P5170 [模板] 类欧几里得算法

传送门

关于 f 

     ( a >= c | b >= c)

 (a < c & b < c)

关于 g

首先有 

暴力拆开发现

但 a < c 的时候有点头凸

考虑   

类似套路的可以推到

关于 h 

跟 f 一样，不想打了

---

#include
using namespace std;
typedef long long ll;
const int Mod = 998244353, inv2 = (Mod+1)/2, inv6 = (Mod+1)/6;
ll mul(ll a, ll b){ return a * b % Mod;}
ll add(ll a, ll b){ return (a + b) >= Mod ? a + b - Mod : a + b; }
ll n, a, b, c; int T; 
struct Node{ ll f, g, h;};
Node Solve(ll a, ll b, ll c, ll n){
	if(c == 0) return (Node){0, 0, 0};
	if(a >= c || b >= c){
		Node tmp = Solve(a % c, b % c, c, n);
		ll f = add(add(mul(mul(a/c, n), mul(n+1, inv2)), mul(b/c, n+1)), tmp.f);
		ll g = mul(mul(a/c, a/c), mul(mul(inv6, n), mul(n+1, n*2+1)));
		g = add(g, mul(mul(b/c, b/c), n+1));
		g = add(g, mul(mul(a/c, b/c), mul(n, n+1)));
		g = add(g, mul(mul(2, a/c), tmp.h));
		g = add(g, mul(mul(2, b/c), tmp.f));
		g = add(g, tmp.g);
		ll h = mul(mul(a/c, n), mul(mul(n+1, n*2+1), inv6));
		h = add(h, mul(mul(b/c, n), (mul(n+1, inv2))));
		h = add(h, tmp.h);
		return (Node){f, g, h};
	}
	ll fn = (a * n + b) / c;
	Node tmp = Solve(c, c - b - 1, a, fn - 1);
	ll f = add(mul(n, fn), Mod - tmp.f);
	ll g = mul(mul(n, fn), fn+1);
	g = add(g, Mod - mul(2, add(tmp.h, tmp.f)));
	g = add(g, Mod - f);
	ll h = mul(n, mul(n+1, fn));
	h = add(h, Mod - add(tmp.f, tmp.g)); h = mul(h, inv2);
	return (Node){f, g, h};
}
int main(){
	scanf("%d", &T);
	while(T--){
		scanf("%lld%lld%lld%lld", &n, &a, &b, &c);
		Node ans = Solve(a, b, c, n);
		cout << ans.f << " " << ans.g <<" " << ans.h << "\n";
	} return 0;
}