题目：
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=∑1≤i≤adi,
sum1=∑1≤i≤adi,
sum2=∑a+1≤i≤a+bdi,
sum2=∑a+1≤i≤a+bdi,
sum3=∑a+b+1≤i≤a+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 (1≤n≤2⋅1051≤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 (1≤di≤1091≤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
5
1 3 1 1 4
Output
5
Input
5
1 3 2 1 4
Output
4
Input
3
4 1 2
Output
0
Note
In the first example there is only one possible splitting which maximizes sum1: [1,3,1],[ ],[1,4]
In the second example the only way to have sum1=4sum1=4 is: [1,3],[2,1],[4]
In the third example there is only one way to split the array:,[ ],[4,1,2],[ ]
题意：
输入一个数组，将其分为三个部分,同时需要做到第一组和第三组数据相等，并且保证最大。
解题思路：
从头尾开始分别查找，然后记录头尾两个数组的总和，当两边的和有相等时，就记录并输出
AC代码：
#include
#define ll long long
using namespace std;
int n;
int pre[200005];
int main()
{
cin>>n;
for(int i=0;i>pre[i];
}
ll l = 0, r = n - 1, sl = pre[0], sr = pre[n-1];
ll ans = 0;
while(l < r){
if(sl == sr)
ans = max(ans,sr);
if(sl < sr){
sl += pre[++l];
}
else{
sr += pre[--r];
}
}
cout<