Topic description
【Question 】
CH recently found a hay buy-one-get-one deal on the Internet. For every bale of hay he buys, he can get a bale of hay of size B (1 ≤ B < A) for free. Note that B < A.
However, there are limits to this sale: the hay that costs money is high-quality, and the hay that is given away for free is inferior. Yet the frugal CH doesn't care about quality: he will buy hay of any quality to save money.
Given N bales of high quality hay, and M bales of defective hay, find out how many bales CH can buy at most. He can buy high quality hay alone without receiving free defective hay; but he cannot buy inferior hay alone.
[Input description]
Line 1: Two integers separated by spaces: N and M.
Lines 2 to N+1: One integer per line, representing the size of a bale of high-quality hay.
Lines N+2 to N+M+1: One integer per line representing the size of a bale of defective hay.
[Output description]
The only line: an integer, indicating the maximum number of hay bales that CH can buy.
【Sample input】
3 4
6
1
3
1
5
3
4
[Sample output]
5
【提示】
样例说明:
共有3捆高质量的干草，尺寸分别是6，1，和3； 共有4捆次品干草，尺寸分别是1，5，3，和4。
显然，FJ可以买到所有高质量的干草。当他买尺寸为6的高质量干草时，他可以获赠尺寸为3的次品。当他买尺寸为3的高质量干草时，他可以获赠尺寸为1的次品。然而，当他买质量为1的高质量干草时，无论如何，他都不可能再获赠次品干草。所以答案就是3+2=5。
1 ≤ A ≤ 1，000，000
1 ≤ N ≤ 10，000
1 ≤ M ≤ 10，000
题解：
这道题十分水，首先高级稻草我们可以全部买齐，每个低级稻草都要配对一个离它最近且比他大的稻草才不会浪费，最后加上能派对上高级稻草的低级稻草的个数
代码(建议自己打)：
#include<cstdio>
#include<cstring>
#include<algorithm>
using
namespace
std;
int
a[21000],b[21000],n,m;
int
main()
{
scanf
(
"%d%d"
,&n,&m);
for
(
int
i=1;i<=n;i++)
scanf
(
"%d"
,&a[i]);
for
(
int
i=1;i<=m;i++)
scanf
(
"%d"
,&b[i]);
sort(a+1,a+n+1);
sort(b+1,b+m+1);
int
ni=n;//ni表示没配对低级稻草的个数
for
(
int
i=m;i>=1;i--)
{
if
(ni==0)
break
;
if
(b[i]<a[ni])ni--;//如果这颗高级稻草配对了一枚低级稻草，那么把他踢出为匹配队列
}
printf
(
"%d\n"
,n+n-ni);
return
0;
}