codeforces 158B Taxi（贪心小水题）

B. Taxi
点击打开题目

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 thei-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 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

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.

题目大意：有n个组，每个组有si人，1=<si<=4的，每辆taxi只能做四个人，并且要求每个组的人坐在一起，求需要的taxi的最小辆数。

贪心策略：先装最接近四个人的，不够四个的再要过来人数少的组拼车。

代码：

#include <iostream>#include<stdlib.h>using namespace std;int cmp(const void *a,const void *b){    return *(int *)b-*(int *)a;}int a[100010];int main(){    int n,i,ans,s,k;    while(cin>>n)    {        for(i=0; i<n; i++)        {            cin>>a[i];        }        ans=0;        qsort(a,n,sizeof(a[0]),cmp);        for(i=0,s=n-1; i<n; i++)        {            if(a[i]==0)                continue;            if(a[i]==4)//四个的先装            {                ans++;            }            else if(a[i]<4)//不足四个再判断            {                k=a[i]+a[s];                if(k>4)                    ans++;                else if(k==4)                {                    ans++;                    a[s]=0;                    s--;                }                else if(k<4)                {                    a[s]=0;                    for(s--;s>=i;s--)//在人数少的组中找到一些组，使得总人数最接近四个                    {                        k+=a[s];                        if(k>4)                        {                            break;                        }                        else if(k==4)                        {                            a[s]=0;                            break;                        }                        else                        {                            a[s]=0;                        }                    }                    ans++;                }            }        }        cout<<ans<<endl;    }    return 0;}