• 注释：本章同余针对$2$2。
• 题面
• 题意：见题面。
• 解决思路：考虑前四个向量$\small (a,b), (a,-b), (-a,b), (-a,-b)$(a,b),(a,−b),(−a,b),(−a,−b)，设共取出$\small n$n个向量，$\small a,-a,b,-b$a,−a,b,−b的个数分别为$x_{a1},x_{a2},y_{b1},y_{b2}$xa1​,xa2​,yb1​,yb2​。
  $\small \begin{cases} \ x_{a1}+x_{a2}=n\\ \ y_{b1}+y_{b2}=n\\ \end{cases}${ xa1​+xa2​=n yb1​+yb2​=n​
  $\small n$n个向量之和为$\small (ax_{a1}-ax_{a2},by_{b1}-by_{b2})$(axa1​−axa2​,byb1​−byb2​)
  设$\small x_1=x_{a1}-x_{a2}$x1​=xa1​−xa2​，$\small y_1=y_{b1}-y_{b2}$y1​=yb1​−yb2​，证明$\small x_1\equiv y_1$x1​≡y1​
  $\small n\equiv p$n≡p，$\small p$p的取值为$\small 0$0或$\small 1$1
  $\small \therefore \begin{cases} \ x_{a1}+x_{a2}\equiv p\\ \ y_{b1}+y_{b2}\equiv p\\ \end{cases}$∴{ xa1​+xa2​≡p yb1​+yb2​≡p​
  $\small \therefore \begin{cases} \ x_{a1}+x_{a2}-2x_{a2}\equiv p\\ \ y_{b1}+y_{b2}-2y_{b2}\equiv p\\ \end{cases}$∴{ xa1​+xa2​−2xa2​≡p yb1​+yb2​−2yb2​≡p​
  $\small \therefore \begin{cases} \ x_{a1}-x_{a2}\equiv p\\ \ y_{b1}-y_{b2}\equiv p\\ \end{cases}$∴{ xa1​−xa2​≡p yb1​−yb2​≡p​
  $\small \therefore x_1\equiv y_1\equiv p$∴x1​≡y1​≡p
  证毕$\small ！！！$！！！
  结论：$\small n$n个向量之和为$\small (ax_1,by_1)$(ax1​,by1​)，$\small x_1\equiv y_1$x1​≡y1​
  同理考虑后四个向量$\small (b,a), (b,-a), (-b,a), (-b,-a)$(b,a),(b,−a),(−b,a),(−b,−a)，设共取出$\small m$m个向量，设$\small m$m个向量之和为$\small (by_2,ax_2)$(by2​,ax2​)，且$\small y_2\equiv x_2$y2​≡x2​
  八个向量之和为$\small (ax_1+by_2,by_1+ax_2)$(ax1​+by2​,by1​+ax2​)，$\small x_1\equiv y_1$x1​≡y1​，$\small y_2\equiv x_2$y2​≡x2​
  $\small \therefore \begin{cases} \ ax_1+by_2=x~~~~~~\small x_1\equiv y_1\\ \ by_1+ax_2=y~~~~~~\small y_2\equiv x_2\\ \end{cases}$∴{ ax1​+by2​=x      x1​≡y1​ by1​+ax2​=y      y2​≡x2​​
  $\small \therefore \begin{cases} \ ax_1+by_2=x~~~~~~\small x_1\equiv y_1\\ \ ax_2+by_1=y~~~~~~\small y_2\equiv x_2\\ \end{cases}$∴{ ax1​+by2​=x      x1​≡y1​ ax2​+by1​=y      y2​≡x2​​
  情况一：$\small x_1\equiv y_1\equiv 0$x1​≡y1​≡0，$\small y_2\equiv x_2\equiv 0$y2​≡x2​≡0
  $\small \therefore \begin{cases} \ ax_1+by_2=x\\ \ ax_2+by_1=y\\ \end{cases}$∴{ ax1​+by2​=x ax2​+by1​=y​
  将同余$\small 0$0的项提出$\small 2$2
  $\small \therefore \begin{cases} \ 2ax_1'+2by_2'=x\\ \ 2ax_2'+2by_1'=y\\ \end{cases}$∴{ 2ax1′​+2by2′​=x 2ax2′​+2by1′​=y​
  有解的条件是$\small (2a,2b)\mid x$(2a,2b)∣x，$\small (2a,2b)\mid y$(2a,2b)∣y
  情况二：$\small x_1\equiv y_1\equiv 0$x1​≡y1​≡0，$\small y_2\equiv x_2\equiv 1$y2​≡x2​≡1
  $\small \therefore \begin{cases} \ ax_1+by_2=x\\ \ ax_2+by_1=y\\ \end{cases}$∴{ ax1​+by2​=x ax2​+by1​=y​
  将同余$\small 0$0的项提出$\small 2$2
  $\small \therefore \begin{cases} \ 2ax_1'+by_2=x\\ \ ax_2+2by_1'=y\\ \end{cases}$∴{ 2ax1′​+by2​=x ax2​+2by1′​=y​
  $\small \therefore \begin{cases} \ 2ax_1'+b(y_2+1)=x+b\\ \ a(x_2+1)+2by_1'=y+a\\ \end{cases}$∴{ 2ax1′​+b(y2​+1)=x+b a(x2​+1)+2by1′​=y+a​
  $\small \therefore \begin{cases} \ 2ax_1'+2by_2'=x+b\\ \ 2ax_2'+2by_1'=y+a\\ \end{cases}$∴{ 2ax1′​+2by2′​=x+b 2ax2′​+2by1′​=y+a​
  有解的条件是$\small (2a,2b)\mid (x+b)$(2a,2b)∣(x+b)，$\small (2a,2b)\mid (y+a)$(2a,2b)∣(y+a)
  同理
  情况三：有解的条件是$\small (2a,2b)\mid (x+a)$(2a,2b)∣(x+a)，$\small (2a,2b)\mid (y+b)$(2a,2b)∣(y+b)
  情况四：有解的条件是$\small (2a,2b)\mid (x+a+b)$(2a,2b)∣(x+a+b)，$\small (2a,2b)\mid (y+a+b)$(2a,2b)∣(y+a+b)
• AC代码

