Codeforces1006C——Three Parts of the Array

You are given an array d1,d2,…,dn consisting of n 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 sum1, the sum of elements of the second part be sum2 and the sum of elements of the third part be sum3. Among all possible ways to split the array you have to choose a way such that sum1=sum3 and sum1 is maximum possible.
More formally, if the first part of the array contains a elements, the second part of the array contains b elements and the third part contains c elements, then:
sum1=∑1≤i≤adi,
sum2=∑a+1≤i≤a+bdi,
sum3=∑a+b+1≤i≤a+b+cdi.
The sum of an empty array is 0.
Your task is to find a way to split the array such that sum1=sum3 and sum1 is maximum possible.
Input
The first line of the input contains one integer n (1≤n≤2⋅105) — the number of elements in the array d.
The second line of the input contains n integers d1,d2,…,dn (1≤di≤109) — the elements of the array d.
Output
Print a single integer — the maximum possible value of sum1, considering that the condition sum1=sum3 must be met.
Obviously, at least one valid way to split the array exists (use a=c=0 and b=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=4 is: [1,3],[2,1],[4].
In the third example there is only one way to split the array: [ ],[4,1,2],[ ].

将数组分为三部分 使得第一和第三部分的和相等 求最大的第一部分和
想法就是用两个数组分别记录前缀和后缀和 然后最后枚举后缀和 二分找相等的前缀和 如果两部分不重叠 即更新答案
刚开始双重for超时了
还有记住数组要开longlong

代码：

#include <cstdio>
#include <algorithm>
using namespace std;
const int MAXN=200050;
int a[MAXN];
long long pre[MAXN];
long long las[MAXN];
int main(void){
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    pre[0]=a[0];
    for(int i=1;i<n;i++){
        pre[i]=pre[i-1]+a[i];
    }
    las[n-1]=a[n-1];
    for(int i=n-2;i>=0;i--){
        las[i]=las[i+1]+a[i];
    }
    long long ans=0;
    for(int i=0;i<n;i++){
        int p=lower_bound(pre,pre+n,las[i])-pre;
        if(p<i && pre[p]==las[i]){
            ans=max(ans,pre[p]);
            continue;
        }
    }
    printf("%lld\n",ans);
    return 0;
}