codeforces 158B Taxi （贪心）

题目：

B. Taxi

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of si friends (1 ≤ si ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of groups of schoolchildren. The second line contains a sequence of integers s1, s2, ..., sn (1 ≤ si ≤ 4). The integers are separated by a space, si is the number of children in the i-th group.

Output

Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.

Sample test(s)

input

5
1 2 4 3 3

output

4

input

8
2 3 4 4 2 1 3 1

output

5

Note

In the first test we can sort the children into four cars like this:

• the third group (consisting of four children),
• the fourth group (consisting of three children),
• the fifth group (consisting of three children),
• the first and the second group (consisting of one and two children, correspondingly).

There are other ways to sort the groups into four cars.

题意分析：

有n组小伙伴，每组都是1到4人，准备做出租车去参加Party。一个出租车只能坐4个人，每个组的小伙伴不能拆分，求需要出租车的数量。

贪心。记录每种一组人数的个数，4个人一组的就直接一辆了。分类讨论在于1人一组和3人一组的个数谁大，如果3人一组的个数多，就拿一人的都填上，然后就需要3人一组的车数，两人一组的就2组一辆车。如果一个人一组的组数多，就先把3人一组的填满，然后再拿1人一组的去补二人一组的。

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;

const int maxn = 100000 + 10;
int a[maxn],b[5];

int main()
{
    int n, i;
    while(~scanf("%d", &n))
    {
        memset(b,0,sizeof(b));
        for(i = 0; i < n; i++)
        {
            scanf("%d", &a[i]);
            b[a[i]]++;
        }
        int ans = b[4] + b[3] + b[2]/2;
        if(b[1] > b[3]) b[1] -= b[3];
        else b[1] = 0;
        b[2] %= 2;
        if(b[2] == 1)
        {
            ans++;
            if(b[1] > 2) b[1] -= 2;
            else b[1] = 0;
        }
        ans += b[1] / 4;
        if(b[1] % 4 != 0) ans++;
        printf("%d\n", ans);
    }
    return 0;
}