题解 | #小名的回答#

1. 从(0,0)走到(a,b)最少要a+b步，
2. 在每一步的移动中总会使得x或y其中一个从奇数变成偶数或从偶数变成奇数。如果起点与终点x，y的性质完全相同或完全相反，需要一共走偶数步，否则一共走奇数步。因此，如果多走，只能多走2n步，才能保持与终点的奇偶性不变。

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

public class Main {

	public static void main(String[] args) {
		InputReader in = new InputReader();
		PrintWriter out = new PrintWriter(System.out);
        int a = in.nextInt(), b = in.nextInt(), n = in.nextInt();
        out.println((n >= (a + b) && (n - (a + b)) % 2 == 0) ? "YES" : "NO");
		out.close();
	}
}
class InputReader{BufferedReader buf;StringTokenizer tok;InputReader(){buf = new BufferedReader(new InputStreamReader(System.in));}boolean hasNext(){while (tok == null || !tok.hasMoreElements()){try{tok = new StringTokenizer(buf.readLine());}catch (Exception e){return false;}}return true;}String next(){if (hasNext())
			return tok.nextToken();return null;}int nextInt(){return Integer.parseInt(next());}long nextLong(){return Long.parseLong(next());}double nextDouble(){return Double.parseDouble(next());}BigInteger nextBigInteger(){return new BigInteger(next());}BigDecimal nextBigDecimal(){return new BigDecimal(next());}}