hdu 4630 No Pain No Game(线段树+离线操作)

No Pain No Game

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1769    Accepted Submission(s): 748

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

 

Author

WJMZBMR

 

Source

2013 Multi-University Training Contest 3

 

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题意： 有N个数。 是 1~N的一个排列。有M个询问， 每次询问一个区间， 问从这个区间中，取两个数的最大的最大公约数。

题解：先把查询按右区间升序排序。在将数组按顺序插入，记录当前这个数的因子出现的位置，假设之前有出现则代表这两个因子出现的

位置之间有两个数的公共约数是它，用线段树维护区间约数最大值就可以。

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<vector>
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define N 50050
#define lc idx<<1
#define rc idx<<1|1

using namespace std;

int n,q,flag;
int a[N],tree[N*4];
int L[N],R[N];
int first[N],ans[N];
vector<int>vec;

struct node {
    int id;
    int l,r;
} Q[N];

bool cmp(node a,node b) {
    if(a.r==b.r)
        return a.l<b.l;
    return a.r<b.r;
}

///求全部因子
void FJ(int x) {
    vec.clear();
    for(int i=1; i*i<=x; i++) {
        if(x%i==0) {
            vec.push_back(i);
            if(x/i!=i)
                vec.push_back(x/i);
        }
    }
}

void push_up(int idx) {
    tree[idx]=max(tree[lc],tree[rc]);
}

void build(int idx,int l,int r) {
    tree[idx]=0;
    if(l==r) {
        return;
    }
    int mid=(l+r)>>1;
    build(lson);
    build(rson);
}

void update(int idx,int l,int r,int x,int v) { //x处的值改为v
    if(l==r) {
        if(tree[idx]<v)
            tree[idx]=v;
        return;
    }
    int mid=(l+r)>>1;
    if(x<=mid)update(lson,x,v);
    else      update(rson,x,v);
    push_up(idx);
}

int query(int idx,int l,int r,int x,int y) {
    if(l>=x&&y>=r) {
        return tree[idx];
    }
    int ans=0;
    int mid=(l+r)>>1;
    if(x<=mid) {
        ans=max(ans,query(lson,x,y));
    }
    if(y>mid) {
        ans=max(ans,query(rson,x,y));
    }
    return ans;
}

void debug() {
    for(int i=0; i<vec.size(); i++) {
        printf("%d ",vec[i]);
    }
    cout<<endl;
}

int main() {
    //freopen("test.in","r",stdin);
    int t;
    scanf("%d",&t);
    while(t--) {
        scanf("%d",&n);
        for(int i=1; i<=n; i++) {
            scanf("%d",&a[i]);
        }
        scanf("%d",&q);
        for(int i=1; i<=q; i++) {
            scanf("%d%d",&Q[i].l,&Q[i].r);
            Q[i].id=i;
        }
        sort(Q+1,Q+1+q,cmp);
        memset(first,0,sizeof first);
        memset(L,0,sizeof L);
        memset(R,0,sizeof R);
        build(1,1,n);
        ///预处理同样有区间的左右区间
        int f=1;
        L[Q[f].r]=f;
        R[Q[f].r]=f;
        for(int i=1; i<=q;) {
            while(Q[i].r==Q[f].r&&i<=q) {
                i++;
            }
            L[Q[f].r]=f;
            R[Q[f].r]=i-1;
            f=i;
        }
        for(int i=1; i<=n; i++) {
            //FJ(a[i]);
            //debug();
            int xx=a[i];
            for(int k=1; k*k<=xx; k++) {
                if(xx%k==0) {
                    if(!first[k]) {
                        first[k]=i;
                    } else {
                        update(1,1,n,first[k],k);
                        first[k]=i;
                    }
                    int kk=xx/k;
                    if(k!=kk) {
                        if(!first[kk]) {
                            first[kk]=i;
                        } else {
                            update(1,1,n,first[kk],kk);
                            first[kk]=i;
                        }
                    }
                }
            }
            int x=L[i],y=R[i];
            if(x==0||y==0)continue;
            for(int j=x; j<=y; j++) {
                int k=Q[j].l;
                if(k==i) {
                    ans[Q[j].id]=0;
                } else {
                    ans[Q[j].id]=query(1,1,n,k,i);
                }
            }
            if(y==q)break;
        }
        for(int i=1; i<=q; i++) {
            printf("%d\n",ans[i]);
        }
    }
    return 0;
}