Coprime Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 217    Accepted Submission(s): 126

Problem Description

Do you know what is called ``Coprime Sequence''? That is a sequence consists of n positive integers, and the GCD (Greatest Common Divisor) of them is equal to 1.
``Coprime Sequence'' is easy to find because of its restriction. But we can try to maximize the GCD of these integers by removing exactly one integer. Now given a sequence, please maximize the GCD of its elements.

 

 

Input

The first line of the input contains an integer T(1≤T≤10), denoting the number of test cases.
In each test case, there is an integer n(3≤n≤100000) in the first line, denoting the number of integers in the sequence.
Then the following line consists of n integers a1,a2,...,an(1≤ai≤109), denoting the elements in the sequence.

 

Output

For each test case, print a single line containing a single integer, denoting the maximum GCD.

 

Sample Input

 

3
3
1 1 1
5
2 2 2 3 2
4
1 2 4 8

 

Sample Output

1

2

2

 

//T 组数据，n 个数，开始时，n 个数 gcd 为 1 ，现可以任意抽走一个数，问剩下数最大 gcd 是多少?

// st 表很好做，O(n*logn) + n * O(n) 的时间

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 using namespace std;
 5 #define MX 100005
 6 
 7 int n,m;
 8 int dp[MX][20];
 9 int mi[MX];
10 int data[MX];
11 
12 int gcd(int a,int b)
13 {return b==0?a:gcd(b,a%b);}
14 
15 void build()
16 {
17     mi[0]=-1;
18     for (int i=1;i<=n;i++)
19     {
20         mi[i]=((i&(i-1))==0) ? mi[i-1]+1:mi[i-1];   //注意位运算优先级很低
21         dp[i][0]=data[i];
22     }
23     for (int k=1;k<=mi[n];k++)
24     {
25         for (int i=1; (i+(1<<k)-1)<=n ;i++)
26         {
27             dp[i][k]=gcd(dp[i][k-1],dp[i+(1<<(k-1))][k-1]);
28         }
29     }
30 }
31 
32 int find_(int a,int b)
33 {
34     int x = mi[b-a+1];
35     return gcd(dp[a][x],dp[b-(1<<x)+1][x]);
36 }
37 
38 int main()
39 {
40     int T;
41     cin>>T;
42     while (T--)
43     {
44         scanf("%d",&n);
45         for (int i=1;i<=n;i++)
46             scanf("%d",&data[i]);
47         build();
48         int ans = 1;
49         for (int i=1;i<=n;i++)
50         {
51             int res;
52             if (i==1)
53                 res = find_(2,n);
54             else if (i==n)
55                 res = find_(1,n-1);
56             else
57                 res = gcd(find_(1,i-1),find_(i+1,n));
58             ans = max(ans,res);
59         }
60         printf("%d\n",ans);
61     }
62 }
View Code

 

 强行上log n 啊，啊哈哈，其实，只要从左向右扫一遍，存储gcd前缀和，再从右向左扫一遍，存gcd后缀和，然后枚举要剔除的数即可

 

posted @ 2017-05-08 19:51  happy_codes  阅读(302)  评论(0)  编辑  收藏  举报