hdu 5902 GCD is Funny

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≤T≤100), indicating the number of test cases. For each test case:

The first line contains an integer n (3≤n≤500) -- the number of integers written on the board. The next line contains n integers: a1,a2,...,an (1≤ai≤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

341 2 3 442 2 2 255 6 2 3 4

 

Sample Output

1 221 2 3
刚开始看是1001 所以，直接两两gcd（）跑了一遍。
结果就阵亡了。。。。。。。。。。。。。。。。
晚上想了想，6 6 10 15 这个例子我出的结果显然不对。
然后我保留两两gcd的结果，继续gcd 。
反正晚上一直wa。跑到700+ms
今天上午再写，才发现，6 10 15 我多跑一边， 多出了个1 
然后记录一下剩余个数，别多跑，就过了。
#include <bits/stdc++.h>using namespace std;const int maxn=1110;int gcd(int x,int y){    if(x%y==0) return y;    else  return gcd(y,x%y);}int num[maxn],a[maxn];int main(){    int T,n;    cin>>T;    while(T--)    {        cin>>n;        memset(num,0,sizeof(num));        for(int i=1;i<=n;i++){            scanf("%d",&a[i]);        }        for(int i=1;i<n;i++)            for(int j=i+1;j<=n;j++)                num[gcd(a[i],a[j])]=1;        int flag=1,cnt=n-1;        while(flag&&cnt>=3)        {            cnt--;            flag=0;            for(int i=1;i<=1000;i++)                if(num[i])                    for(int j=1;j<=n;j++)                    {                        int t=gcd(i,a[j]);                        if(num[t]==0)                        {                            num[t]=1;                            flag=1;                        }                    }        }        int t=0;        for(int i=1;i<=1000;i++)            if(num[i])            {                if(t)                    printf(" %d",i);                else{                    printf("%d",i);                    t=1;                }            }        printf("\n");    }}