Repair the Wall （贪心）

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 ?

InputThe 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). 
OutputFor 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

题目意思：Kitty的墙裂了，出现了一个长为L宽为1的矩形，幸好他身边有一些同样宽为1的矩形积木，但是这些积木的长度不确定，问你最少需要多少块
积木就能将墙补好。

解题思路：简单的贪心题目，贪心策略，将积木按长度从大到小排列，每次都先取最长的，直到所加的积木的长度大于墙缝即可，不需要恰好等于。

上代码：

 #include<stdio.h>
 #include<algorithm>
 using namespace std;
 int my_comp(int a,int b)
 {
     if(a>b)
         return ;
     else
         return ;
 }
 int main()
 {
     int i,l,n,sum,count;
     int a[];
     while(scanf("%d%d",&l,&n)!=EOF)
     {
         sum=;
         count=;
         for(i=;i<n;i++)
         {
             scanf("%d",&a[i]);
         }
         sort(a,a+n,my_comp);
         for(i=;i<n;i++)
         {
             sum=sum+a[i];
             count++;
             if(sum>=l)
             {
                 break;
             }
         }
         if(i>=n)
             printf("impossible\n");
         else
             printf("%d\n",count);

     }
     return ;
 }