HDU 5054 Alice and Bob——BestCoder Round #11(div.2)

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Alice and Bob

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Bob and Alice got separated in the Square, they agreed that if they get separated, they'll meet back at the coordinate point (x, y). Unfortunately they forgot to define the origin of coordinates and the coordinate axis direction. Now, Bob in the lower left corner of the Square, Alice in the upper right corner of the the Square. Bob regards the lower left corner as the origin of coordinates, rightward for positive direction of axis X, upward for positive direction of axis Y. Alice regards the upper right corner as the origin of coordinates, leftward for positive direction of axis X, downward for positive direction of axis Y. Assuming that Square is a rectangular, length and width size is N * M. As shown in the figure:

Bob and Alice with their own definition of the coordinate system respectively, went to the coordinate point (x, y). Can they meet with each other ?
Note: Bob and Alice before reaching its destination, can not see each other because of some factors (such as buildings, time poor).

Input

There are multiple test cases. Please process till EOF. Each test case only contains four integers : N, M and x, y. The Square size is N * M, and meet in coordinate point (x, y). ( 0 < x < N <= 1000 , 0 < y < M <= 1000 ).

Output

If they can meet with each other, please output "YES". Otherwise, please output "NO".

Sample Input

10 10 5 510 10 6 6

Sample Output

YESNO

Source

BestCoder Round #11 (Div. 2)

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出题人的解题思路：

两个人的坐标系不同，如果都走到（x，y）能够碰面的话，只有一种可能：在广场矩形的中心位置。即： 2*x == N 并且 2*y == M。

题意：如下图所示，Bob在广场的左下角，对于他来说，往右是X轴正方向，往上是Y轴正方向；而Alice在广场的右上角，对于她来说，往左是X轴正方向，往下是Y轴正方向。他们要在各自的坐标系中到达某点(x,y)，问是否存在这样的点，两人能够相遇

解题思路：我们以Bob所在点为坐标原点建立坐标系，那么

化简可得

因此我们只需判断一下给定的n、m、x、y是否满足上式即可

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<stack>#include<math.h>#include<vector>#include<map>#include<set>#include<stdlib.h>#include<cmath>#include<string>#include<algorithm>#include<iostream>#define exp 1e-10using namespace std;const int N = 105;const int inf = 2147483647;const int mod = 2009;int main(){    int n,m,x,y;    while(~scanf("%d%d%d%d",&n,&m,&x,&y))    {        if(n==2*x&&m==2*y)            puts("YES");        else            puts("NO");    }    return 0;}

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