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 a
₁, a
₂, …, 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 a
₁, a
₂, …, 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

  1 /*
  2  *  Author:kuangbin
  3  *  1010.cpp
  4  */
  5 
  6 #include <stdio.h>
  7 #include <algorithm>
  8 #include <string.h>
  9 #include <iostream>
 10 #include <map>
 11 #include <vector>
 12 #include <queue>
 13 #include <set>
 14 #include <string>
 15 #include <math.h>
 16 using namespace std;
 17 
 18 const int MAXN = 50010;
 19 int c[MAXN];
 20 int n;
 21 int lowbit(int x)
 22 {
 23     return x&(-x);
 24 }
 25 void add(int i,int val)
 26 {
 27     while(i <= n)
 28     {
 29         c[i] = max(c[i],val);
 30         i += lowbit(i);
 31     }
 32 }
 33 int Max(int i)
 34 {
 35     int s = 0;
 36     while(i > 0)
 37     {
 38         s = max(s,c[i]);
 39         i -= lowbit(i);
 40     }
 41     return s;
 42 }
 43 
 44 int a[MAXN];
 45 int b[MAXN];
 46 int ans[MAXN];
 47 
 48 struct Node
 49 {
 50     int l,r;
 51     int index;
 52 }node[MAXN];
 53 
 54 bool cmp(Node a,Node b)
 55 {
 56     return a.l > b.l;
 57 }
 58 
 59 int main()
 60 {
 61     //freopen("in.txt","r",stdin);
 62     //freopen("out.txt","w",stdout);
 63     int T;
 64     int m;
 65     int l,r;
 66     scanf("%d",&T);
 67     while(T--)
 68     {
 69         scanf("%d",&n);
 70         for(int i = 1;i <= n;i++)
 71             scanf("%d",&a[i]);
 72         scanf("%d",&m);
 73         for(int i = 0;i < m;i++)
 74         {
 75             scanf("%d%d",&node[i].l,&node[i].r);
 76             node[i].index = i;
 77         }
 78         sort(node,node+m,cmp);
 79         int i = n;
 80         int j = 0;
 81         memset(b,0,sizeof(a));
 82         memset(c,0,sizeof(c));
 83         while(j < m)
 84         {
 85             while(i > 0 && i>= node[j].l)
 86             {
 87                 for(int k =1;k*k <= a[i];k++)
 88                 {
 89                     if(a[i]%k == 0)
 90                     {
 91                         if(b[k]!=0)
 92                         {
 93                             add(b[k],k);
 94                         }
 95 
 96                         b[k] = i;
 97                         if(k != a[i]/k)
 98                         {
 99                             if(b[a[i]/k]!=0)
100                             {
101                                 add(b[a[i]/k],a[i]/k);
102                             }
103 
104                             b[a[i]/k]=i;
105                         }
106                     }
107                 }
108                 i--;
109             }
110             while(j < m && node[j].l > i)
111             {
112                 ans[node[j].index]=Max(node[j].r);
113                 j++;
114             }
115         }
116         for(int i = 0;i < m;i++)
117             printf("%d\n",ans[i]);
118     }
119     return 0;
120 }