hdu6025 Coprime Sequence(2017女生赛)

Problem Description

Do you know what is called ``Coprime Sequence''? That is a sequence consists ofn 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

331 1 152 2 2 3 241 2 4 8

 

Sample Output

122

 

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<cstdio>#include<cstring>#include<algorithm>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<n; i++)            scanf("%d",&a[i]);        sort(a,a+n);        int flag=0;        int v1=a[0],v=0;        for(int i=1; i<n; i++)        {            if(v1==1&&flag==0&&i==1) //特判第一位为1否；            {                flag=1;                v1=a[i];                continue;            }            v=__gcd(a[i],v1);            if(i==n-1&&flag==0)  //特判最后一位；            {                if(v<v1) v=v1;                flag=1;                break;            }            if((v==1)&&flag==0)  //特判结果为1            {                flag=1;                continue;            }            else if(__gcd(v1,a[i])<__gcd(v1,a[i+1])&&i+1<n&&flag==0)            {                flag=1;                continue;            }        }        v1=v;        int ans=v;//        printf("------------%d\n",ans);        v1=a[n-1],v=0;        flag=0;        for(int i=n-2; 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<v1) v=v1;                flag=1;                break;            }            if(v==1&&flag==0) //            {                flag=1;                continue;            }            else if(__gcd(a[i],v1)<__gcd(a[i-1],v1)&&i-1>=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<cstdio>#include<cstring>#include<algorithm>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;}