GCD is Funny

GCD is Funny

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

Alex has invented a new game for fun. There are $n$ integers at a board and he performs the following moves repeatedly:

1. He chooses three numbers $a$, $b$ and $c$ written at the board and erases them.
2. He chooses two numbers from the triple $a$, $b$ and $c$ and calculates their greatest common divisor, getting the number $d$ ($d$ maybe $gcd(a,b)$, $gcd(a,c)$ or $gcd(b, c)$).
3. He writes the number $d$ to the board two times.

It can be seen that after performing the move $n-2$ times, there will be only two numbers with the same value left on the board. Alex wants to know which numbers can left on the board possibly. Can you help him?

 

Input

There are multiple test cases. The first line of input contains an integer $T$ $(1 le T le 100)$, indicating the number of test cases. For each test case:

The first line contains an integer $n$ $(3 le n le 500)$ -- the number of integers written on the board. The next line contains $n$ integers: $a_1, a_2, ..., a_n$ $(1 le a_i le 1000)$ -- the numbers on the board.

 

Output

For each test case, output the numbers which can left on the board in increasing order.

 

Sample Input

3 4 1 2 3 4 4 2 2 2 2 5 5 6 2 3 4

 

Sample Output

1 2 2 1 2 3

分析：大意：给你>=3个数，每次取3个数去掉，添上其中两个数的gcd的2遍，最后剩两个相同数是多少？

　　　仔细分析可知第一次取两个数gcd，去掉另一个数，以后就可以任取剩下n-3个数作gcd；

　　　所以就是要求2~n-1个数的gcd，模拟即可；

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, ls[rt]
#define Rson mid+1, R, rs[rt]
#define sys system("pause")
#define intxt freopen("in.txt","r",stdin)
const int maxn=1e3+10;
using namespace std;
int gcd(int p,int q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
inline ll read()
{
    ll x=0;int f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,k,t,a[maxn],ok[maxn];
queue<pii>p;
int main()
{
    int i,j;
    scanf("%d",&t);
    while(t--)
    {
        memset(ok,0,sizeof(ok));
        while(!p.empty())p.pop();
        scanf("%d",&n);
        rep(i,1,n)scanf("%d",&a[i]);
        rep(i,1,n)rep(j,i+1,n)
        {
            k=gcd(a[i],a[j]);
            if(!ok[k])ok[k]=1,p.push(mp(k,2));
        }
        while(!p.empty())
        {
            pii q=p.front();
            p.pop();
            if(q.se==n-1)break;
            rep(i,1,n)
            {
                k=gcd(q.fi,a[i]);
                if(!ok[k])ok[k]=1,p.push(mp(k,q.se+1));
            }
        }
        bool flag=false;
        rep(i,1,1000)
        {
            if(ok[i])
            {
                if(flag)printf(" %d",i);
                else printf("%d",i),flag=true;
            }
        }
        printf("
");
    }
    //system("Pause");
    return 0;
}