//优化
#pragma GCC optimize(2)
//C
#include<string.h>
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
//C++
//#include<unordered_map>
#include<algorithm>
#include<iostream>
#include<istream>
#include<iomanip>
#include<climits>
#include<cstdio>
#include<string>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
//宏定义
#define N 1010
#define DoIdo main
//#define scanf scanf_s
#define it set<ll>::iterator
//定义+命名空间
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 998244353;
const ll INF = 1e18;
const int maxn = 1e6 + 10;
using namespace std;
//全局变量
//函数区
ll max(ll a, ll b) { return a > b ? a : b; }
ll min(ll a, ll b) { return a < b ? a : b; }
ll gcd(ll a, ll b) { return !b ? a : gcd(b, a % b); }
//主函数
int DoIdo() {
 
    ios::sync_with_stdio(false);
    cin.tie(NULL), cout.tie(NULL);
 
    int T;
    cin >> T;
 
    while (T--) {
        ll a, b, x, y;
        cin >> a >> b >> x >> y;
 
        ll g = gcd(2 * a, 2 * b);
        bool jg = 0;
        if (x % g == 0 && y % g == 0) jg = 1;
        if ((x + a) % g == 0 && (y + b) % g == 0) jg = 1;
        if ((x + b) % g == 0 && (y + a) % g == 0) jg = 1;
        if ((x + a + b) % g == 0 && (y + a + b) % g == 0) jg = 1;
        if (jg) cout << "Y" << endl;
        else cout << "N" << endl;
    }
    return 0;
}
//分割线---------------------------------QWQ
/*
 
 
 
*/