【Codeforces 466C】Number of Ways

【链接】 我是链接,点我呀:)
【题意】

让你把数组分成3个连续的部分 每个部分的和要一样 问你有多少种分法

【题解】

先处理出来num[i] 表示i..n这里面有多少个j 满足aft[j] = aft[i]/2 这aft[i]=a[j]+a[j+1]..+a[n] 然后for从1..n 看看pre[i]*2=aft[i+1]是否成立。 如果成立的话那么答案就加上num[i+1]

【代码】

import java.io.*;
import java.util.*;


public class Main {
	
	static int N = (int)5e5;
	static InputReader in;
	static PrintWriter out;
		
	public static void main(String[] args) throws IOException{
		in = new InputReader();
		out = new PrintWriter(System.out);
		//code start from here
		new Task().solve(in, out);
		out.close();
	}
	
	static class Task{
		public void solve(InputReader in,PrintWriter out) {
			int n = in.nextInt();
			int []a = new int[N+10];
			long []pre = new long[N+10];
			long []aft = new long[N+10];
			int []num = new int[N+10];
			for (int i = 1;i <= n;i++) a[i] = in.nextInt();
			for (int i = 1;i <= n;i++) pre[i] = pre[i-1]+a[i];
			for (int i = n;i >= 1;i--) aft[i] = aft[i+1]+a[i];
			Hashtable<Long, Integer> dic = new Hashtable<Long,Integer>();
			for (int i = n;i >= 1;i--) {
				if (aft[i]%2==0) {
					long temp = aft[i]/2;
					if (dic.get(temp)!=null)
						num[i] = dic.get(temp);
				}
				Integer temp1 = dic.get(aft[i]);
				if (temp1==null) {
					dic.put( aft[i], 1);
				}else {
					dic.put(aft[i], temp1+1);
				}
			}
			long ans = 0;
			for (int i = 1;i <=n;i++) {
				long cur = pre[i];
				cur = pre[n]-cur;
				if (cur!=pre[i]*2) {
					continue;
				}
				ans = ans + num[i+1];
			}
			out.println(ans);
		}
	}

	

	static class InputReader{
		public BufferedReader br;
		public StringTokenizer tokenizer;
		
		public InputReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
			tokenizer = null;
		}
		
		public String next(){
			while (tokenizer==null || !tokenizer.hasMoreTokens()) {
				try {
				tokenizer = new StringTokenizer(br.readLine());
				}catch(IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		
		public int nextInt() {
			return Integer.parseInt(next());
		}
	}
}