Codeforces Round #297 (Div. 2)C. Ilya and Sticks 贪心-优快云博客

Codeforces Round #297 (Div. 2)C. Ilya and Sticks

Time Limit: 2 Sec Memory Limit: 256 MB
Submit: xxx Solved: 2xx

题目连接

http://codeforces.com/contest/525/problem/C

Description

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 l_i.

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 l_i (2 ≤ l_i ≤ 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 Input

Input

4
2 4 4 2

Input

4
2 2 3 5

Input

4
100003 100004 100005 100006

Sample Output

Output

10000800015

HINT

题意：

给你一堆棍子，然后选择一些棍子来组成多个矩形，这些棍子有一个特点，长度为l的可以当成长度为l-1的用，问你最多能够成的矩形总面积是多少！

总面积是多少！

唔，因为很重要，所以说三遍，喵～

题解：

用一个flag[l]记录长度为l的棍子有多少个，flag1[l]记录长度为l+1的棍子有多少个，然后从大往小扫，如果发现flag[l]+flag1[l]=>2的时候，那就把长度为l的当成一条边，然后再i++再继续扫一下这个长度，优先减flag1[l]的数量。

注意，当减flag[l]的时候，flag1[l-1]也会减少~

就这样边扫边维护就好啦~

唔，我感觉我说的不是很清楚……

还是看我呆呆的代码吧

~\(≧▽≦)/~啦啦啦，这道题完啦

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 4000100
#define mod 10007
#define eps 1e-9
//const int inf=0x7fffffff;   //无限大
const int inf=0x3f3f3f3f;
/*

*/
//**************************************************************************************
inline ll read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int flag[maxn];
int flag2[maxn];
map<int,int> s;
int main()
{
    int n;
    n=read();
    int m=0;
    for(int i=0;i<n;i++)
    {
        int x=read();
        flag[x]++;
        flag2[x-1]++;
        m=max(m,x);
    }
    ll ans2=0;
    ll ans1=0;
    ll ans=0;
    for(int i=m;i>=0;i--)
    {
        if(flag[i]+flag2[i]>=2)
        {
            if(ans1==0)
            {
                ans1=i;
                if(flag2[i]==0)
                {
                    flag[i]-=2;
                    flag2[i-1]-=2;
                }
                else if(flag2[i]==1)
                {
                    flag2[i]-=1;
                    flag[i]-=1;
                    flag2[i-1]-=1;
                }
                else
                    flag2[i]-=2;
            }
            else
            {
                if(flag2[i]==0)
                {
                    flag[i]-=2;
                    flag2[i-1]-=2;
                }
                else if(flag2[i]==1)
                {
                    flag2[i]-=1;
                    flag[i]-=1;
                    flag2[i-1]-=1;
                }
                else
                    flag2[i]-=2;
                ans+=ans1*i;
                ans1=0;
            }
            i++;
        }
    }
    cout<<ans<<endl;
    return 0;
}