C. Lucky Tickets
time limit per test
memory limit per test
input
output
Vasya thinks that lucky tickets are the tickets whose numbers are divisible by 3. He gathered quite a large collection of such tickets but one day his younger brother Leonid was having a sulk and decided to destroy the collection. First he tore every ticket exactly in two, but he didn’t think it was enough and Leonid also threw part of the pieces away. Having seen this, Vasya got terrified but still tried to restore the collection. He chose several piece pairs and glued each pair together so that each pair formed a lucky ticket. The rest of the pieces Vasya threw away reluctantly. Thus, after the gluing of the 2t pieces he ended up with t
When Leonid tore the tickets in two pieces, one piece contained the first several letters of his number and the second piece contained the rest.
Vasya can glue every pair of pieces in any way he likes, but it is important that he gets a lucky ticket in the end. For example, pieces 123and 99 can be glued in two ways: 12399 and 99123.
What maximum number of tickets could Vasya get after that?
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of pieces. The second line contains n space-separated numbers ai(1 ≤ ai ≤ 108) — the numbers on the pieces. Vasya can only glue the pieces in pairs. Even if the number of a piece is already lucky, Vasya should glue the piece with some other one for it to count as lucky. Vasya does not have to use all the pieces. The numbers on the pieces an on the resulting tickets may coincide.
Output
Print the single number — the maximum number of lucky tickets that will be able to be restored. Don't forget that every lucky ticket is made of exactly two pieces glued together.
Examples
input
3
123 123 99
output
1
input
6
1 1 1 23 10 3
output
1
把所有的数字对3取模，分别记录0, 1, 2的个数为h0, h1, h2, 那么说有h0个数已经是三的倍数那么他们两两组合形成h0/2张卡片，取模结果为1的有h1张，结果为2的有h2张，它们能形成min(h1, h2)张卡片
#include <bits/stdc++.h>
#define maxn 200005
using namespace std;
typedef long long ll;
int main(){
int n, a;
int k0 = 0, k1 = 0, k2 = 0;
scanf("%d", &n);
for(int i = 0; i < n; i++){
scanf("%d", &a);
if(a % 3 == 0)
k0++;
else if(a % 3 == 1)
k1++;
else
k2++;
}
printf("%d\n", k0 / 2 + min(k1, k2));
return 0;
}