就是等价于一个4个元2个方程判断解的存在性。。。
然后乱搞吧。。。（貌似叫裴蜀定理？）
1 /************************************************************** 2 Problem: 2299 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:148 ms 7 Memory:804 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <algorithm> 12 13 using namespace std; 14 typedef long long ll; 15 ll T, a, b, x, y, d; 16 17 inline ll read(){ 18 ll x = 0, sgn = 1; 19 char ch = getchar(); 20 while (ch < '0' || ch > '9'){ 21 if (ch == '-') sgn = -1; 22 ch = getchar(); 23 } 24 while (ch >= '0' && ch <= '9'){ 25 x = x * 10 + ch - '0'; 26 ch = getchar(); 27 } 28 return sgn * x; 29 } 30 31 ll gcd(ll a, ll b){ 32 return a ? gcd(b % a, a) : b; 33 } 34 35 bool check(ll a, ll b){ 36 return a % d == 0 && b % d == 0; 37 } 38 39 int main(){ 40 T = read(); 41 while (T--){ 42 a = read(), b = read(), x = read(), y = read(); 43 d = gcd(a, b) * 2; 44 if (check(x, y) || check(x - a, y - b) || check(x - b, y - a) || check(x - a - b, y - a - b)) 45 printf("Y "); 46 else printf("N "); 47 } 48 return 0; 49 }