Alice and Bob

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 747    Accepted Submission(s): 549

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 5
10 10 6 6

Sample Output

YES
NO

题意:Bob(0,0) 和 ALice(n,m) 要在(x,y)点汇合。问是否可能？

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <map>
using namespace std;
int main()
{
    int a,b,c,d;
    while(scanf("%d%d%d%d",&a,&b,&c,&d)!=EOF){
        if(a-c==c&&b-d==d) printf("YES\n"); ///Bob要从 0,0 走到 (x,y) Alice 要从 (n,m)走到(n-x,m-y)
                                            /// x = n - x,y = m-y;
        else printf("NO\n");
    }
    return 0;
}