Problem:
Score: 200 points

Problem Statement

Takahashi, who works at DISCO, is standing before an iron bar.
The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters.

Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length.
However, this may not be possible as is, so he will do the following operations some number of times before he does the cut:

Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan).

Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen.

Find the minimum amount of money needed before cutting the bar into two parts with the same length.

Constraints

2 \leq N \leq 200000

1 \leq A_i \leq 2020202020

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 A_3 ... A_N

Output

Print an integer representing the minimum amount of money needed before cutting the bar into two parts with the same length.

Sample Input 1

3
2 4 3

Sample Output 1

3

The initial lengths of the sections are [2, 4, 3] (in millimeters). Takahashi can cut the bar equally after doing the following operations for 3 yen:

Shrink the second section from the left. The lengths of the sections are now [2, 3, 3].

Shrink the first section from the left. The lengths of the sections are now [1, 3, 3].

Shrink the second section from the left. The lengths of the sections are now [1, 2, 3], and we can cut the bar at the second notch from the left into two parts of length 3 each.

Sample Input 2

12
100 104 102 105 103 103 101 105 104 102 104 101

Sample Output 2

0

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.NoSuchElementException;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeSet;
 
 
public class Main {
 
	static long MOD = 1_000_000_007;
	static int INF = 1_000_000_007;
	//static long INF = Long.MAX_VALUE;
	//二項係数関連ここから-----
	static boolean isPreprocessing = false;	//二項係数を計算するための前処理を行っているかどうかのフラグ
	static int MAX = 666667;				//MAXまで階乗やそれらの逆元のテーブルを保持する。
	static long[] fac;	//fac[i] := i! % MOD
	static long[] inv;	//inv[i] := mod. MODにおける i の逆元
	static long[] finv;	//finv[i] := mod. MODにおける i! の逆元(invの累積"積")
	//二項係数関連おわり------
	
	static int[] roots;		//Union-find木用
	static int[] ans;
	//static List<ArrayList<Integer>> G;	//グラフ用

	static List<List<E>> G = new ArrayList<>();	//グラフ用
	static List<E> Edges = new ArrayList<>();	//グラフ用
	static int V;								//グラフ用頂点の数
	static int E;								//グラフ用辺の数
	
	
	public static void main(String[] args) throws Exception {
		FastScanner scan = new FastScanner();
		//int N = scan.nextInt();
		//long L = scan.nextLong();
		//char[] A = scan.next().toCharArray();
		int N = scan.nextInt();
		long[] A = new long[N];
		long[] B = new long[N];
		long sum = 0;
		for(int i=0 ;i<N; i++){
			A[i] = scan.nextLong();
			if(i==0){
				B[i] = A[i];
			}else{
				B[i] = B[i-1] + A[i];
			}
			sum += A[i];
		}
		long ans = Long.MAX_VALUE;
		for(int i=0; i<N; i++){
			long tmp = Math.abs(2*B[i]-sum);
			ans = Math.min(ans, tmp);
		}
		
		System.out.println(ans);
		
	}
	
	//２点間の距離
	public static double distance(double X1, double Y1, double X2, double Y2){
		return Math.sqrt((X1-X2)*(X1-X2) + (Y1-Y2)*(Y1-Y2));
	}
	
	//階乗
	public static int fact(int a){
		if(a == 0 || a == 1)return 1;
		return a*fact(a-1);
	}
	
	
	//約数の配列を返す
	static int[] divisor(int num){
		Set<Integer> divisors = new TreeSet<>();
		int tmp=1;
		while(tmp*tmp<=num){
			if(num%tmp == 0){
				divisors.add(tmp);
				divisors.add(num/tmp);
			}
			tmp++;
		}
		Integer[] rtn = new Integer[divisors.size()];
		divisors.toArray(rtn);
		int[] ans = new int[rtn.length];
		for(int i=0; i<rtn.length; i++){
			ans[i] = rtn[i];
		}
		return ans;
	}
	
	//GCD最大公約数を返す
	static long gcd(long a, long b){
		if(b == 0) return a;
		return gcd(b, a%b);
	}
	
