hdu 4630 No Pain No Game(树状数组）

No Pain No Game

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1809    Accepted Submission(s): 775

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

 

 

题意：给定一个n个元素的数列，对于每组查询l,r，在区间［l,r］内的最大公约数是多少。

转载一个题解，说得比较清楚：http://blog.csdn.net/qq564690377/article/details/9674355

 

/*************************************************************************
    > File Name: code/hdu/4630.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年08月11日 星期二 17时47分01秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef unsigned long long ULL;
const int inf = 0x7fffffff;
const int N=5E4+7;
int a[N];
int n,q;
int b[N],c[N],ans[N];
struct NODE
{
    int l,r;
    int id;
}Q[N];

bool cmp(NODE a,NODE b)
{
    if (a.l>b.l) return true;
    return false;
}
int lowbit( int x)
{
    return x&(-x);
}

void update ( int x,int delta)
{
    for ( int i  = x; i <= n ; i = i + lowbit(i))
    {
    c[i] = max(c[i],delta);
    }
}
int sum( int x)
{
    int res  = 0;
    for ( int i = x ; i >= 1 ; i = i - lowbit(i))
    {
    res = max(res,c[i]);
    }
    return res;
}
int main()
{
    int T;
    cin>>T;
    while (T--)
    {
    memset(c,0,sizeof(c));
    memset(b,0,sizeof(b));
    scanf("%d",&n);
    for ( int i = 1 ; i <= n ; i++)
    {
        scanf("%d",&a[i]);
    }
    scanf("%d",&q);
    for ( int i = 0 ; i < q ; i++ )
    {
        scanf("%d %d",&Q[i].l,&Q[i].r);
        Q[i].id = i;
    }
    sort(Q,Q+q,cmp);
    int i = n;
    int j = 0;
    while (j<q)
    {
        
        while (i>0&&i>=Q[j].l)
        {
        for ( int k = 1 ; k*k<=a[i]; k++)  //k为约数
        {
            if (a[i]%k==0)
            {
            if (b[k]!=0)
            {
                update(b[k],k);
            }
            b[k] = i;  //b[k]纪录的是约数k最后一次出现的位置
            if (k!=a[i]/k)
            {
                if (b[a[i]/k])
                {
                update (b[a[i]/k],a[i]/k);
                }
                b[a[i]/k]=i;
                }

            }
        }
        i--;
        }
        while (j<q&&Q[j].l>i)
        {
        ans[Q[j].id]=sum(Q[j].r);
        j++;
        }
    
    }
    for ( int i = 0 ; i < q ; i ++)
    {
        printf("%d
",ans[i]);
    }
    }
  
    return 0;
}