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[50002];
struct node
{
    int l,r,len;
    int maxn;
}f[50002*4];
struct node2
{
    int st,ed;
    int i,val;
}a[50002];
int date[50002];
int pre[50002];

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=1;i<=50000;i++)
        for(int j=i;j<=50000;j=j+i)
        yz[j].push_back(i);
}
void build(int l,int r,int n)
{
    int mid = (l+r)/2;
    f[n].l = l;
    f[n].r = r;
    f[n].len = f[n].r-f[n].l +1;
    f[n].maxn = 0;
    if(l==r) return;
    build(l,mid,n*2);
    build(mid+1,r,n*2+1);
}
void update(int wz,int num,int n)
{
    int mid=(f[n].l + f[n].r)/2;
    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*2);
    else if(mid<wz) update(wz,num,n*2+1);
    f[n].maxn = f[n*2].maxn>f[n*2+1].maxn ? f[n*2].maxn : f[n*2+1].maxn;
}
int query(int l,int r,int n)
{
    int mid=(f[n].l + f[n].r)/2;
    if(f[n].l == l && f[n].r == r)
    {
        return f[n].maxn;
    }
    if(mid>=r) return query(l,r,n*2);
    else if(mid<l) return query(l,r,n*2+1);
    else return max(query(l,mid,n*2),query(mid+1,r,n*2+1));
}
int main()
{
    int T , n , m;
    init();
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        build(1,n,1);
        for(int i=1;i<=n;i++)
            scanf("%d",&date[i]);
        scanf("%d",&m);
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d",&a[i].st,&a[i].ed);
            a[i].i = i;
            a[i].val = 0;
        }
        sort(a+1,a+1+m,cmp);
        memset(pre,0,sizeof(pre));
        int k = 1;
        for(int i=1;i<=n;i++)
        {
            int ans = yz[date[i]].size();
            for(int j=0;j<ans;j++)
            {
                int temp = yz[date[i]][j];
                if(pre[temp])
                    update(pre[temp],temp,1);
                pre[temp] = i;
            }
            while(a[k].ed == i && k<=m)
            {
                a[k].val = query(a[k].st,a[k].ed,1);
                k++;
            }
        }
        sort(a+1,a+1+m,cmp1);
        for(int i=1;i<=m;i++)   printf("%d
",a[i].val);
    }
    return 0;
}