CF_525C_IlyaAndSticks

C. Ilya and Sticks

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

In the evening, after the contest Ilya was bored, and he really felt like maximizing. He remembered that he had a set of
n sticks and an instrument. Each stick is characterized by its length
li.

Ilya decided to make a rectangle from the sticks. And due to his whim, he decided to make rectangles in such a way that maximizes their total area. Each stick is used in making at most one rectangle, it is possible that some of sticks remain unused. Bending
sticks is not allowed.

Sticks with lengths a1,
a2,
a3 and
a4 can make a rectangle if the following properties are observed:

a1 ≤ a2 ≤ a3 ≤ a4

a1 = a2

a3 = a4

A rectangle can be made of sticks with lengths of, for example,
3 3 3 3 or 2 2 4 4. A rectangle cannot be made of, for example, sticks
5 5 5 7.

Ilya also has an instrument which can reduce the length of the sticks. The sticks are made of a special material, so the length of each stick can be reduced by at most one. For example, a stick with length
5 can either stay at this length or be transformed into a stick of length
4.

You have to answer the question — what maximum total area of the rectangles can Ilya get with a file if makes rectangles from the available sticks?

Input
The first line of the input contains a positive integer
n (1 ≤ n ≤ 105) — the number of the available sticks.

The second line of the input contains n positive integers
li (2 ≤ li ≤ 106) — the
lengths of the sticks.

Output
The first line of the output must contain a single non-negative integer — the maximum total area of the rectangles that Ilya can make from the available sticks.

Sample test(s)

Input

4
2 4 4 2

Output

8

Input

4
2 2 3 5

Output

0

Input

4
100003 100004 100005 100006

Output

10000800015

简单的模拟，另外贪心

优先取边长最长的两组组成的长方形一定比其他取法的长方形面积大

注意结果是total 不是只构成一个矩形

a*b+c*d其大小关系为a>b>c>d

若改成a*c+b*d

上下做差a*(b-c)+d*(c-b)

(a-d)(b-c)显然是个正值

若改成a*d+b*c

上下做差a*(b-d)+c*(d-b)

(a-c)(b-d)显然也是正值，因此方案一最大

其他的只需要记下数就可以了

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;

typedef long long LL;
const int M=1e6+5;
int len[M];

int main()
{
int n;
int maxl;
int max1,max2;
int le;
long long tot;
while(scanf("%d",&n)!=EOF)
{
memset(len,0,sizeof(len));
tot=0;
maxl=max1=max2=0;
for(int i=0;i<n;i++)
{
scanf("%d",&le);
len[le]++;
maxl=max(maxl,le);
}
for(int i=maxl;i>=2;i--)
{
while(len[i]+len[i+1]>=2)//每次找两边
{
if(len[i+1]==1)
{
len[i+1]--;
len[i]--;
}
else
len[i]-=2;
if(max1==0)
max1=i;
else
{
tot+=(LL)max1*i; //注意数据范围
max1=0;
}
}
}
printf("%I64d\n",tot);
}
return 0;
}