HDU 2124 Repair the Wall （贪心）

最新推荐文章于 2022-12-07 14:44:37 发布

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最新推荐文章于 2022-12-07 14:44:37 发布

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在一场台风过后，Kitty发现她的木制墙壁出现了一道裂缝。为了修复这道长度为L英寸的裂缝，Kitty从邻居那里得到了N块同样宽度为1英寸的木块。通过切割这些木块或直接使用，目标是以最少的木块数量完成墙壁的修复。

Repair the Wall

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3498 Accepted Submission(s): 1724

Problem Description

Long time ago , Kitty lived in a small village. The air was fresh and the scenery was very beautiful. The only thing that troubled her is the typhoon.

When the typhoon came, everything is terrible. It kept blowing and raining for a long time. And what made the situation worse was that all of Kitty's walls were made of wood.

One day, Kitty found that there was a crack in the wall. The shape of the crack is
a rectangle with the size of 1×L (in inch). Luckly Kitty got N blocks and a saw(锯子) from her neighbors.
The shape of the blocks were rectangle too, and the width of all blocks were 1 inch. So, with the help of saw, Kitty could cut down some of the blocks(of course she could use it directly without cutting) and put them in the crack, and the wall may be repaired perfectly, without any gap.

Now, Kitty knew the size of each blocks, and wanted to use as fewer as possible of the blocks to repair the wall, could you help her ?

Input

The problem contains many test cases, please process to the end of file( EOF ).
Each test case contains two lines.
In the first line, there are two integers L(0<L<1000000000) and N(0<=N<600) which
mentioned above.
In the second line, there are N positive integers. The i ^th integer Ai(0<Ai<1000000000 ) means that the i ^th block has the size of 1×Ai (in inch).

Output

For each test case , print an integer which represents the minimal number of blocks are needed.
If Kitty could not repair the wall, just print "impossible" instead.

Sample Input

Sample Output

  
   2
impossible

Author

linle

Source

HDU 2007-10 Programming Contest

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题意：用最小的木块数，补墙，木块可分割

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
bool cmp(int a,int b)
{
	return a>b;
}
int main()
{
	int l,m,n,i,j,k,a[11000];
	while(scanf("%d%d",&l,&m)!=EOF)
	{
		for(i=0;i<m;i++)
		scanf("%d",&a[i]);
		sort(a,a+m,cmp);
		int flag=0;
		int ans=0;
		for(i=0;i<m;i++)
		{
			if(l>a[i])
			{
				l=l-a[i];
				ans++;
			}
			else 
			{
				l=0;
				ans++;
				break;
			}
		}
		if(i==m)
		printf("impossible\n");
		else
		{
			printf("%d\n",ans);
		}
	}
	return 0;
}