	//Bellman-Ford法
	//辺情報とstartの頂点を指定すると、すべての頂点への最短コストを返す（負の重みの辺もOK）
	//(問題に応じて要変化)1→Vのパスに負の閉路がある場合はindex:Vの要素に-1を入れて返す。
	static long[] bellmanFord(List<E> Edges, int from){
		long[] rtn = new long[V+1];
		Arrays.fill(rtn, INF);
		rtn[from] = 0;
		int count = 0;
		
		for(int i=0; i<V-1; i++){
			boolean update = false;
			for(E e : Edges){
				if(rtn[e.from] != INF && rtn[e.to] > rtn[e.from] + e.weight){
					rtn[e.to] = rtn[e.from] + e.weight;
					update = true;
				}
			}
			if(!update)break;
		}	
		//閉経路判定
		boolean[] is_update = new boolean[V];
		Arrays.fill(is_update, false);
		for(int i=0; i<V; i++){
			for(E e : Edges){
				if(rtn[e.from] != INF && rtn[e.to] > rtn[e.from] + e.weight){
					rtn[e.to] = rtn[e.from] + e.weight;
					is_update[e.to] = true;
				}
			}
		}
		
		if(is_update[V-1]){
			rtn[V] = -1;
		}
		return rtn;
	}
	
	//Kruskal法
	//最小全域木を構成したときのコストの総和を返す
	static long kruskal(List<E> Edges){
		Collections.sort(Edges);
		UnionFind uni = new UnionFind(V);
		long res = 0L;
		for (E e : Edges){
			if(!uni.same(e.from, e.to)){
				uni.join(e.from, e.to);
				res += e.weight;
			}
		}
		return res;
	}
	
	//Dijkstra法
	//辺情報とstartの頂点を指定すると、すべての頂点への最短コストを返す
	static long[] dijkstra(List<List<E>> G, int from){
		int N = G.size();
		long[] rtn = new long[N];
		Arrays.fill(rtn, INF);
		rtn[from] = 0;
		Queue<V> que = new PriorityQueue<>();
		
		que.add(new V(from,0));
        while(!que.isEmpty()){
            V doneNode = que.poll();
            for(E e: G.get(doneNode.num)){
                  long alt = rtn[doneNode.num] + e.weight;
                  if(rtn[e.to] > alt){
                        rtn[e.to] = alt;
                        que.add(new V(e.to, alt));
                  }     
            }
        }
		return rtn;
	}
	
	//グラフ用：辺
	static class E implements Comparable<E>{
		int from;
		int to;
		long weight;
		public E(int from, int to, long weight){
			this.from = from;
			this.to = to;
			this.weight = weight;
		}
		@Override
		public int compareTo(E o) {
			// TODO Auto-generated method stub
			return Long.compare(this.weight, o.weight);
		}
	}
	//グラフ用：頂点
	static class V implements Comparable<V>{
		int num;
		long cost;
		public V(int num, long cost){
			this.num = num;
			this.cost = cost;
		}
		//コストの低い順に並べる(PriorityQueue用)
		@Override
		public int compareTo(V o) {
			return Long.compare(this.cost, o.cost);
		}
	}
	
	//Union-Find木
	static class UnionFind{
		private int[] roots;
		public UnionFind(int N){
			this.roots = new int[N];
			for(int i=0; i<N; i++){
				this.roots[i] = i;
			}
		}
		
		public boolean same(int a, int b){
			if(root(a) == root(b)){
				return true;
			}else{
				return false;
			}
		}
		
		public int root(int a){
			if(a == this.roots[a])return a;
			return this.roots[a] = root(this.roots[a]);
		}
		
		public void join(int a, int b){
			if(root(a) == root(b))return;
			this.roots[root(b)] = root(a);
		}
		
		public int[] getRoots(){
			return this.roots;
		}
	}
 
	//素因数分解をする(index_0:素数 index_1:個数)
	static List<long[]> factorization(long M){
		List<long[]> primes = new ArrayList<>();
 
		long now_prime = 2;
		long now_M = M;
		while(now_prime*now_prime<=M){
			if(now_M%now_prime == 0){
				now_M /= now_prime;
				long[] prime = new long[2];
				prime[0] = now_prime;
				prime[1] = 1;
				while(now_M%now_prime == 0){
					now_M /= now_prime;
					prime[1]++;
				}
				primes.add(prime);
				
			}
			if(now_prime%2 == 0){
				now_prime++;
			}else{
				now_prime += 2;
			}
		}
		if (now_M>1 || M==1){
			long[] prime = new long[2];
			prime[0] = now_M;
			prime[1] = 1;
			primes.add(prime);
		}
		
		return primes;
	}
	
	//２つのint配列を連結させる
	static int[] joinArray(int[] A, int[] B){
		int[] tmp = new int[A.length + B.length];
		System.arraycopy(A, 0, tmp, 0, A.length);
		System.arraycopy(B, 0, tmp, A.length, B.length);
		return tmp;
	}
	
