HDU4630-No Pain No Game(离线，线段树)

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

题目大意：

给一个序列，然后q次询问[l,r]询问两两gcd最大值是多少（不能是同一个数），无解输出0；

解题思路：

想总结一下，感觉自己好菜啊。

两两取gcd，最后的值肯定是该数的约数。

也就是说，数是可以一个一个加入的。

现将约数分解。

一个数被加入。如果他的约数k上一次在p处，那么询问左端点<=p的话答案就至少值为k。

就离线处理一下。

保证加入的点来更新答案。

我觉得和HH的项链有点像。

代码：

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define lll spc<<1
 #define rrr spc<<1|1
 using std::max;
 struct qst{
     int l,r;
     int no;
     int ans;
     void Insert(int i)
     {
         no=i;
         ans=;
         scanf("%d%d",&l,&r);
         return ;
     }
 }q[];
 int maxs[];
 int a[];
 int lst[];
 int n,Q;
 int T;
 bool cmp(qst x,qst y)
 {
     return x.r<y.r;
 }
 bool cmq(qst x,qst y)
 {
     return x.no<y.no;
 }
 void Reset(void)
 {
     memset(lst,,sizeof(lst));
     memset(maxs,,sizeof(maxs));
     return ;
 }
 void pushup(int spc)
 {
     maxs[spc]=max(maxs[lll],maxs[rrr]);
     return ;
 }
 void update(int spc,int l,int r,int v,int pos)
 {
     if(l==r)
     {
         maxs[spc]=max(v,maxs[spc]);
         return ;
     }
     int mid=(l+r)>>;
     if(pos<=mid)
         update(lll,l,mid,v,pos);
     else
         update(rrr,mid+,r,v,pos);
     pushup(spc);
     return ;
 }
 int query(int l,int r,int ll,int rr,int spc)
 {
     if(l>rr||ll>r)
         return -0x3f3f3f3f;
     if(ll<=l&&r<=rr)
         return maxs[spc];
     int mid=(l+r)>>;
     return max(query(l,mid,ll,rr,lll),query(mid+,r,ll,rr,rrr));
 }
 int main()
 {
     scanf("%d",&T);
     while(T--)
     {
         Reset();
         scanf("%d",&n);
         for(int i=;i<=n;i++)
             scanf("%d",&a[i]);
         scanf("%d",&Q);
         for(int i=;i<=Q;i++)
             q[i].Insert(i);
         std::sort(q+,q+Q+,cmp);
         int j=;
         for(int i=;i<=n;i++)
         {
             for(int k=;k*k<=a[i];k++)
             {
                 if(a[i]%k==&&lst[k])
                     update(,,n,k,lst[k]);
                 if(a[i]%k==&&lst[a[i]/k]&&k!=a[i]/k)
                     update(,,n,a[i]/k,lst[a[i]/k]);
                 if(a[i]%k==)
                     lst[k]=lst[a[i]/k]=i;
             }
             while(q[j].r==i)
             {
                 if(j>Q)
                     break;
                 if(q[j].l>=q[j].r)
                     q[j].ans=;
                 else
                     q[j].ans=query(,n,q[j].l,q[j].r,);
                 j++;
             }
         }
         std::sort(q+,q+Q+,cmq);
         for(int i=;i<=Q;i++)
             printf("%d\n",q[i].ans);
     }
     return ;
 }