Problem Statement

Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i 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≤a_i≤10⁹
• a_i is an integer.

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.

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.

前缀和查询

#include <iostream>

#include <cstdio>

#include <cstring>

#include <queue>

#include <cmath>

#include <vector>

#include <algorithm>

using namespace std;

#define lowbit(x) (x&(-x))

#define max(x,y) (x>y?x:y)

#define min(x,y) (x<y?x:y)

#define MAX 100000000000000000

#define MOD 1000000007

#define PI 3.141592653589793238462

#define INF 1000000000

#define mem(a) (memset(a,0,sizeof(a)))

typedef long long ll;

ll x,sum[],n,ans,minn;

int main()

{

    scanf("%lld",&n);

    sum[]=;

    for(ll i=;i<=n;i++)

    {

        scanf("%lld",&x);

        sum[i]=sum[i-]+x;

    }

    ans=sum[n];

    minn=abs(*sum[]-sum[n]);

    for(ll i=;i<=n-;i++)

    {

        minn=min(minn,abs(*sum[i]-sum[n]));

    }

    printf("%lld\n",minn);

    return ;

}