标签：
Repair the Wall
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3076 Accepted Submission(s): 1504
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 ith integer Ai(0<Ai<1000000000 ) means that the ith 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
Author
linle
Source
Recommend
威士忌
注意：使用int变量存储木块的长度和会超出int的范围。
1 #include <cstdio>
2 #include <algorithm>
3 using namespace std;
4 const int maxn = 650;
5 long long a[maxn];
6 int main()
7 {
8 long long L , sum, ans;
9 int n;
10 while (scanf("%lld%d", &L, &n) == 2)
11 {
12 sum = 0; ans = 0;
13 for (int i = 0; i < n; i++)
14 scanf("%lld", &a[i]);
15 sort(a, a + n);
16 int i = n - 1;
17 while (sum < L&&i >=0)
18 {
19 ans++; sum += a[i--];
20 }
21 if (sum < L) printf("impossible\n");
22 else printf("%lld\n", ans);
23 }
24 }
HDU 2124 Repair the Wall
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原文地址：http://www.cnblogs.com/cumulonimbus/p/5169939.html
