HDU 4630 No Pain No Game 线段树 和 hdu3333有共同点

No Pain No Game

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1465    Accepted Submission(s):
631

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, ..., 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 a₁, a₂, ..., 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

Source

2013
Multi-University Training Contest 3

 

 /**
 题意：求解区间[ L , R ]   max gcd() ;

 如果只求一次，那么我们可以这样做。
 把每个数字的因子刷一遍，满足>=2的最大就是结果。
 如果多次，可以这样求解。
 在区间[ L , R ] 如果素因子出现，更新它上次出现的位置的最大值，
 保存当前出现的位置。第一次出现不更新，只保存。

 当枚举到R的时候，我们就可以在[ L, R ]中找最大值即可。

 所以这一题就可以对r 排序，然后进行更新。
 **/
 #include<iostream>
 #include<stdio.h>
 #include<cstring>
 #include<cstdlib>
 #include<algorithm>
 #include<vector>
 using namespace std;

 vector<int>yz[];
 struct node
 {
     int l,r,len;
     int maxn;
 }f[*];
 struct node2
 {
     int st,ed;
     int i,val;
 }a[];
 int date[];
 int pre[];

 bool cmp(struct node2 n1,struct node2 n2)
 {
     return n1.ed<n2.ed;
 }
 bool cmp1(struct node2 n1,struct node2 n2)
 {
     return n1.i<n2.i;
 }
 void init()
 {
     for(int i=;i<=;i++)
         for(int j=i;j<=;j=j+i)
         yz[j].push_back(i);
 }
 void build(int l,int r,int n)
 {
     int mid = (l+r)/;
     f[n].l = l;
     f[n].r = r;
     f[n].len = f[n].r-f[n].l +;
     f[n].maxn = ;
     if(l==r) return;
     build(l,mid,n*);
     build(mid+,r,n*+);
 }
 void update(int wz,int num,int n)
 {
     int mid=(f[n].l + f[n].r)/;
     if(f[n].l == wz && f[n].r == wz)
     {
         if(f[n].maxn<num)
             f[n].maxn = num;
         return;
     }
     if(mid>=wz) update(wz,num,n*);
     else if(mid<wz) update(wz,num,n*+);
     f[n].maxn = f[n*].maxn>f[n*+].maxn ? f[n*].maxn : f[n*+].maxn;
 }
 int query(int l,int r,int n)
 {
     int mid=(f[n].l + f[n].r)/;
     if(f[n].l == l && f[n].r == r)
     {
         return f[n].maxn;
     }
     if(mid>=r) return query(l,r,n*);
     else if(mid<l) return query(l,r,n*+);
     else return max(query(l,mid,n*),query(mid+,r,n*+));
 }
 int main()
 {
     int T , n , m;
     init();
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d",&n);
         build(,n,);
         for(int i=;i<=n;i++)
             scanf("%d",&date[i]);
         scanf("%d",&m);
         for(int i=;i<=m;i++)
         {
             scanf("%d%d",&a[i].st,&a[i].ed);
             a[i].i = i;
             a[i].val = ;
         }
         sort(a+,a++m,cmp);
         memset(pre,,sizeof(pre));
         int k = ;
         for(int i=;i<=n;i++)
         {
             int ans = yz[date[i]].size();
             for(int j=;j<ans;j++)
             {
                 int temp = yz[date[i]][j];
                 if(pre[temp])
                     update(pre[temp],temp,);
                 pre[temp] = i;
             }
             while(a[k].ed == i && k<=m)
             {
                 a[k].val = query(a[k].st,a[k].ed,);
                 k++;
             }
         }
         sort(a+,a++m,cmp1);
         for(int i=;i<=m;i++)   printf("%d\n",a[i].val);
     }
     return ;
 }