codeforces158B Taxi(贪心)

After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that thei-th group consists ofs_i friends (1 ≤ s_i ≤ 4), and they want toGo 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 integerss₁, s₂, ..., s_n (1 ≤ s_i ≤ 4). The integers are separated by a space,s_i is the number of children in thei-th group.

Output

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

Sample test(s)

Input

51 2 4 3 3

Output

4

Input

82 3 4 4 2 1 3 1

Output

5

这个题就是贪心的一个运用，每个队有１到４人，每个车可以容纳４个人，每各队必须坐一辆车，但是一个车可以坐几个不同的队伍只要不超过４个人．所以这题的思路就是先进行由大到小的排序，接着最大的和最小的加起来和４比较，如果大于４，队伍减一，车数加一，如果小于４的话，队伍减２，车数加一．再加一下最小的前面的，如果仍旧可以想加，队伍再减一，车数不变．

#include<iostream>#include<algorithm>using namespace std;int a[100010];int cmp(int a,int b){return a>b;}int main(){int n;cin>>n;for(int i=0;i<n;i++)   cin>>a[i];sort(a,a+n,cmp);//for(int i=0;i<n;i++)  cout<<a[i]<<" ";int min=0,sum=0;for(int i=0;i<n;i++){int sum=a[i]+a[n-1];if(sum<=4){min++;n--;while(sum+a[n-1]<=4){n--;sum+=a[n-1];}}else min++;}cout<<min<<endl;return 0;}