CF——Taxi

本文主要是介绍CF——Taxi，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

Description

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 s_i friends (1 ≤ s_i ≤ 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, ..., s_n (1 ≤ s_i ≤ 4). The integers are separated by a space, s_i 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 Input

5
1 2 4 3 3

4

8
2 3 4 4 2 1 3 1

5

Hint
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.
题解：有N组人打出租，出租车最多4个人，且每组人都要在同一个车里，问至少需要多少车
#include <string.h>
#include <stdio.h>
#include <malloc.h>
int main()
{
int n,m,i,j,yu = 0,yu2 = 0;
int has[5];
memset(has,0,sizeof(has));
scanf("%d",&n);
while(n--)
{
scanf("%d",&m);
has[m]++;
}
int cha = has[3] - has[1];
if(cha == 0)
i = has[3];
else if(cha > 0)
i = has[1] + cha;
else if(cha < 0)
{
cha = -cha;
yu = cha % 4;
cha = cha / 4;
i = has[3] + cha;
}
yu2 = has[2] % 2;
int shang = has[2] / 2;
j = shang;
int q;
if(((yu == 1 || yu == 2) && yu2 == 1) || (yu != 0 && yu2 == 0) || (yu == 0 && yu2 == 1))
q = 1;
else if(yu == 3 && yu2 == 1)
q = 2;
else
q = 0;
printf("%d\n",i+j+has[4] + q);
return 0;
}

这篇关于CF——Taxi的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！