Educational Codeforces Round 21 D.Array Division(二分)

最新推荐文章于 2022-07-08 21:41:28 发布

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最新推荐文章于 2022-07-08 21:41:28 发布

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原文链接：http://www.cnblogs.com/ECJTUACM-873284962/p/6870147.html

探讨如何通过移动数组中的一个元素来实现将数组分割为两个部分，使得两部分元素之和相等的问题。采用二分查找优化算法实现。

Vasya has an array a consisting of positive integer numbers. Vasya wants to divide this array into two non-empty consecutive parts (the prefix and the suffix) so that the sum of all elements in the first part equals to the sum of elements in the second part. It is not always possible, so Vasya will move some element before dividing the array (Vasya will erase some element and insert it into an arbitrary position).

Inserting an element in the same position he was erased from is also considered moving.

Can Vasya divide the array after choosing the right element to move and its new position?

Input

The first line contains single integer n (1 ≤ n ≤ 100000) — the size of the array.

The second line contains n integers a₁, a₂... a_n (1 ≤ a_i ≤ 10⁹) — the elements of the array.

Output

Print YES if Vasya can divide the array after moving one element. Otherwise print NO.

Examples

Input

3 
1 3 2

Output

YES

Input

5 
1 2 3 4 5

Output

NO

Input

5 
2 2 3 4 5

Output

YES

Note

In the first example Vasya can move the second element to the end of the array.

In the second example no move can make the division possible.

In the third example Vasya can move the fourth element by one position to the left.

题目链接：http://codeforces.com/contest/808/problem/D

题意：在数组中移动一个数使得分组可以分割成两个数组使得两个数组之和相等

分析：

首先分两种情况
1. 往后面移动一个数到前面 (维护前缀和看看后面有没有符合条件的数)
2. 移掉一个数 (移掉第i个数看看连续的数能不能符合条件)

用二分做，观摩观摩

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int N=100010;
 5 ll s[N],a[N],sum;
 6 int n;
 7 bool find(int l,int r,ll x)
 8 {
 9     while(l<=r)
10     {
11         int mid=(l+r)/2;
12         if(s[mid]==x)
13             return 1;
14         if(s[mid]<x)
15             l=mid+1;
16         else r=mid-1;
17     }
18    return 0;
19 }
20 int main()
21 {
22     scanf("%d",&n);
23     for(int i=1;i<=n;i++)
24     {
25         scanf("%lld",&a[i]);
26         sum+=a[i];
27         s[i]=s[i-1]+a[i];//求前缀和
28     }
29     if(sum&1)
30     {
31         cout<<"NO"<<endl;
32         return 0;
33     }
34     sum/=2;
35     for(int i=1;i<=n;i++)
36     {
37         if(find(i+1,n,sum+a[i])||find(1,i-1,sum-a[i]))
38         {
39             cout<<"YES"<<endl;
40             return 0;
41         }
42     }
43     cout<<"NO"<<endl;
44     return 0;
45 }