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HDU4630:No Pain No Game(線段樹)

編輯：關於C++

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

Author WJMZBMR
Source 2013 Multi-University Training Contest 3

題意：求區間內的兩兩gcd裡最大的
思路：我們可以對於每個數求出因子，而在一個區間內出現超過兩次的那麼必然是一個gcd,這樣我們只需要把所有出現超過兩次的因子加入線段樹中，並更新即可對於查詢，我們可以先按r從小到大排序，離線處理所有答案

#include 
#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

#define lson 2*i
#define rson 2*i+1
#define ls l,mid,lson
#define rs mid+1,r,rson
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 50005
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define mpa make_pair
#define lowbit(x) (x&-x)
const int mod = 10007;

struct node
{
    int l,r,id;
} op[N];
int tree[N<<2],a[N],ans[N];

int cmp(node a,node b)
{
    return a.rmid) ans2=query(L,R,rs);
        return max(ans1,ans2);
    }
}

int n,m;
int pre[N];

int main()
{
    int t,i,j,k,x;
    scanf(%d,&t);
    while(t--)
    {
        scanf(%d,&n);
        build();
        for(i = 1; i<=n; i++)
            scanf(%d,&a[i]);
        scanf(%d,&m);
        for(i = 0; i