Drazil and Date

最新推荐文章于 2021-08-31 16:58:41 发布

A_Happy_ACMer

最新推荐文章于 2021-08-31 16:58:41 发布

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Drazil在没有方向感的情况下，从坐标(0,0)出发，尝试到达Varda所在的(a,b)位置，通过随机行走的方式，探究是否能在指定步数s内成功会面。

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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".

Sample test(s)

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: $\text{[math]}$ .

已知：从0,0点到n,m点走过了num步，期间可以任意方向地走，求有没有可能走到。

因为可以任意方向的走，所以会有很多种情况，但是可以注意到，两点之间有一个最短的路线，那就是n+m的和，剩下的步数，无论有多少，都要能够整除2，因为有走过去的道路，就要有走回来的道路，而且给定的步数一定要大于等于最小的步数才行。

注意：n,m的值有可能是负数，所以在计算最短步数的时候一定要换算成正整数。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std;

int main()
{
    __int64 n,m,i,j,k,num;
    while(~scanf("%I64d%I64d%I64d",&n,&m,&num))
    {
        if(n < 0)
            n = -n;
        if(m < 0)
            m = -m;
        k = n + m;
        if(num < k)
        {
            printf("No\n");
        }
        else
        {
            i = num - k;
            if(i%2==0)
                printf("Yes\n");
            else
                printf("No\n");
        }
    }
    return 0;
}