Text Reverse
Time limit 1000 ms
Memory limit 262144 kB
Source Codeforces Round #292 (Div. 2)
Tags math *1100
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Problem Description
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: .
问题链接：
CodeForces - 515A
问题简述：
Drazil是个糊涂鬼，说自己进行了多少步的位移到了Varda的家，Varda并不相信，希望我们证明Drazil说的话是否正确。
问题分析：
判断奇偶性，正负号的转换。
程序说明：
先给V家的位移加个绝对值。如果a+b>s,则D不可能在s步数内到达V家；如果位移差为0（即s-a-b=0）或者位移差为正偶数，则能在s步内到达V家。
#include <iostream>
using namespace std;
int main()
{
int a,b,s;
cin >> a>>b>>s;
if (a < 0)
a = -a;
if (b < 0)
b = -b;
if (a+b>s)
cout << "NO" << endl;
else if((s - a - b) == 0 | (s - a - b) % 2 == 0)
cout << "YES" << endl;
else cout << "NO" << endl;
return 0;
}