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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 a 1, a 2, …, a n.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 1, a 2, …, a n.
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
求区间任意两个数的最大gcd。做了几个gcd的题目，貌似都是离线处理的。
对于两个数的gcd，是因为这两个数都有相同的约数x，这样这两个数的gcd才有可能是x。这样我们用O（nsqrt n）的时间复杂度处理处所有的数约数。假如a[k]的一个约数是x，x上一次出现的位置是i。那么在i~k这一段的gcd有可能就会变成x。按着这个思路去更新，线段树去求gcd最大值。询问离线处理，当现在处理的位置恰好是某一次询问的右端点的话，就可以直接通过线段树求出最大值gcd来了。
代码如下：
#include<bits/stdc++.h>
#define ll long long
using namespace std;const int maxx=5e4+100;
struct node{int l;int r;int sum;
}p[maxx<<2];
struct Node{int l;int r;int id;bool operator<(const Node &a)const{return a.r>r;}
}b[maxx];
vector<int> g[maxx];
int a[maxx],ans[maxx];
int n,m;
/*------------预处理出所有数的约数-------------*/
inline void init()
{for(int i=1;i<=maxx;i++)for(int j=i;j<=maxx;j+=i)g[j].push_back(i);
}
/*---------------线段树----------------*/
inline void pushup(int cur)
{p[cur].sum=max(p[cur<<1].sum,p[cur<<1|1].sum);
}
inline void build(int l,int r,int cur)
{p[cur].l=l;p[cur].r=r;p[cur].sum=0;if(l==r) return ;int mid=l+r>>1;build(l,mid,cur<<1);build(mid+1,r,cur<<1|1);
}
inline void update(int pos,int v,int cur)
{int L=p[cur].l;int R=p[cur].r;if(L==R){p[cur].sum=max(p[cur].sum,v);return ;}int mid=L+R>>1;if(pos<=mid) update(pos,v,cur<<1);else update(pos,v,cur<<1|1);pushup(cur);
}
inline int query(int l,int r,int cur)
{int L=p[cur].l;int R=p[cur].r;if(l<=L&&R<=r) return p[cur].sum;int mid=L+R>>1;if(r<=mid) return query(l,r,cur<<1);else if(l>mid) return query(l,r,cur<<1|1);else return max(query(l,mid,cur<<1),query(mid+1,r,cur<<1|1));
}
int main()
{init();int t,x;scanf("%d",&t);while(t--){map<int,int> mpp;scanf("%d",&n);for(int i=1;i<=n;i++) scanf("%d",&a[i]);scanf("%d",&m);for(int i=1;i<=m;i++) scanf("%d%d",&b[i].l,&b[i].r),b[i].id=i;sort(b+1,b+1+m);build(1,n,1);for(int i=1,j=1;i<=m&&j<=n;j++){for(int k=0;k<g[a[j]].size();k++){x=g[a[j]][k];if(mpp[x]!=0) update(mpp[x],x,1);mpp[x]=j;}for(i;i<=m&&j==b[i].r;i++) ans[b[i].id]=query(b[i].l,b[i].r,1);}for(int i=1;i<=m;i++) printf("%d\n",ans[i]);}return 0;
}
努力加油a啊，(o)/~
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