hdu 5054

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;
}