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三模數NTT

假設三個模數算出的答案分別爲$a_1,a_2,a_3$

模數爲$m_1,m_2,m_3$

假設真實的乘積爲$ans$

根據中國剩餘定理，求出$ans \equiv a_1\times inv(m_2,m_1) + a_2 \times inv(m_1,m_2) (\mod m_1m_2)$

其中$inv(x,p)$表示$x$在模$p$意義下的逆元

令$t = inv(m_2,m_1) + a_2 \times inv(m_1,m_2)$，則$ans = km_1m_2 + t$

而我知道$ans<m_1m_2m_3$，所以$km_1m_2 + t<m_1m_2m_3$，我又知道$t,k$都是非負數，所以$k<m_3$

所以兩邊同時對$m_3$取模

$a_3 \equiv km_1m_2 + t (\mod m_3)$

得到$k = (a_3-t) \times inv(m_1m_2,m_3)$

題解

讀完題直接寫出指數型生成函數的乘積

$\prod (1+\frac{x}{1!}+\frac{x^2}{2!}+\dots+\frac{x^{a_i}}{a_i!} )$

然後這樣還多算了前導$0$的，所以讓$a[0]--$，再做一次多項式乘法

代碼

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
ll MOD=1e9+7;
ll mod;
struct EasyMath
{
    ll prime[maxn], phi[maxn], mu[maxn];
    bool mark[maxn];
    ll fastpow(ll a, ll b, ll c)
    {
        ll t(a%c), ans(1ll);
        for(;b;b>>=1,t=t*t%c)if(b&1)ans=ans*t%c;
        return ans;
    }
    void exgcd(ll a, ll b, ll &x, ll &y)
    {
        if(!b){x=1,y=0;return;}
        ll xx, yy;
        exgcd(b,a%b,xx,yy);
        x=yy, y=xx-a/b*yy;
    }
    ll inv(ll x, ll p)  //p是素數
    {return fastpow(x%p,p-2,p);}
    ll inv2(ll a, ll p)
    {
        ll x, y;
        exgcd(a,p,x,y);
        return (x+p)%p;
    }
    void shai(ll N)
    {
        ll i, j;
        for(i=2;i<=N;i++)mark[i]=false;
        *prime=0;
        phi[1]=mu[1]=1;
        for(i=2;i<=N;i++)
        {
            if(!mark[i])prime[++*prime]=i, mu[i]=-1, phi[i]=i-1;
            for(j=1;j<=*prime and i*prime[j]<=N;j++)
            {
                mark[i*prime[j]]=true;
                if(i%prime[j]==0)
                {
                    phi[i*prime[j]]=phi[i]*prime[j];
                    break;
                }
                mu[i*prime[j]]=-mu[i];
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
            }
        }
    }
    ll mult(ll a, ll b, ll c)
	{
		ll t=a, ans=0;
		for(;b;b>>=1,t=(t+t)%c)if(b&1)(ans+=t)%=c;
		return ans;
	}
    ll CRT(vector<ll> a, vector<ll> m) //要求模數兩兩互質
    {
        ll M=1, ans=0, n=a.size(), i;
        for(i=0;i<n;i++)M*=m[i];
        for(i=0;i<n;i++)
		{
			ll x = mult(a[i],M/m[i],M);
			ll y = mult(x,inv2(M/m[i],m[i]),M);
			(ans+=y)%=M;
		}
        return ans;
    }
}em;
struct NTT
{
    ll n;
    vector<ll> R;
    void init(ll bound)    //bound是積多項式的最高次冪
    {
        ll L(0);
        for(n=1;n<=bound;n<<=1,L++);
        R.resize(n);
        for(ll i=0;i<n;i++)R[i]=(R[i>>1]>>1)|((i&1)<<(L-1));
    }
    void ntt(vector<ll>& a, int opt)
    {
        ll i, j, k, wn, w, x, y, inv(em.fastpow(n,mod-2,mod));
        for(i=0;i<n;i++)if(i>R[i])swap(a[i],a[R[i]]);
        for(i=1;i<n;i<<=1)
        {
            if(opt==1)wn=em.fastpow(3,(mod-1)/(i<<1),mod);
            else wn=em.fastpow(3,(mod-1-(mod-1)/(i<<1)),mod);
            for(j=0;j<n;j+=i<<1)
                for(w=1,k=0;k<i;k++,w=w*wn%mod)
                {
                    x=a[k+j], y=a[k+j+i]*w%mod;
                    a[k+j]=(x+y)%mod, a[k+j+i]=(x-y)%mod;
                }
        }
        if(opt==-1)for(i=0;i<n;i++)(a[i]*=inv)%=mod;
    }
};
struct Poly
{
    vector<ll> v;
    ll n;	//n是最高次項的次數
    Poly(ll N){v.resize(N+1);n=N;}
    Poly(const Poly& p){v=p.v; n=p.n;}
    void resize(ll N){n=N; v.resize(N+1);}
    ll& operator[](ll id){return v[id];}
    void show()
    {
        printf("n=%lld\n",n);
        ll i; rep(i,0,n-1)printf("%lldx^%lld + ",(v[i]+mod)%mod,i);
        printf("%lldx^%lld\n",(v[n]+mod)%mod,n);
    }
};
Poly operator+(Poly A, Poly B)
{
    ll i;
    Poly C(max(A.n,B.n));
	A.resize(C.n), B.resize(C.n);
    rep(i,0,C.n)C[i]=(A[i]+B[i])%MOD;
    return C;
}
Poly operator*(Poly A, Poly B)
{
    NTT ntt;
    ll i, n=A.n+B.n, m1=998244353, m2=1004535809, m3=469762049;
    Poly a(0), b(0);
    
