Problem:
Score : 300 points

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 \leq N \leq 2 \times 10^5

-10^{9} \leq a_i \leq 10^{9}

a_i is an integer.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_{N}

Output

Print the answer.

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.

Solution in C (GCC 5.4.1) (judge verdict: Accepted)


#include <stdio.h>
 


long long int minx(long long int v[],int n){
	int i,j,k;
	long long int tmp;
	tmp=v[0];
	for(i=1;i<=n-1;i++){
		if(tmp>v[i])
		tmp=v[i];}
		return tmp;}
int maxx(int v[],int n){
	int i,j,k,tmp;
	tmp=v[0];
	for(i=1;i<=n-1;i++){
		if(tmp<v[i])
		tmp=v[i];}
		return tmp;}
int max(int n1,int n2,int n3){
	int tmp;
	tmp=n1;
	if(n2>tmp)
	tmp=n2;
	if(n3>tmp)
	tmp=n3;
	return tmp;}
int max2(int n1,int n2){
	int tmp;
	tmp=n1;
	if(n2>tmp)
	tmp=n2;
	return tmp;}
int min3(int n1,int n2,int n3){
	int tmp;
	tmp=n1;
	if(n2<tmp)
	tmp=n2;
	if(n3<tmp)
	tmp=n3;
	return tmp;}
int min2(int n1,int n2){
	int tmp;
	tmp=n1;
	if(n2<tmp)
	tmp=n2;
	return tmp;}
long long int abs(long long int n){
	if(n>=0)
	return n;
	else return(-1)*n;
}
int main(void)
{int N,A,M,B,j,i,k,H,W,l,m,flag;
	long long int sum,min;
	scanf("%d",&N);
	long long int a[200000];
	long long int x[200000];
	long long int diff[200000];
	flag=0;
	sum=0;
	for(i=0;i<=N-1;i++){
		scanf("%lld",&a[i]);}
	x[0]=a[0];
	for(j=0;j<=N-1;j++){
		sum+=a[j];}
	for(l=0;l<=N-2;l++){
			x[l+1]=x[l]+a[l+1];
			diff[l]=abs(sum-2*x[l]);}
	printf("%lld",minx(diff,N-1));



	return 0;
}