Codeforces 651B Beautiful Paintings(贪心策略—两个优先队列实现)

B. Beautiful Paintings

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

There are n pictures delivered for the new exhibition. The i-th
painting has beauty ai.
We know that a visitor becomes happy every time he passes from a painting to a more beautiful one.

We are allowed to arranged pictures in any order. What is the maximum possible number of times the visitor may become happy while passing all pictures from first to last? In other words, we are allowed to rearrange elements of a in
any order. What is the maximum possible number of indices i (1 ≤ i ≤ n - 1),
such that ai + 1 > ai.

Input

The first line of the input contains integer n (1 ≤ n ≤ 1000) —
the number of painting.

The second line contains the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000),
where ai means the beauty
of the i-th painting.

Output

Print one integer — the maximum possible number of neighbouring pairs, such that ai + 1 > ai,
after the optimal rearrangement.

Examples

input

5
20 30 10 50 40

output

4

input

4200 100 100 200

output

2

Note

In the first sample, the optimal order is: 10, 20, 30, 40, 50.

In the second sample, the optimal order is: 100, 200, 100, 200.

题意：游客在欣赏画时，通常会因为当前这幅画比前一幅画美丽而获得一点幸福度。  给出n副画作的美丽值，现在要将他们排成一行。求出游客能获得的最大幸福值。

题解：贪心策略。从小到大排序，按顺序站队列。比当前值大的就站进去，和当前值一样大的就存入一个新的容器。处理到最后一个就按照上面方法处理那个存重复值的新容器。以此方式循环下去，知道所有的数都站进了自己的位置。 按照上述思想，我们可以用两个优先队列实现。很简单

代码如下：

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int main()
{
int n,i,a,b;
while(scanf("%d",&n)!=EOF)
{
priority_queue<int, vector<int>, greater<int> >q1;
priority_queue<int, vector<int>, greater<int> >q2;
for(i=0;i<n;++i)
{
scanf("%d",&a);
q1.push(a);
}
int ans=0;
while((!q1.empty())||(!q2.empty()))
{
if(!q1.empty())
{
a=q1.top();
q1.pop();
}
while(!q1.empty())
{
b=q1.top();
q1.pop();
if(b>a)
{
ans++;
a=b;
}
else
q2.push(b);
}
if(!q2.empty())
{
a=q2.top();
q2.pop();
}
while(!q2.empty())
{
b=q2.top();
q2.pop();
if(b>a)
{
ans++;
a=b;
}
else
q1.push(b);
}
}
printf("%d\n",ans);
}
return 0;
}