HDU 5902 GCD is Funny 数学

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本文介绍了一款名为GCDisFunny的游戏，玩家需从N个数中选三个数，再从中选两个数求最大公约数(GCD)，将结果放回序列重复操作N-2次，最终留下两个相同的数。文章提供了题目的背景、输入输出样例及解答思路。

GCD is Funny

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=5902

Description

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

He chooses three numbers a, b and c written at the board and erases them.
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)).
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

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

Hint

题意

给你n个数，然后选出三个数出来，然后再从这三个数中选择两个数做GCD，然后再扔两个GCD回到原序列。

一直重复N-2次，最后显然只会剩下两个相同的数，问你这个数是多少。

题解：

感觉好神啊……
这道题是某次BC的出题事故= =
答案是所有size>=2的子集的gcd
模拟n-2次暴力去搞一搞就好了。
不用考虑新增加的数，因为没有意义，你总会从之前的数里面求GCD得到。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1005;
int n;
int can[maxn];
int a[maxn];
int gcd(int a,int b){
    return b==0?a:gcd(b,a%b);
}
void solve()
{
    memset(can,0,sizeof(can));
    scanf("%d",&n);
    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++){
            can[gcd(a[i],a[j])]=1;
        }
    }
    int num = n-3;
    bool flag = true;
    while(num>=1&&flag){
        num--;
        flag = false;
        for(int i=1;i<=1000;i++){
            if(can[i]){
                for(int j=1;j<=n;j++){
                    int p = gcd(i,a[j]);
                    if(!can[p]){
                        can[p]=1;
                        flag = true;
                    }
                }
            }
        }
    }
    int first = 0;
    for(int i=1;i<=1000;i++){
        if(can[i]){
            if(first==0)printf("%d",i),first=1;
            else printf(" %d",i);
        }
    }
    printf("\n");
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)solve();
    return 0;
}