题目链接
C. Three Parts of the Array
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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
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],[ ].
算法分析：
本题的一个难点解决就是，如果sum1==sum2，是否要往下进行，进行的话不相等怎么办，所以用一个ans保存。
代码实现：
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
using namespace std;
const double eps = 1e-8;
typedef long long LL;
typedef unsigned long long ULL;
const int INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN=5010;
const int MAXM=100010;
using namespace std;
int a[200005];
int main()
{int n;while(scanf("%d",&n)!=EOF){ for(int i=1;i<=n;i++){scanf("%d",&a[i]);}long long ans=0;long long sum1=a[1];long long sum2=a[n];int l=1,r=n;for(;l<r;){if(sum1==sum2) ans=sum1;if(sum1<sum2){ l++;sum1+=a[l];}else{ r--;sum2+=a[r];}}cout<<ans<<endl;}return 0;
}