Splitting Pile

Splitting Pile

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Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer ai written on it.

They will share these cards. First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards. Here, both Snuke and Raccoon have to take at least one card.

Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively. They would like to minimize |x−y|. Find the minimum possible value of |x−y|.

Constraints

• 2≤N≤2×10⁵
• −10⁹≤ai≤10⁹
• ai is an integer.

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Input

Input is given from Standard Input in the following format:

N
a1 a2 … aN

Output

Print the answer.

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Sample Input 1

6
1 2 3 4 5 6

Sample Output 1

1

If Snuke takes four cards from the top, and Raccoon takes the remaining two cards, x=10, y=11, and thus |x−y|=1. This is the minimum possible value.

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Sample Input 2

2
10 -10

Sample Output 2

20

Snuke can only take one card from the top, and Raccoon can only take the remaining one card. In this case, x=10, y=−10, and thus |x−y|=20.

题意：一共有n张牌，A拿走前面的至少一张，最多只剩一张，B拿走剩下的，问他们各自的牌的和的绝对值差最小是多少？

直接暴力得了，把所有能拿的情况都列出来，然后排个序ok

 #include <iostream>
 #include <stdio.h>
 #include <string.h>
 #include <queue>
 #include <algorithm>
 #include <vector>
 using namespace std;
 #define LL long long
 #define MX 200010

 LL a[MX];
 LL ans[MX];

 int main()
 {
     int n;
     scanf("%d",&n);
     LL sum=;
     for(int i=;i<n;i++)
     {
         cin>>a[i];
         sum+=a[i];
     }
     LL x = a[];
     LL y = sum-a[];
     ans[]=abs(x-y);
     for (int i=;i<n-;i++)
     {
         x+=a[i];
         y-=a[i];
         ans[i]=abs(x-y);
     }
     sort(ans,ans+n-);
     cout<<ans[]<<endl;
     return ;
 }