cf刷题---贪心算法应用-B. Taxi_all possible digits time limit per test3 seconds-优快云博客

探讨一组学校儿童如何有效使用最少数量的出租车前往聚会，每个出租车最多可载四名乘客，确保同一组的孩子们能同乘一辆车。通过算法优化，实现资源的有效分配。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

B. Taxi
time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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.

Examples
inputCopy
5
1 2 4 3 3
outputCopy
4
inputCopy
8
2 3 4 4 2 1 3 1
outputCopy
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.

#include<bits/stdc++.h>
using namespace std;

int n,num=0,sum=0,a[100005];//num为车辆数

bool cmp(int a,int b){return a>b;}

int main()
{
    cin>>n;
    for(int i=0;i<n;++i)
    {
		cin>>a[i];
	}
    sort(a,a+n,cmp);									//降序排列
    for(int i=0;i<n;i++)
	{
													//这里注意小于n，即前面往后移，后面往前贪心。
        sum=a[i]+a[n-1];
       // cout<<sum<<endl;
        if(sum<=4)
		{	
														//如果sum不大于4，说明可能还有几个组的人可以容纳进一辆车
            num++;n--;									//车数先加1，队伍减1，再来看看能否再容纳一个组的人数
            while(sum+a[n-1]<=4){sum+=a[n-1];n--;}			//如果还可以容纳，那就n--；
        }
        else num++;	
		//cout<<num<<endl;										//如果sum比4大，说明a[i]这个组的人可以单独坐一辆车
    }
    //cout<<num<<endl;
    return 0;
}