C. GCD Table

The GCD table?G?of size?n?×?n?for an array of positive integers?a?of length?n?is defined by formula

Let us remind you that the greatest common divisor (GCD) of two positive integers?x?and?y?is the greatest integer that is divisor of both?xand?y, it is denoted as?. For example, for array?a?=?{4,?3,?6,?2}?of length 4 the GCD table will look as follows:

Given all the numbers of the GCD table?G, restore array?a.

Input

The first line contains number?n?(1?≤?n?≤?500) — the length of array?a. The second line contains?n2?space-separated numbers — the elements of the GCD table of?G?for array?a.

All the numbers in the table are positive integers, not exceeding?109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array?a.

Output

In the single line print?n?positive integers — the elements of array?a. If there are multiple possible solutions, you are allowed to print any of them.

Sample test(s) input 4
2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2 output 4 3 6 2 input 1
42 output 42 input 2
1 1 1 1 output 1 1


思路：

　　設數列X: a11, a12,...., ann;
由于gcd(a,b)<=min(a,b);
ans[N]存放已經選中的數,即array中一定存在的數；?
首先從X中找到最大的一個值aij，然后對ans[N]中的每一個數，得到g = gcd(aij, ans[i]),?
由于table矩陣是對稱的，所以從X中刪除2個值為 g 的數值！
最后將aij放入ans中！不斷重復此過程，知道ans中數字個數為n；

#include<iostream> #include<cstring> #include<cstdio> #include<cmath> #include<map> #include<set> #include<algorithm> #define N 505 using namespace std;int n; map<int, int, greater<int> >mp;//key按照由大到小排序 int gcd(int a, int b){return b==0 ? a : gcd(b, a%b); }int ans[N];int main(){cin>>n;int nn = n*n;for(int i=0; i<nn; ++i){int x;cin>>x;mp[x]++;}int len = 0;for(map<int, int, greater<int> >::iterator it=mp.begin(); it!=mp.end();){if(it->second == 0){//不為0，說明這個數還是array中的數字++it;continue; }--it->second;for(int i=0; i<len; ++i){int gcdn = gcd(it->first, ans[i]);mp[gcdn]-=2;}ans[len++] = it->first;} for(int i=0; i<n; ++i){if(i!=0) cout<<" ";cout<<ans[i];}cout<<endl;return 0; }

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