ACM(gzhu周赛题）H - Problem H
原题目：
H - Problem H
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 ≤ 2·10*9) 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);
题目简述：
两个好朋友要见面，a的家在笛卡尔坐标系的（0，0），b的点在（a，b）。a要去找b玩，但是a是个路痴，他会乱走，但是最终能走到b的家。值得注意的是a可能会经过b的家而不知道，继续去走。但最终一定能找到b。找到b后a会告诉b自己走了多少步，b要求我们来写一个程序判断a有没有说错。
题目分析：
一开始看到这个题目，我觉得很麻烦，因为需要考虑的情况似乎很多。比如：走的步数是坐标和的倍数，刚好走的步数是坐标和。b的坐标就定的是（0，0），那就要走出去偶数步或者不动。但是多思考一会后发现，只要走的步数减去坐标和是偶数就可以了。当然前提的走的步数一定要大于或者等于坐标和。
还要注意的是坐标可能是负数，我们可以直接用一个取绝对值的函数来将坐标直接全部变正来进行计算就好了，毕竟步数不可能为负。
AC代码如下：
#include
#include
using namespace std;
int main()
{
long a, b, s;
while (cin >> a >> b >> s)
{
a = abs(a);
b = abs(b);
if (a + b <= s&& (s - (a + b)) % 2 == 0)
cout << "YES" << endl;
else
cout << "NO" << endl;
}
return 0;
}