hdu6025 Coprime Sequence(2017女生赛)

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

 

Source

2017中国大学生程序设计竞赛 - 女生专场 

题意：给你n个数字，你可以任意删除一个数字使得剩下的数字的gcd最大，输出最后最大的gcd；

思路：我的思路太暴力，就是正反各一遍特判；gcd后的值会更小或者相等、1和任意数的gcd的值都是1；

           另外一种思路就是：正反各存一下起始位置到这个位置的gcd值，然后枚举删点，O(n)的复杂度；

第一种思路对应的代码：

#include
#include
#include
using namespace std;
const int maxn=100005;

int a[maxn];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        for(int i=0; i=0; i--)
        {
            if(v1==1&&flag==0&&i==n-2)//
            {
                flag=1,v1=a[i];
                continue;
            }
            v=__gcd(a[i],v1);
            if(i==0&&flag==0) //
            {
                if(v=0&&flag==0)
            {
                flag=1;
                continue;
            }
            v1=v;

        }
//        printf("---------------%d\n",v);
        ans=max(ans,v);
        printf("%d\n",ans);
    }
    return 0;
}

第二种思路的代码：

#include
#include
#include
using namespace std;
const int maxn=100005;
typedef long long ll;
const ll tomod=1e9+7;

int a[maxn];
int gcd1[maxn],gcd2[maxn];//正着来一遍，反着来一遍；
int main()
{
    int T,n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        gcd1[0]=gcd2[n+1]=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            if(i==1) gcd1[i]=a[i];
            else  gcd1[i]=__gcd(gcd1[i-1],a[i]);
        }

        for(int i=n;i>=1;i--)
        {
            if(i==n) gcd2[i]=a[i];
            else gcd2[i]=__gcd(gcd2[i+1],a[i]);
        }

        int ans=max(gcd1[n],gcd2[1]);
        for(int i=1;i<=n;i++)
        {
            ans=max(ans,__gcd(gcd1[i-1],gcd2[i+1]));
        }
        printf("%d\n",ans);
    }
    return 0;
}