hdu-4630-No Pain No Game-（树状数组，离线处理）

Problem Description

Life is a game,and you lose it,so you suicide.

But you can not kill yourself before you solve this problem:

Given you a sequence of number a1, a2, ..., an.They are also a permutation of 1...n.

You need to answer some queries,each with the following format:

If we chose two number a,b (shouldn't be the same) from interval [l, r],what is the maximum gcd(a, b)? If there's no way to choose two distinct number(l=r) then the answer is zero.

 

Input

First line contains a number T(T <= 5),denote the number of test cases. Then follow T test cases. For each test cases,the first line contains a number n(1 <= n <= 50000). The second line contains n number a[sub]1[/sub], a[sub]2[/sub], ..., a[sub]n[/sub]. The
third line contains a number Q(1 <= Q <= 50000) denoting the number of queries. Then Q lines follows,each lines contains two integer l, r(1 <= l <= r <= n),denote a query.

 

Output

For each test cases,for each query print the answer in one line.

 

Sample Input

1
10
8 2 4 9 5 7 10 6 1 3
5
2 10
2 4
6 9
1 4
7 10

 

Sample Output

5
2
2
4
3

参考大神kuangbin的文章：http://www.cnblogs.com/kuangbin/archive/2013/07/30/3225627.html

题目给出n个数的序列，这个序列是1~n的一个全排列，然后查询是给出一个区间[l,r],让求出区间内任意两个数的最大公约数中的最大值。

让求区间内的gcd的最大值，可以求出区间内的所有数的约数，而那个最大的且是至少两个数的公约数的数就是这个区间的最大gcd，可以用树状数组存放

代码：

//tree[i]是i处的最大约数
#include<iostream>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<queue>
#include<cstring>
#include<map>
using namespace std;
#define M 50005
int n;
int tree[M];
int a[M],b[M];
int ans[M];
struct
c028
node {
int l,r,id;
bool operator < (const node &obj) const
{
return l>obj.l;
}
}p[M];
inline int lowbit(int i)
{
return i&(-i);
}
void add(int i,int v)
{
while(i<=n)
{
tree[i]=max(tree[i],v);
i+=lowbit(i);
}
}
int sum(int i)
{
int res=-1;
while(i>0)
{
res=max(res,tree[i]);
i-=lowbit(i);
}
return res;
}
int main()
{
int T,i,j,m;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
scanf("%d",&m);
for(i=1;i<=m;i++)
{
scanf("%d%d",&p[i].l,&p[i].r);
p[i].id=i;
}
sort(p+1,p+1+m); //对区间按照l从大到小排序，
memset(tree,0,sizeof(tree));
memset(b,0,sizeof(b));
i=n; j=1;
while(j<=m) //处理区间
{
while(i>0&&i>=p[j].l) //循环点
{
for(int k=1;k*k<=a[i];k++) //找到a[i]的每一个约数，不断更新tree使其中存的是最大的约数
{
if(a[i]%k==0)
{
if(b[k]!=0) //k不是第一次出现，也就是保证k至少是两个数的约数
{
add(b[k],k);
}
b[k]=i;
if(k!=a[i]/k)
{
if(b[a[i]/k]!=0)
{
add(b[a[i]/k],a[i]/k);
}
b[a[i]/k]=i;
}
}
}
i--;
}
while(j<=m&&p[j].l>i)
{
ans[p[j].id]=sum(p[j].r);
j++;
}
}
for(i=1;i<=m;i++)
printf("%d\n",ans[i]);
}
return 0;
}