New Year Snowmen CodeForces - 140C

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.

Example

Input
7
1 2 3 4 5 6 7
Output
2
3 2 1
6 5 4
Input
3
2 2 3
Output
0

题解：每次取数量最多的三个，记录，然后数量减一，再重新取数量最多的三个。正确性不知道怎么证明。
 1 // ConsoleApplication1.cpp: 定义控制台应用程序的入口点。
 2 //
 3 
 4 #include "stdafx.h"
 5 #include<map>
 6 #include<queue>
 7 #include<cstdio>
 8 #include<cstring>
 9 #include<iostream>
10 #include<algorithm>
11 using namespace std;
12 
13 const int maxn = 100005;
14 
15 int n;
16 int vec[maxn][3];
17 
18 struct node {
19     int x, num;
20     node(int x, int num) :x(x), num(num) {};
21     bool operator<(const node& i)const {
22         if (num == i.num) return x < i.x;
23         else return num < i.num;
24     }
25 };
26 
27 int main()
28 {
29     while (scanf_s("%d", &n) != EOF) {
30         map<int, int> p;
31         int temp;
32         for (int i = 0; i < n; i++) {
33             scanf_s("%d", &temp);
34             p[temp]++;
35         }
36         priority_queue<node> q;
37         map<int, int>::iterator it;
38         for (it = p.begin(); it != p.end(); it++) q.push(node(it->first, it->second));
39         int t = 0;
40         while (!q.empty()) {
41             if (q.size() < 3) break;
42             node a = q.top(); q.pop();
43             node b = q.top(); q.pop();
44             node c = q.top(); q.pop();
45             vec[t][0] = a.x; vec[t][1] = b.x; vec[t][2] = c.x;
46             t++;
47             if (a.num - 1 != 0) q.push(node(a.x, a.num - 1));
48             if (b.num - 1 != 0) q.push(node(b.x, b.num - 1));
49             if (c.num - 1 != 0) q.push(node(c.x, c.num - 1));
50         }
51         cout << t << endl;
52         for (int i = 0; i < t; i++) {
53             sort(vec[i], vec[i] + 3);
54             cout << vec[i][2] << " " << vec[i][1] << " " << vec[i][0] << endl;
55         }
56     }
57     return 0;
58 }

 

posted @ 2018-03-03 10:28  天之道，利而不害  阅读(283)  评论(0)  编辑  收藏  举报