    ntt.init(n);
    A.resize(ntt.n-1), B.resize(ntt.n-1);
	
	mod = m1;
    a=A, b=B;
    ntt.ntt(a.v,1), ntt.ntt(b.v,1);
    Poly C(ntt.n-1);
    rep(i,0,C.n)C[i]=(a[i]*b[i])%mod;
    ntt.ntt(C.v,-1);
    C.resize(n);
	
	mod = m2;
	a=A, b=B;
    ntt.ntt(a.v,1), ntt.ntt(b.v,1);
    Poly D(ntt.n-1);
    rep(i,0,D.n)D[i]=(a[i]*b[i])%mod;
    ntt.ntt(D.v,-1);
    D.resize(n);

    mod = m3;
	a=A, b=B;
    ntt.ntt(a.v,1), ntt.ntt(b.v,1);
    Poly E(ntt.n-1);
    rep(i,0,E.n)E[i]=(a[i]*b[i])%mod;
    ntt.ntt(E.v,-1);
    E.resize(n);
	
	Poly F(n);
	rep(i,0,n)
    {
        ll m12 = m1*m2;
        ll a1=(C[i]+m1)%m1, a2=(D[i]+m2)%m2, a3=(E[i]+m3)%m3;
        ll t = ( em.mult(a1*m2,em.inv(m2,m1),m12) + em.mult(a2*m1,em.inv(m1,m2),m12) ) % m12;
        ll k = (a3-t%m3+m3)*em.inv(m12,m3) %m3;
        F[i] = ( k%MOD * (m12%MOD) %MOD + t ) %MOD;
    }
	
    return F;
}
int main()
{
    ll T=read(), kase;
    rep(kase,1,T)
    {
        ll n=read(), i, j, ans=0;
        vector<ll> a(5), fact(n+1), _fact(n+1);
        rep(i,0,4)a[i]=min(n,read());
        fact[0]=_fact[0]=1;
        rep(i,1,n)fact[i]=fact[i-1]*i%MOD, _fact[i]=em.inv(fact[i],MOD);
        
        Poly* p[5];
        rep(i,0,4)
        {
            p[i] = new Poly(a[i]);
            rep(j,0,p[i]->n)p[i]->v[j]=_fact[j];
        }
        Poly P(0);
        P[0]=1;
        rep(i,1,4)
        {
            P = P*(*(p[i]));
            if(P.n>n)P.resize(n);
        }
        P.resize(n);
        auto P1 = P*(*(p[0]));
        if(a[0])p[0]->resize(a[0]-1);
        auto P2 = P*(*(p[0]));

        (ans = P1[n]*fact[n] - (a[0]!=0)*P2[n-1]*fact[n-1] )%=MOD;
        (ans+=MOD)%=MOD;
        
        printf("Case #%lld: %lld\n",kase,ans);
    }
    return 0;
}