Codeforces-158B

最新推荐文章于 2021-07-25 11:32:42 发布

原创最新推荐文章于 2021-07-25 11:32:42 发布 · 423 阅读

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本文解析了Codeforces平台上的158B题目，该题要求根据给定的学生分组情况计算最少需要多少辆能容纳四人的车。通过详细分析各种学生组合情况，并给出了一段AC代码作为实现示例。

题目意思就是给你不同学生人数的组，要求一个组的人必须要在一辆车上，问你要几辆车，模拟一下

题目链接：http://codeforces.com/problemset/problem/158/B

AC代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define MIN(x,y) x<y?x:y
#define MAX(x,y) x>y?x:y
using namespace std;
const int N=110000;
int a[N];
bool cmp(int x,int y)
{
	return x>y;
}
int main()
{
	int i,j,n,sum,p,pos,four,three,two,one;
	while(~scanf("%d",&n))
	{
		sum=0;
		four=0;
		three=0;
		two=0;
		one=0;
		pos=0;
		for(i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
			if(a[i]==1)
			one++;
			if(a[i]==2)
			two++;
			if(a[i]==3)
			three++;
			if(a[i]==4)
			four++;
		}
	//	printf("%d %d %d %d\n",one,two,three,four);
		sum+=four;
		if(one<three)
		{
			sum+=one;
			three=three-one;
			sum+=three;
			if(two%2==0)
			{
				sum+=(two/2);
			}
			else
			{
				sum+=(two/2+1);
			}
		}
		else if(three<one)
		{
			if(three!=0)
			{
				sum+=three;
				one=one-three;
			}
			if(one<two)
			{
				if(one%2==0)
				{
					one=one/2;
					two+=one;
					if(two%2==0)
					{
						sum+=two/2;
					}
					else
					{
						sum+=(two/2+1);
					}
				}
				else
				{
					one=one/2;
					two+=one;
					if(two%2==0)
					{
						sum+=two/2+1;
					}
					else
					{
						sum+=(two/2+1);
					}
					
				}
			}
			else if(one>two)
			{
				if(one%2==0)
				{
					one=one/2;
					two+=one;
					if(two%2==0)
					{
						sum+=two/2;
					}
					else
					{
						sum+=(two/2+1);
					}
				//	printf("%d\n",sum);
				}
				else
				{
					one=one/2;
					two+=one;
					if(two%2==0)
					{
						sum+=two/2+1;
					}
					else
					{
						sum+=(two/2+1);
					}
				}
			}
			else if(one==two)
			{
				sum+=one;
			}
		}
		else if(three==one)
		{
			sum+=one;
			if(two%2==0)
			{
				sum+=(two/2);
			}
			else
			{
				sum+=(two/2+1);
			}
		}
		printf("%d\n",sum);
	}
	return 0;
}