codeforces 899A Splitting in Teams题解

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本文介绍了一种算法问题，旨在通过组合单人和双人学生小组来形成尽可能多的三人团队参加比赛。提供了问题背景、输入输出说明及示例，并附带了C语言实现代码。

题目链接：点击打开链接

A. Splitting in Teams

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There were n groups of students which came to write a trainingcontest. A group is either one person who can write the contest with anyoneelse, or two people who want to write the contest in the same team.

The coach decided to form teams of exactly three people for this training.Determine the maximum number of teams of three people he can form. It ispossible that he can't use all groups to form teams. For groups of two, eitherboth students should write the contest, or both should not. If two studentsfrom a group of two will write the contest, they should be in the same team.

Input

The first line contains single integer n (2 ≤ n ≤ 2·10⁵) — the number ofgroups.

The second line contains a sequence of integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 2), where a_i isthe number of people in group i.

Output

Print the maximum number of teams of three people the coach can form.

Examples

Input

4
1 1 2 1

Output

Input

2
2 2

Output

Input

7
2 2 2 1 1 1 1

Output

Input

3
1 1 1

Output

Note

In the first example the coach can form one team. For example, he can takestudents from the first, second and fourth groups.

In the second example he can't make a single team.

In the third example the coach can form three teams. For example, he cando this in the following way:

The first group (of two people) and the seventh group (of one person),
The second group (of two people) and the sixth group (of one person),
The third group (of two people) and the fourth group (of one person).

大意：给定一系列数据（只有1和2）将一个1和一个2组合或者三个1组合，求最大的组合方式可以有多少种。

分析：先把2都和1组合，再判断剩下的1可以组成多少组。（水题）

代码如下：

#include <stdio.h>
#include <string.h>
int main() {
    int n;
    while (scanf("%d",&n) != EOF) {
        int a,count1 = 0, count2 = 0;
        while (n--){
            scanf("%d",&a);
            if (a ==1) count1++;
            else if(a == 2)count2++;
        }
        int countn =count1 < count2 ? count1 : count2;
        if (count1 >count2)
            countn +=(count1 - count2) / 3;
        printf("%d\n",countn);
    }
}