Codeforces Round #498 (Div. 3)

Posted moomcake

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了Codeforces Round #498 (Div. 3)相关的知识,希望对你有一定的参考价值。

 

You are given an array d1,d2,,dnd1,d2,…,dn consisting of nn integer numbers.

Your task is to split this array into three parts (some of which may be empty) in such a way that each element of the array belongs to exactly one of the three parts, and each of the parts forms a consecutive contiguous subsegment (possibly, empty) of the original array.

Let the sum of elements of the first part be sum1sum1 , the sum of elements of the second part be sum2sum2 and the sum of elements of the third part be sum3sum3 . Among all possible ways to split the array you have to choose a way such that sum1=sum3sum1=sum3 and sum1sum1 is maximum possible.

More formally, if the first part of the array contains aa elements, the second part of the array contains bb elements and the third part contains cc elements, then:

sum1=1iadi,sum1=∑1≤i≤adi, sum2=a+1ia+bdi,sum2=∑a+1≤i≤a+bdi, sum3=a+b+1ia+b+cdi.sum3=∑a+b+1≤i≤a+b+cdi.

The sum of an empty array is 00 .

Your task is to find a way to split the array such that sum1=sum3sum1=sum3 and sum1sum1 is maximum possible.

Input

The first line of the input contains one integer nn (1n21051≤n≤2⋅105 ) — the number of elements in the array dd .

The second line of the input contains nn integers d1,d2,,dnd1,d2,…,dn (1di1091≤di≤109 ) — the elements of the array dd .

Output

Print a single integer — the maximum possible value of sum1sum1 , considering that the condition sum1=sum3sum1=sum3 must be met.

Obviously, at least one valid way to split the array exists (use a=c=0a=c=0 and b=nb=n ).

Examples
Input
Copy
5
1 3 1 1 4
Output
Copy
5
Input
Copy
5
1 3 2 1 4
Output
Copy
4
Input
Copy
3
4 1 2
Output
Copy
0
Note

In the first example there is only one possible splitting which maximizes sum1sum1 : [1,3,1],[ ],[1,4][1,3,1],[ ],[1,4] .

In the second example the only way to have sum1=4sum1=4 is: [1,3],[2,1],[4][1,3],[2,1],[4] .

In the third example there is only one way to split the array: [ ],[4,1,2],[ ][ ],[4,1,2],[ ] .

题意:将长度为n的集合划分为3个集合要求sum1==sum3,并且求出在满足条件的基础上sum1的最大值,sum1,sum2均可为空集。

分析:我们可以sum1从前往后加,sum3从后往前加,从满足条件(sum1==sum3)中取最大值就OK。

 1 #include <cstdio>
 2 #include <cstring>
 3 #include<algorithm>
 4 #include<iostream>
 5 #include<map>
 6 #include<stack>
 7 #include<cmath>
 8 typedef long long ll;
 9 using namespace std;
10 const int MAXN=1e6+10;
11 int str[MAXN];
12 int  m,n;
13 int main()
14 {
15     cin>>m;
16     for(int i=0;i<m;i++)
17     {
18         cin>>str[i];
19     }
20     int l=1,r=m-2;
21     ll kk=0;
22     ll ans1=str[0],ans2=str[m-1];
23     while(l<=r)
24     {
25         if(ans1<ans2)
26         {
27             ans1+=str[l++];
28         }
29         else if(ans2<ans1)
30         {
31             ans2+=str[r--];
32         }
33         else if(ans1==ans2)
34         {
35 
36             kk=ans1;
37             ans1+=str[l++];
38             if(r>l)//在此一定判断一下是否越界,我死在这一次
39             ans2+=str[r--];
40         }
41     }
42     if(ans1==ans2&&m>1)//当m<2时不满足条件,ans=0
43         kk=max(ans1,kk);
44     cout<<kk<<endl;
45 }

 

以上是关于Codeforces Round #498 (Div. 3)的主要内容,如果未能解决你的问题,请参考以下文章

Codeforces Round #498 (Div. 3) 简要题解

Codeforces Round #498 (Div. 3)

CodeForces Round #498 div3

Codeforces Round #498 (Div. 3)

Codeforces Round #498 (Div. 3) E. Military Problem (DFS)

Codeforces Round #498 (Div. 3) E. Military Problem (DFS)