类欧几里得

对于给定的元$a,b,c,n$

设$f(i)=\lfloor\frac{ai+b} {c}\rfloor$

求

Part1

$a\ge c$ 或 $b\ge c$

$a<c$ 且 $b<c$

其中

每次操作交换了$a,c$然后再次取模，复杂度是$O(\log n)$

递归边界是$a=0$

Part2

$G(a,b,c,n)=\sum_0^nf(i)^2$

$a\ge c$ 或 $b\ge c$

设$\lfloor \frac{a} {c}\rfloor=x,\lfloor \frac{b} {c}\rfloor=y$

$a<c$ 且 $b<c$

Part3

$H(a,b,c,n)=\sum_0^ni\cdot f(i)$

$a\ge c$ 或 $b\ge c$

$a<c$ 且 $b<c$

设$f’(i)=\lfloor\frac{ci-b+c-1} {a}\rfloor$

$H(a,b,c,n)=\frac{f(n)n(n+1)-G(c,-b+c-1,a,f(n)-1)-F(c,-b+c-1,a,f(n)-1)} {2}$

Tip:

这个多个函数的情况，如果直接递归写复杂度极高

所以需要把访问到的所有状态存下来递推，就能保证复杂度$O(\log n)$

模板题传送门

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#include<bits/stdc++.h>
using namespace std;

#define Mod1(x) ((x>=P)&&(x-=P))
#define Mod2(x) ((x<0)&&(x+=P))

#define reg register
typedef long long ll;
#define rep(i,a,b) for(reg int i=a,i##end=b;i<=i##end;++i)
#define drep(i,a,b) for(reg int i=a,i##end=b;i>=i##end;--i)

#define pb push_back
template <class T> inline void cmin(T &a,T b){ ((a>b)&&(a=b)); }
template <class T> inline void cmax(T &a,T b){ ((a<b)&&(a=b)); }

char IO;
int rd(){
	int s=0;
	int f=0;
	while(!isdigit(IO=getchar())) if(IO=='-') f=1;
	do s=(s<<1)+(s<<3)+(IO^'0');
	while(isdigit(IO=getchar()));
	return f?-s:s;
}

const int N=1010,P=998244353;

ll qpow(ll x,ll k=P-2) {
	ll res=1;
	for(;k;k>>=1,x=x*x%P) if(k&1) res=res*x%P;
	return res;
}

const ll Inv2=(P+1)/2,Inv6=qpow(6);

//ll F(ll a,ll b,ll c,ll n);
//ll G(ll a,ll b,ll c,ll n);
//ll H(ll a,ll b,ll c,ll n);

ll D2(ll n){
	n%=P;
	return n*(n+1)/2%P;
}
ll D3(ll n){
	n%=P;
	return n*(n+1)%P*(2*n+1)%P*Inv6%P;
}

ll sa[N],sb[N],sc[N],sn[N];
ll F[N],G[N],H[N];
int cnt;

void PreCalc(ll a,ll b,ll c,ll n){
	sa[++cnt]=a; sb[cnt]=b; 
	sc[cnt]=c; sn[cnt]=n;
	if(a==0) return;
	if(a>=c || b>=c) PreCalc(a%c,b%c,c,n);
	else {
		ll t=(a*n+b)/c;
		PreCalc(c,-b+c-1,a,t-1);
	}
}


/*ll F(ll a,ll b,ll c,ll n){
	if(a==0) return (b/c)%P*((n+1)%P)%P;
	if(a>=c || b>=c) {
		ll ans=(F(a%c,b%c,c,n)+D2(n)*(a/c)%P+(n+1)%P*(b/c))%P;
		ans=(ans%P+P)%P;
		return ans;
	}
	ll t=(a*n+b)/c;
	ll ans=t%P*n%P-F(c,-b+c-1,a,t-1);
	ans=(ans%P+P)%P;
	return ans;
}

ll G(ll a,ll b,ll c,ll n){
	if(a==0) return (b/c)*(b/c)%P*(n+1)%P;
	if(a>=c || b>=c) {
		ll x=a/c%P,y=b/c%P;
		ll ans=(G(a%c,b%c,c,n)+2*x*H(a%c,b%c,c,n)+2*y*F(a%c,b%c,c,n))%P;
		ans=(ans+D3(n)*x%P*x%P+(n+1)%P*y%P*y%P+2*D2(n)*x%P*y%P)%P;
		ans=(ans%P+P)%P;
		return ans;
	}
	ll t=(a*n+b)/c;
	ll ans=(t%P)*(t%P)%P*(n%P)%P-2*H(c,-b+c-1,a,t-1)-F(c,-b+c-1,a,t-1)%P;
	ans=(ans%P+P)%P;
	return ans;
}

ll H(ll a,ll b,ll c,ll n){
	if(a==0) return D2(n)%P*(b/c)%P;
	if(a>=c || b>=c) {
		ll ans=(H(a%c,b%c,c,n)+(a/c)*D3(n)+D2(n)*(b/c))%P;
		ans=(ans%P+P)%P;
		return ans;
	}
	ll t=(a*n+b)/c;
	ll ans=(t%P*D2(n)%P*2%P-G(c,-b+c-1,a,t-1)-F(c,-b+c-1,a,t-1))%P;
	ans=(ans*Inv2%P+P)%P;
	return ans;
}*/

int main(){
	rep(kase,1,rd()) {
		int n=rd(),a=rd(),b=rd(),c=rd();
		
		cnt=0;
		PreCalc(a,b,c,n);
		F[cnt+1]=G[cnt+1]=H[cnt+1]=0;

		drep(i,cnt,1) {
			ll a=sa[i],b=sb[i],c=sc[i],n=sn[i];
			if(a==0) F[i]=(b/c)%P*((n+1)%P)%P;
			else if(a>=c || b>=c) {
				ll ans=(F[i+1]+D2(n)*(a/c)%P+(n+1)%P*(b/c))%P;
				ans=(ans%P+P)%P;
				F[i]=ans;
			} else {
				ll t=(a*n+b)/c;
				ll ans=t%P*n%P-F[i+1];
				ans=(ans%P+P)%P;
				F[i]=ans;
			}

			if(a==0) G[i]=(b/c)*(b/c)%P*(n+1)%P;
			else if(a>=c || b>=c) {
				ll x=a/c%P,y=b/c%P;
				ll ans=(G[i+1]+2*x*H[i+1]+2*y*F[i+1])%P;
				ans=(ans+D3(n)*x%P*x%P+(n+1)%P*y%P*y%P+2*D2(n)*x%P*y%P)%P;
				ans=(ans%P+P)%P;
				G[i]=ans;
			} else {
				ll t=(a*n+b)/c;
				ll ans=(t%P)*(t%P)%P*(n%P)%P-2*H[i+1]-F[i+1];
				ans=(ans%P+P)%P;
				G[i]=ans;
			}

			if(a==0) H[i]=D2(n)%P*(b/c)%P;
			if(a>=c || b>=c) {
				ll ans=(H[i+1]+(a/c)*D3(n)+D2(n)*(b/c))%P;
				ans=(ans%P+P)%P;
				H[i]=ans;
			} else {
				ll t=(a*n+b)/c;
				ll ans=(t%P*D2(n)%P*2%P-G[i+1]-F[i+1])%P;
				ans=(ans*Inv2%P+P)%P;
				H[i]=ans;
			}
		}
		printf("%lld %lld %lld\n",F[1],G[1],H[1]);
	}
}