Codeforces Round #498 (Div. 3)

Posted 2020-12-27 moomcake

 

You are given an array ${\mathrm{d1,d2,\dots ,dnd1,d2,\dots ,dn\; consisting\; of\; nn\; integer\; numbers.}}^{}$

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

More formally, if the first part of the array contains $\mathrm{aa\; elements,\; the\; second\; part\; of\; the\; array\; contains\; bb\; elements\; and\; the\; third\; part\; contains\; cc\; elements,\; then:}$

$$\mathrm{sum1=\sum 1\le i\le adi,sum1=\sum 1\le i\le adi,\; sum2=\sum a+1\le i\le a+bdi,sum2=\sum a+1\le i\le a+bdi,\; sum3=\sum a+b+1\le i\le a+b+cdi.sum3=\sum a+b+1\le i\le a+b+cdi.}$$

The sum of an empty array is $\mathrm{00\; .}$

Your task is to find a way to split the array such that $\mathrm{sum1=sum3sum1=sum3\; and\; sum1sum1\; is\; maximum\; possible.}$

Input

The first line of the input contains one integer $\mathrm{nn\; (1\le n\le 2\cdot 1051\le n\le 2\cdot 105\; )\; —\; the\; number\; of\; elements\; in\; the\; array\; dd\; .}$

The second line of the input contains $\mathrm{nn\; integers\; d1,d2,\dots ,dnd1,d2,\dots ,dn\; (1\le di\le 1091\le di\le 109\; )\; —\; the\; elements\; of\; the\; array\; dd\; .}$

Output

Print a single integer — the maximum possible value of $\mathrm{sum1sum1\; ,\; considering\; that\; the\; condition\; sum1=sum3sum1=sum3\; must\; be\; met.}$

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

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 $\mathrm{sum1sum1\; :\; [1,3,1],[ ],[1,4][1,3,1],[ ],[1,4]\; .}$

In the second example the only way to have $\mathrm{sum1=4sum1=4\; is:\; [1,3],[2,1],[4][1,3],[2,1],[4]\; .}$

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 }