HDU 4630 No Pain No Game（2013多校3 1010题 离线处理+树状数组求最值）

No Pain No Game

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17    Accepted Submission(s): 5

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

 

Recommend

zhuyuanchen520

 

题目给出n个数，每个数的范围是1~n的。

n<=50000;

然后查询m次，m<=50000

每次查询[l,r]区间内，两个数的gcd的最大值.

n个数，如果把n个数的约数全部写出来。查询[l,r]之间的gcd的最大值，就相当于找一个最大的数，使得这个数是[l,r]之间至少是两个的约数。

对于一个数n,在sqrt(n)内可以找出所有约数。

我的做法是对查询进行离线处理。

将每个查询按照 l 从大到小排序。

然后 i 从 n~0 ,表示从后面不断扫这些数。

对于数a[i],找到a[i]的所有约数，对于约数x,在x上一次出现的位置加入值x.

这样的查询的时候，只要差值前 r 个数的最大值就可以了。

看代码吧，不解释了。

这么水竟然想了这么久，非常sad

 /*
  *  Author:kuangbin
  *  1010.cpp
  */

 #include <stdio.h>
 #include <algorithm>
 #include <string.h>
 #include <iostream>
 #include <map>
 #include <vector>
 #include <queue>
 #include <set>
 #include <string>
 #include <math.h>
 using namespace std;

 const int MAXN = ;
 int c[MAXN];
 int n;
 int lowbit(int x)
 {
     return x&(-x);
 }
 void add(int i,int val)
 {
     while(i <= n)
     {
         c[i] = max(c[i],val);
         i += lowbit(i);
     }
 }
 int Max(int i)
 {
     int s = ;
     while(i > )
     {
         s = max(s,c[i]);
         i -= lowbit(i);
     }
     return s;
 }

 int a[MAXN];
 int b[MAXN];
 int ans[MAXN];

 struct Node
 {
     int l,r;
     int index;
 }node[MAXN];

 bool cmp(Node a,Node b)
 {
     return a.l > b.l;
 }

 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int T;
     int m;
     int l,r;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d",&n);
         for(int i = ;i <= n;i++)
             scanf("%d",&a[i]);
         scanf("%d",&m);
         for(int i = ;i < m;i++)
         {
             scanf("%d%d",&node[i].l,&node[i].r);
             node[i].index = i;
         }
         sort(node,node+m,cmp);
         int i = n;
         int j = ;
         memset(b,,sizeof(a));
         memset(c,,sizeof(c));
         while(j < m)
         {
             while(i >  && i>= node[j].l)
             {
                 for(int k =;k*k <= a[i];k++)
                 {
                     if(a[i]%k == )
                     {
                         if(b[k]!=)
                         {
                             add(b[k],k);
                         }

                         b[k] = i;
                         if(k != a[i]/k)
                         {
                             if(b[a[i]/k]!=)
                             {
                                 add(b[a[i]/k],a[i]/k);
                             }

                             b[a[i]/k]=i;
                         }
                     }
                 }
                 i--;
             }
             while(j < m && node[j].l > i)
             {
                 ans[node[j].index]=Max(node[j].r);
                 j++;
             }
         }
         for(int i = ;i < m;i++)
             printf("%d\n",ans[i]);
     }
     return ;
 }