Codeforces 342 A. Xenia and Divisors

A. Xenia and Divisors

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Xenia the mathematician has a sequence consisting of n (n is
divisible by 3) positive integers, each of them is at most 7. She wants to split the sequence into groups of three so that for each group of three a, b, c the
following conditions held:

a < b < c;

a divides b, b divides c.

Naturally, Xenia wants each element of the sequence to belong to exactly one group of three. Thus, if the required partition exists, then it has

groups
of three.

Help Xenia, find the required partition or else say that it doesn't exist.

Input

The first line contains integer n (3 ≤ n ≤ 99999) —
the number of elements in the sequence. The next line contains n positive integers, each of them is at most 7.

It is guaranteed that n is divisible by 3.

Output

If the required partition exists, print

groups
of three. Print each group as values of the elements it contains. You should print values in increasing order. Separate the groups and integers in groups by whitespaces. If there are multiple solutions, you can print any of them.

If there is no solution, print -1.

Sample test(s)

input

6
1 1 1 2 2 2

output

-1

input

6
2 2 1 1 4 6

output

1 2 4
1 2 6

题目大意：
给你一个数m，然后有m个数arr[i],每个数都<=7 && >=1,让你找满足以下关系的数（3个一组3个一组）
1)a<b<c;
2)a能被b整除，b能被c整除；
解题思路：
首先想到满足关系的三个数
1 2 4 1 2 61 3 6
然后再找不属于这三个数的条件，只要去掉不满足条件的就ok了，
不满足条件的:
1）1 的个数必须是m/3；
2）不能有5和7；
3）2 的个数必须大于等于4的个数
4) 6的个数必须大于等于3的个数
5）2
的个数 + 3的个数 == 6的个数 + 4的个数
然后上代码：

/**
Author: ITAK

Date: 2015-09-29
**/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;

#define MM(a) memset(a,0,sizeof(a))

typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 100000+5;
const int mod = 1000000007;
const double eps = 1e-7;

int cmp(int a, int b)
{
    return a > b;
}
int gcd(int a, int b)
{
    if(b == 0)
        return a;
    else
        return gcd(b, a%b);
}
int arr[maxn];
int data[10];
int main()
{
    int m;
    scanf("%d",&m);
    MM(data);
    for(int i=0; i<m; i++)
    {
        scanf("%d",&arr[i]);
        data[arr[i]]++;
    }
    m /= 3;
    if(data[1]!=m || data[5] | data[7] || data[2]<data[4] ||
       data[6]<data[3] || data[2]+data[3]!=data[4]+data[6])
        puts("-1");
    else
    {
        for(int i=1; i<=data[4]; i++)
            puts("1 2 4");
        for(int i=1; i<= data[2]-data[4]; i++)
            puts("1 2 6");
        for(int i=1; i<=data[3]; i++)
            puts("1 3 6");
    }
    return 0;
}