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  • codeforces 158B

    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 thei-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 integerss1, s2, ..., sn (1 ≤ si ≤ 4). The integers are separated by a space, si 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
    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.


    key:

    Choose the 3 and 1 as much as possible. And let 2 combine with themselves. If there are more 1s and 2s, let two 1s combine with 2, and every four 1s take same taxi.

    #include<stdio.h>
    #include<string.h>
    int main()
    {
    	
    	int a1,a2,a3;
    	int i,n,c,l;
    	while(scanf("%d",&n)!=EOF)
    	{
    		i=0;
    		c=0;a1=0,a2=0,a3=0;
    		while(i<n)
    		{
    			scanf("%d",&l);
    			if(l==4) c++;
    			else if(l==1) a1++;
    			else if(l==2) a2++;
    			else if(l==3) a3++;
    			i++;
    		}
    		if(a3<a1)
    		{
    			c+=a3;
    			a1-=a3;
    			a3=0;
    		}
    		else
    		{
    			c+=a3;
    			a1=0;a3=0;
    		}
    		c+=a2/2;
    		if(a2%2)
    		{
    			c++;a1-=2;
    		}
    		if(a1>0)
    		{
    			c+=a1/4;
    			if(a1%4) c++;
    		}
    		printf("%d
    ",c);
    	}
    	return 0;
    }


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  • 原文地址:https://www.cnblogs.com/leejuen/p/5547495.html
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