Drazil and Date

最新推荐文章于 2021-10-20 13:17:06 发布

weixin_44006014

最新推荐文章于 2021-10-20 13:17:06 发布

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本文链接：https://blog.youkuaiyun.com/weixin_44006014/article/details/85171638

探讨Drazil从(0,0)出发到(a,b)的最短路径及随机行走到达目标的可能性，分析最少步数与实际步数的关系。

Drazil and Date

Someday, Drazil wanted to go on date with Varda. Drazil and Varda live on Cartesian plane. Drazil's home is located in point (0, 0) and Varda's home is located in point (a, b). In each step, he can move in a unit distance in horizontal or vertical direction. In other words, from position (x, y) he can go to positions (x + 1, y), (x - 1, y), (x, y + 1) or (x, y - 1).

Unfortunately, Drazil doesn't have sense of direction. So he randomly chooses the direction he will go to in each step. He may accidentally return back to his house during his travel. Drazil may even not notice that he has arrived to (a, b) and continue travelling.

Luckily, Drazil arrived to the position (a, b) successfully. Drazil said to Varda: "It took me exactly s steps to travel from my house to yours". But Varda is confused about his words, she is not sure that it is possible to get from (0, 0) to (a, b) in exactly s steps. Can you find out if it is possible for Varda?

Input

You are given three integers a, b, and s ( - 109 ≤ a, b ≤ 109, 1 ≤ s ≤ 2·109) in a single line.

Output

If you think Drazil made a mistake and it is impossible to take exactly s steps and get from his home to Varda's home, print "No" (without quotes).

Otherwise, print "Yes".

Examples

Input

5 5 11

Output

No

Input

10 15 25

Output

Yes

Input

0 5 1

Output

No

Input

0 0 2

Output

Yes

Note

In fourth sample case one possible route is: （0，0）——（0，1）——（0，0）。

问题链接：https://vjudge.net/problem/CodeForces-515A

问题分析：

1.最小步数为a+b,所走的步数必须要大于或等于最小步数，否则直接判断为“NO”；

2.当所走步数大于最小步数时，要确保多出来的步数是2的倍数，即走出去了还要走回来，来回步数是2的倍数；

3.注意要把a,b转化成整数，因为题目中说Drazil没有方向感，即可能走到x轴或y轴的负方向

4.注意数据大小

#include<iostream>
using namespace std;
int main()
{
	long long a, b, s;
	cin >> a >> b >> s;
	if (a < 0)a = -a;
	if (b < 0)b = -b;
	if (a + b > s)cout << "NO";
	else if (( s- a - b) % 2 == 0)cout << "YES";
	else cout << "NO";
}