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

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

題目連接

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 8 Output 0 Output 10000800015

HINT

?

?

題意：

給你一堆棍子，然后選擇一些棍子來組成多個矩形，這些棍子有一個特點，長度為l的可以當成長度為l-1的用，問你最多能夠成的矩形總面積是多少！

總面積是多少！

總面積是多少！

唔，因為很重要，所以說三遍，喵～

題解：

用一個flag[l]記錄長度為l的棍子有多少個，flag1[l]記錄長度為l+1的棍子有多少個，然后從大往小掃，如果發現flag[l]+flag1[l]=>2的時候，那就把長度為l的當成一條邊，然后再i++再繼續掃一下這個長度，優先減flag1[l]的數量。

注意，當減flag[l]的時候，flag1[l-1]也會減少~

就這樣邊掃邊維護就好啦~

唔，我感覺我說的不是很清楚……

還是看我呆呆的代碼吧

~\(≧▽≦)/~啦啦啦，這道題完啦

代碼：

?

//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; //無限大 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;}elseflag2[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;}elseflag2[i]-=2;ans+=ans1*i;ans1=0;}i++;}}cout<<ans<<endl;return 0; }

?

轉載于:https://www.cnblogs.com/qscqesze/p/4371809.html

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