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.

 

题意：n个数每次选三个数删除，取其中两个数将gcd放回去两次，问最后剩的数可能是多少。

 

由于一开始取3个数要删掉一个，所以只要求n-1个的gcd。

 

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int a[1010] , dp[1010];
int gcd(int a , int b) {
    while(b)
    {
        int t = a % b;
        a = b;
        b = t;
    }
    return a;
}
int main()
{
    int t;
    scanf("%d" , &t);
    while(t--) {
        int n;
        scanf("%d" , &n);
        memset(dp , 0 , sizeof(dp));
        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++) {
                dp[gcd(a[i] , a[j])] = 1;
            }
        }
        int flag = 1;
        for(int i = 1 ; ; i++) {
            if(flag == 0 || i >= n - 2)
                break;
            flag = 0;
            for(int j = 1 ; j <= 1000 ; j++) {
                for(int l = 1 ; l <= n ; l++) {
                    int gg = gcd(a[l] , j);
                    if(dp[j] && !dp[gg]) {
                        flag = 1;
                        dp[gg] = 1;
                    }
                }
            }
        }
        int temp = 0;
        for(int i = 1 ; i <= 1000 ; i++) {
            if(!temp) {
                if(dp[i]) {
                    printf("%d" , i);
                    temp = 1;
                }
            }
            else {
                if(dp[i])
                    printf(" %d" , i);
            }
        }
        printf("
");
    }
    return 0;
}