hdu5902 GCD is Funny

GCD is Funny

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 354    Accepted Submission(s): 83

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

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

 

BC这一场这道题真是惨烈……MyRoom只有一人过的……什么时候BC第一题这么难过了……

本来是想直接两两GCD去重输出，因为当时想的是a, b, c, e, f

取a, b, c，那么gcd(a, b) = d，丢掉c，最后得到d, d, e, f，然后再取

d, d, e，那么gcd(d, d) = d，丢掉e，最后得到d, d, f，然后再gcd(d, d) = d，

丢掉f，最后得到d, d

也就是说任选两对都可以最后得到这两对的gcd作为最后结果，数据太水，pretest过了就没多想

后来仔细想想，谁跟你说d, d, e的时候一定选gcd(d, d) = d丢掉e的……也可能gcd(d, e) = x，丢掉d啊……这样结果就可能变成3个数，4个数或者更多个数的GCD了

这样的话也就是说在我原来的想法中，每一次求gcd都有可能做一次gcd(d, e) = x，从而变成多一个数的gcd，也就是说最后的答案应该是序列中任选i个数的gcd（2 <= i <= n - 1），因为第一次操作无论如何都会直接丢弃一个数，所以最多只能求出来n - 1个数的GCD

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 500;
int T, n, a[maxn + 5];
bool g[1000 + 5]; //用来标记某gcd是否之前出现过，用于去重

int gcd(int m, int n) {
	if (!n) return m;
	return gcd(n, m % n);
}

int main()
{
	cin >> T;
	while (T--) {
		scanf("%d", &n);
		for (int i = 0; i < n; i++) {
			scanf("%d", &a[i]);
		}
		memset(g, false, sizeof(g));
		for (int i = 0; i < n; i++) {
			for (int j = i + 1; j < n; j++) {
				g[gcd(a[i], a[j])] = true;
			}
		}
		bool flag = true;
		//需要跑n - 3次，因为第一步的操作相当于去掉了3个元素，然后把gcd的初始信息存储在g中了
		//不能多也不能少，多了的话就有可能计算出n个数的gcd了,但最多只能是n - 1个的，因为无论如何第一步操作会直接丢弃一个数
		int cnt = 1;
		while (flag && cnt <= n - 3) {
			cnt++;
			flag = false;
			for (int i = 1; i <= 1000; i++) {
				if (g[i]) {
					for (int j = 0; j < n; j++) {
						if (!g[gcd(a[j], i)]) {
							flag = true;
							g[gcd(a[j], i)] = true;
						}
					}
				}
			}
		}
		bool first = true;
		for (int i = 1; i <= 1000; i++) {
			if (g[i]) {
				if (first) {
					printf("%d", i);
					first = false;
				}
				else {
					printf(" %d", i);
				}
			}
		}
		puts("");
	}
	return 0;
}