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问题 D: Repair the Wall
时间限制: 1 Sec 内存限制: 32 MB题目描述
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.
输入
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 ith integer Ai(0<Ai<1000000000 ) means that the ith block has the size of 1×Ai (in inch).
输出
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.
样例输入
2 2
12 11
14 3
27 11 4
109 5
38 15 6 21 32
5 3
1 1 1
样例输出
1
1
5
impossible
#include<stdio.h>
#include<algorithm>
using namespace std;
bool cmp(int a,int b)
{return a>b;
}
int main()
{int l,n;while(scanf("%d%d",&l,&n)!=EOF){int an[605]={0},sum=0,num=0;for(int i=0;i<n;i++){scanf("%d",&an[i]);sum+=an[i];}if(sum<l){printf("impossible\n");continue;}if(sum==l){printf("%d\n",n);continue;}sum=0;sort(an,an+n,cmp);for(int i=0;i<n;i++){sum+=an[i];if(sum>=l){printf("%d\n",i+1);break;} }}return 0;
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