CF 342A(Xenia and Divisors-考虑有限情况)

A. Xenia and Divisorstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard 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

  61 1 1 2 2 2

  output

  -1

  input

  62 2 1 1 4 6

  output

  1 2 41 2 6

  
  

  可能的序列：

  d[1].1 2 4

  d[2].1 2 6

  d[3].1 3 6

  
  

  红色部分贪心都懂得。。。

  
  

  #include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<ctime>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Lson (x<<1)#define Rson ((x<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define MAXN (99999+10)long long mul(long long a,long long b){return (a*b)%F;}long long add(long long a,long long b){return (a+b)%F;}long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}typedef long long ll;int n,a[MAXN]={0};int main(){//freopen("Divisors.in","r",stdin);cin>>n;For(i,n){int p;cin>>p;a[p]++;}if (a[5]||a[7]) {puts("-1");return 0;}static int d[4]={0};if (a[4]) {a[1]-=a[4];a[2]-=a[4];d[1]=a[4];a[4]=0;if (a[1]<0||a[2]<0) {puts("-1");return 0;}}//For(i,7) cout<<a[i]<<' ';cout<<endl;if (a[3]) {a[1]-=a[3];a[6]-=a[3];d[3]=a[3];a[3]=0;if (a[1]<0||a[6]<0) {puts("-1");return 0;}}if (a[1]==a[2]&&a[2]==a[6]){d[2]=a[1];}else {puts("-1");return 0;}For(i,d[1]) puts("1 2 4");For(i,d[2]) puts("1 2 6");For(i,d[3]) puts("1 3 6");return 0;}