sort.All Rights Reserved
//

#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 1e6+10;
int n,q,a[maxn],ans[maxn],c[maxn],vis[maxn];
vector<pair<int,int> >G[maxn];
struct node{
int l,r,id;
bool operator<(const node &rhs){
return r<rhs.r;
}
}Q[maxn];
int lowbit(int x){
return x&(-x);
}
void update(int i,int v){
while(i<=n){
c[i] += v;
i += lowbit(i);
}
}
int query(int i){
int ret = 0;
while(i>0){
ret += c[i];
i -= lowbit(i);
}
return ret;
}
int main()
{
while(scanf("%d%d",&n,&q)!=EOF)
{
for(int i=1; i<=n; i++) scanf("%d",&a[i]);
for(int i=1; i<=n; i++) G[i].clear();
memset(c,0,sizeof(c));
memset(vis,0,sizeof(vis));
for(int i=1; i<=n; i++){
int x = a[i];
int y = i;
for(int j=0; j<G[i-1].size(); j++){
int curg = __gcd(x,G[i-1][j].first);
if(x != curg){
G[i].push_back(make_pair(x,y));
x = curg;
y = G[i-1][j].second;
}
}
G[i].push_back(make_pair(x,y));
}
for(int i=1; i<=q; i++){
scanf("%d%d",&Q[i].l,&Q[i].r);
Q[i].id = i;
}
sort(Q+1,Q+q+1);
int now = 0;
for(int i=1; i<=q; i++){
while(now<Q[i].r){
now++;
for(int j=0; j<G[now].size(); j++){
int x = G[now][j].first;
int y = G[now][j].second;
if(vis[x]) update(vis[x],-1);
vis[x] = y;
update(vis[x],1);
}
}
ans[Q[i].id] = query(now) - query(Q[i].l-1);
}
for(int i=1; i<=q; i++){
printf("%d\n",ans[i]);
}
}
return 0;
}