Problem Description
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
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 ith integer Ai(0<Ai<1000000000 ) means that the
ith block has the size of 1×Ai (in inch).
Output
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
Source
#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 ;
}