1 second
256 megabytes
standard input
standard output
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?
You are given three integers a, b, and s ( - 109 ≤ a, b ≤ 109, 1 ≤ s ≤ 2·109) in a single line.
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".
5 5 11
No
10 15 25
Yes
0 5 1
No
0 0 2
Yes 居然WA了两发。。。。第一次忘了判断是偶数时能否到达，第二次忘了终点坐标可以是负的，我。。。。
#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <string>
#include <algorithm>
#include <cctype>
#include <queue>
#include <map>
using namespace std; int main(void)
{
int a,b,t; cin >> a >> b >> t;
if((t - (abs(a) + abs(b))) % )
cout << "No" << endl;
else if(t >= (abs(a) + abs(b)))
cout << "Yes" << endl;
else
cout << "No" << endl; return ;
}