程序師世界是廣大編程愛好者互助、分享、學習的平台,程序師世界有你更精彩!
首頁
編程語言
C語言|JAVA編程
Python編程
網頁編程
ASP編程|PHP編程
JSP編程
數據庫知識
MYSQL數據庫|SqlServer數據庫
Oracle數據庫|DB2數據庫
 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> 158B - Taxi

158B - Taxi

編輯:C++入門知識

158B - Taxi


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.

    import java.util.*;
    
    public class Main {
        public static void main(String[] args) {
            
            Scanner sc=new Scanner(System.in);
            int n=sc.nextInt();
            int n1=0,n2=0,n3=0,count=0;
            for(int i=0;i=n1) {count+=n3;count+=(n2+1)/2;}
            else {
                count+=n3+n2/2;n1-=n3; 
                if(n2%2==0&&n1>0) count+=(n1-1)/4+1;
                else count+=(n1+1)/4+1;
                    }
            System.out.println(count);
        }
    }


  1. 上一頁:
  2. 下一頁:
Copyright © 程式師世界 All Rights Reserved