As meticulous Gerald sets the table and caring Alexander sends the postcards, Sergey makes snowmen. Each showman should consist of three snowballs: a big one, a medium one and a small one. Sergey’s twins help him: they’ve already made n snowballs with radii equal to r1, r2, …, rn. To make a snowman, one needs any three snowballs whose radii are pairwise different. For example, the balls with radii 1, 2 and 3 can be used to make a snowman but 2, 2, 3 or 2, 2, 2 cannot. Help Sergey and his twins to determine what maximum number of snowmen they can make from those snowballs.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of snowballs. The next line contains n integers — the balls’ radii r1, r2, …, rn (1 ≤ ri ≤ 109). The balls’ radii can coincide.
Output
Print on the first line a single number k — the maximum number of the snowmen. Next k lines should contain the snowmen’s descriptions. The description of each snowman should consist of three space-separated numbers — the big ball’s radius, the medium ball’s radius and the small ball’s radius. It is allowed to print the snowmen in any order. If there are several solutions, print any of them.
Examples
Input
7
1 2 3 4 5 6 7
Output
2
3 2 1
6 5 4
Input
3
2 2 3
Output
0
整理博客才发现，这一周的题目和c++ stl函数库关联很大，说实在的，好好利用真的很节省时间。
题目大意就是给你一串数然后找能够找出多少个由大到小的三个数。（好绕口）
本意上就是一个贪心了。出现次数多的就要多次利用，利用优先队列先进先出的特性。把出现次数多的数字排在前面。首先还应该对数据离散化。用到map函数，好懵逼。
代码如下：
#include<bits/stdc++.h>
using namespace std;
map<int,int>::iterator it;//map函数it指针
const int maxx=1e5+10;
struct node{
int k,v;
node(int _v, int _k):v(_v),k(_k){}
bool operator <(const node &r)const {
return k < r.k;
}
};
map<int, int> mp;
priority_queue<node> que;//优先队列和结构体的结合。
int a[maxx];
int b[maxx];
int c[maxx];
int n;
int main()
{
cin>>n;
for(int i=0;i<n;i++)
{
int x;
cin>>x;
mp[x]++;
}
for(it=mp.begin();it!=mp.end();it++)
{
que.push(node(it->first,it->second));
}
if(que.size()<3) cout<<0<<endl;
else
{
int k=0;
while(que.size()>=3)
{
node xx=que.top();que.pop();
node yy=que.top();que.pop();
node zz=que.top();que.pop();
a[k]=xx.v;b[k]=yy.v;c[k++]=zz.v;
xx.k--;yy.k--;zz.k--;
if(xx.k) que.push(xx);
if(yy.k) que.push(yy);
if(zz.k) que.push(zz);
}
cout<<k<<endl;
for(int i=0;i<k;i++)
{
if(a[i]<b[i]) swap(a[i],b[i]);
if(a[i]<c[i]) swap(a[i],c[i]);
if(b[i]<c[i]) swap(b[i],c[i]);
cout<<a[i]<<" "<<b[i]<<" "<<c[i]<<endl;
}
}
}
努力加油a啊，(o)/~