hdu 5869 Different GCD Subarray Query(gcd+树状数组）

Different GCD Subarray Query

[align=center]Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 114    Accepted Submission(s): 29
[/align]

[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

 

[align=left]Source[/align]
2016 ACM/ICPC Asia Regional Dalian Online

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：考虑固定左端点的不同GCD值，只有不超过logA种, 所以事件点只有nlogA个. 那么离线处理, 按照区间右端点排序从小到大处理询问,

用一个树状数组维护每个GCD值的最大左端点位置即可. 复杂度是O(nlogAlogn).

这份题解里有两个难点：

1、如何快速的离线化处理出固定的左端点的gcd；

2   如何用树状数组维护最大左端点，又如何求出答案？？

第一个问题：

很显然，若是平常的离散处理，时间复杂度为n^2;

for(int i=1;i<=n;i++)
{
int x=num[i],y=i;
for(int j=0;j<gg[i-1].size();j++)
{
int cc=__gcd(gg[i-1][j].first,x);
if(x!=cc)
{
gg[i].push_back(make_pair(x,y));
x=cc,y=gg[i-1][j].second;
}
}
gg[i].push_back(make_pair(x,y));
}这个方式我也是第一次见，将固定的左端点和得出的gcd同存在右端点，这样的确可以做到nlongA的时间离线处理。
而且这种方式存储的gcd值是从小到大的，所以不会有重存的情况出现。

主要是我不太习惯在vector 数组里用pair ,真是汗颜。。。

如何维护最大左端点呢？？

首先对查询进行右端点排序。

从左到右遍历一遍节点的gcd值。

并在树状数组里面更新相同gcd的最由端点。

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=100100;
int c[maxn];
vector< pair<int,int> >gg[maxn];
int num[maxn],vis[maxn*10],res[maxn];
struct node
{
int l,r,id;
bool operator<(const node&p)const
{
return r<p.r;
}
}A[maxn];
int lowbit(int x)
{
return x&(-x);
}
void add(int x,int value)
{
while(x<maxn)
{
c[x]+=value;
x+=lowbit(x);
}
}
int sum(int x)
{
int ans=0;
while(x)
{
ans+=c[x];
x-=lowbit(x);
}
return ans;
}
int main()
{
int n,q;
while(scanf("%d%d",&n,&q)!=EOF)
{
for(int i=1;i<=n;i++)
{
scanf("%d",&num[i]);
gg[i].clear();
}
memset(c,0,sizeof(c));
for(int i=1;i<=n;i++)
{
int x=num[i],y=i;
for(int j=0;j<gg[i-1].size();j++)
{
int cc=__gcd(gg[i-1][j].first,x);
if(x!=cc)
{
gg[i].push_back(make_pair(x,y));
x=cc,y=gg[i-1][j].second;
}
}
gg[i].push_back(make_pair(x,y));
}
for(int i=1;i<=q;i++)
{
int l,r;
scanf("%d%d",&l,&r);
A[i].l=l,A[i].r=r;
A[i].id=i;
}
sort(A+1,A+q+1);
int len=1;
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++)
{
for(int j=0;j<gg[i].size();j++)
{
int c1=gg[i][j].first;
int c2=gg[i][j].second;
if(vis[c1])
add(vis[c1],-1);
vis[c1]=c2;
add(c2,1);
}
while(A[len].r==i)
{
res[A[len].id]=sum(A[len].r)-sum(A[len].l-1);
len++;
}
}
for(int i=1;i<=q;i++)
printf("%d\n",res[i]);
}
return 0;
}