HDU2124 Repair the Wall（贪心）

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

5 3
3 2 1
5 2
2 1

Sample Output

2
impossible

Author

linle

Source

HDU 2007-10 Programming Contest

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
const int maxn=+;
bool cmp(int a,int b)
{
    return a>b;
}
int main()
{
    int len,n,i,res;
    int a[maxn];
    while(scanf("%d%d",&len,&n)!=EOF)
    {
        for(i=;i<n;i++)scanf("%d",&a[i]);
        sort(a,a+n,cmp);
        res=;
        for(i=;i<n;i++)
        {
           if(a[i]>=len)
           {
               res++;
               len-=a[i];
               break;
           }
           else
           {
               len-=a[i];
               res++;
           }
           }
           if(len>)printf("impossible\n");
           else printf("%d\n",res);
    }
    return ;
}