	//２つのint配列が等しいかどうか判定する
	static boolean arrayEqual(int[] A, int[] B){
		if(A.length != B.length)return false;
		for(int i=0; i<A.length; i++){
			if(A[i] != B[i])return false;
		}
		return true;
	}
	
	//int配列をindex順に見て、要素の重複があった場合その要素と間の要素を削除する12123→23
	static int[] calc(int[] A){
		boolean[] tmp = new boolean[A.length];
		Arrays.fill(tmp, true);
		int fetch = 0;
		while(fetch<A.length-1){
			boolean isExist = false;
			for(int i=fetch+1; i<A.length;i++){
				if(A[fetch] == A[i]){
					for(int j=fetch;j<=i;j++){
						tmp[j]=false;
					}
					fetch = i+1;
					isExist = true;
					break;
				}
			}
			if(!isExist)fetch++;
		}
		List<Integer> list = new ArrayList<>();
		for(int i=0; i<A.length; i++){
			if(tmp[i])list.add(A[i]);
		}
		
		int[] foo = new int[list.size()];
		for(int i=0; i<list.size(); i++){
			foo[i] = list.get(i);
		}
		/*
		for(boolean t : tmp)
			System.out.print(t+" ");
		*/
		return foo;
	}
		
	//aとbの最小公倍数を返す
	static long lcm (long a, long b) {
		return ((a/gcd(a,b))*b);
	}
	
	//二項係数用の前処理
	static void COMinit(){
		//二項係数演算用前処理
		fac = new long[MAX+1];
		fac[0] = 1L;
		fac[1] = 1L;
		inv = new long[MAX+1];
		inv[1] = 1L;
		finv = new long[MAX+1];
		finv[0] = 1L;
		finv[1] = 1L;
		for(int i=2;i<MAX;i++){
			fac[i] = fac[i-1] * i % MOD;
			inv[i] = MOD - inv[(int)(MOD%i)] * (MOD/i) % MOD;
			finv[i] = finv[i-1] * inv[i] % MOD;
		}
		isPreprocessing = true;
	}
	
	//二項係数i_C_jをMODで割った余りを返す
	static long COM(int i, int j){
		if(i<j)return 0;
		if(i<0 || j<0)return 0;
		if(!isPreprocessing)COMinit();
		return fac[i] * (finv[j] * finv[i-j] % MOD) % MOD;
	}
	
	//入力
	//https://qiita.com/p_shiki37/items/a0f6aac33bf60f5f65e4
	static class FastScanner {
	    private final InputStream in = System.in;
	    private final byte[] buffer = new byte[1024];
	    private int ptr = 0;
	    private int buflen = 0;
	    private boolean hasNextByte() {
	        if (ptr < buflen) {
	            return true;
	        }else{
	            ptr = 0;
	            try {
	                buflen = in.read(buffer);
	            } catch (IOException e) {
	                e.printStackTrace();
	            }
	            if (buflen <= 0) {
	                return false;
	            }
	        }
	        return true;
	    }
	    private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
	    private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
	    public boolean hasNext() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++; return hasNextByte();}
	    public String next() {
	        if (!hasNext()) throw new NoSuchElementException();
	        StringBuilder sb = new StringBuilder();
	        int b = readByte();
	        while(isPrintableChar(b)) {
	            sb.appendCodePoint(b);
	            b = readByte();
	        }
	        return sb.toString();
	    }
	    public long nextLong() {
	        if (!hasNext()) throw new NoSuchElementException();
	        long n = 0;
	        boolean minus = false;
	        int b = readByte();
	        if (b == '-') {
	            minus = true;
	            b = readByte();
	        }
	        if (b < '0' || '9' < b) {
	            throw new NumberFormatException();
	        }
	        while(true){
	            if ('0' <= b && b <= '9') {
	                n *= 10;
	                n += b - '0';
	            }else if(b == -1 || !isPrintableChar(b)){
	                return minus ? -n : n;
	            }else{
	                throw new NumberFormatException();
	            }
	            b = readByte();
	        }
	    }
	    public int nextInt() {
	        long nl = nextLong();
	        if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
	        return (int) nl;
	    }
	    public double nextDouble() { return Double.parseDouble(next());}
	}
}