1 条题解

  • 0
    @ 2025-8-24 21:41:40

    自动搬运

    查看原文

    来自洛谷,原作者为

    avatar λᴉʍ
    大家好,我是刚学OI的萌新

    搬运于2025-08-24 21:41:40,当前版本为作者最后更新于2019-06-12 11:55:14,作者可能在搬运后再次修改,您可在原文处查看最新版

    自动搬运只会搬运当前题目点赞数最高的题解,您可前往洛谷题解查看更多

    以下是正文


    我喜欢唱♂跳♂rap♂篮球

    要求的是:i=0kCmiCnmkiiL\sum_{i=0}^kC_m^iC_{n-m}^{k-i}i^L

    这个iLi^L很烦,就把第二类斯特林数的式子套进去

    i=0kCmiCnmkiiL\sum_{i=0}^kC_m^iC_{n-m}^{k-i}i^L

    $\sum_{i=0}^kC_m^iC_{n-m}^{k-i}\sum_{j=0}^iC_{i}^j\begin{Bmatrix}L\\j\end{Bmatrix}j!$

    $\sum_{j=0}^k\begin{Bmatrix}L\\j\end{Bmatrix}j!\sum_{i=j}^kC_m^iC_{n-m}^{k-i}C_{i}^j$

    后面\sum三个组合数好像很不好搞,但是CmiCij=CmjCmjijC_{m}^{i}C_{i}^{j}=C_{m}^{j}C_{m-j}^{i-j},可以拆出一个与ii无关的组合数

    $\sum_{j=0}^k\begin{Bmatrix}L\\j\end{Bmatrix}j!C_{m}^{j}\sum_{i=j}^kC_{m-j}^{i-j}C_{n-m}^{k-i}$

    把式子化的好看一点,发现可以套范德蒙德卷积(i=0kCniCmki=Cn+mk\sum_{i=0}^kC_{n}^{i}C_{m}^{k-i}=C_{n+m}^k

    $\sum_{j=0}^k\begin{Bmatrix}L\\j\end{Bmatrix}j!C_{m}^{j}\sum_{i=0}^kC_{m-j}^{k-i-j}C_{n-m}^{i}$

    $\sum_{j=0}^k\begin{Bmatrix}L\\j\end{Bmatrix}j!C_{m}^{j}C_{n-j}^{k-j}$

    注意上面循环ii的上界实际上是min(k,L,m)=O(L)\min(k,L,m)=O(L)

    求出n=Ln=L的一行第二类斯特林数,每次询问就可以O(L)O(L)

    把组合数全拆出来约分就洛谷rk1了= =

    #include<bits/stdc++.h>
    #define il inline
    #define vd void
    #define mod 998244353
    #define poly std::vector<int>
    typedef long long ll;
    il ll gi(){
        ll x=0,f=1;
        char ch=getchar();
        while(!isdigit(ch))f^=ch=='-',ch=getchar();
        while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
        return f?x:-x;
    }
    il int pow(int x,int y){
        int ret=1;
        while(y){
            if(y&1)ret=1ll*ret*x%mod;
            x=1ll*x*x%mod;y>>=1;
        }
        return ret;
    }
    #define maxn 524289
    poly pA,pB;
    int rev[maxn],_lstN,P[maxn],iP[maxn];
    il vd ntt(int*A,int N,int t){
        for(int i=0;i<N;++i)if(rev[i]>i)std::swap(A[i],A[rev[i]]);
        for(int o=1;o<N;o<<=1){
            int W=t?P[o]:iP[o];
            for(int*p=A;p!=A+N;p+=o<<1)
                for(int i=0,w=1;i<o;++i,w=1ll*w*W%mod){
                    int t=1ll*w*p[i+o]%mod;
                    p[i+o]=(p[i]-t+mod)%mod;p[i]=(p[i]+t)%mod;
                }
        }
        if(!t){
            int inv=pow(N,mod-2);
            for(int i=0;i<N;++i)A[i]=1ll*A[i]*inv%mod;
        }
    }
    int N,lg;
    il vd setN(int n){
        N=1,lg=0;
        while(N<n)N<<=1,++lg;
        if(N!=_lstN)for(int i=0;i<N;++i)rev[i]=(rev[i>>1]>>1)|((i&1)<<lg-1);
    }
    il vd ntt(poly&a,int t){
        static int A[maxn];
        for(int i=0;i<a.size();++i)A[i]=a[i];memset(A+a.size(),0,4*(N-a.size()));
        ntt(A,N,t);
        a.resize(N);
        for(int i=0;i<N;++i)a[i]=A[i];
        int s=a.size();while(s&&!a[s-1])--s;
        a.resize(s);
    }
    il poly mul(poly a,poly b,int newn=-1){
        if(newn==-1)newn=a.size()+b.size()-1;
        setN(a.size()+b.size()-1);
        ntt(a,1),ntt(b,1);
        for(int i=0;i<N;++i)a[i]=1ll*a[i]*b[i]%mod;
        ntt(a,0);a.resize(newn);
        return a;
    }
    il poly operator+(poly a,const poly&b){
        if(a.size()<b.size())a.resize(b.size());
        for(int i=0;i<a.size();++i)if(i<b.size())a[i]=(a[i]+b[i])%mod;
        return a;
    }
    il poly operator-(poly a,const poly&b){
        if(a.size()<b.size())a.resize(b.size());
        for(int i=0;i<a.size();++i)if(i<b.size())a[i]=(a[i]-b[i]+mod)%mod;
        return a;
    }
    il poly operator*(poly a,int b){
        for(auto&i:a)i=1ll*i*b%mod;
        return a;
    }
    il poly qiudao(poly a){
        for(int i=0;i<a.size()-1;++i)a[i]=1ll*a[i+1]*(i+1)%mod;
        a.erase(a.end()-1);
        return a;
    }
    il poly jifen(poly a){
        a.insert(a.begin(),0);
        for(int i=1;i<a.size();++i)a[i]=1ll*a[i]*pow(i,mod-2)%mod;
        return a;
    }
    il poly getinv(poly a){
        if(a.size()==1)return poly(1,pow(a[0],mod-2));
        int n=a.size(),m=a.size()+1>>1;
        poly _a(m);
        for(int i=0;i<m;++i)_a[i]=a[i];
        poly b=getinv(_a);
        setN(n+m*2-2);
        ntt(a,1);ntt(b,1);
        for(int i=0;i<N;++i)a[i]=1ll*a[i]*b[i]%mod*b[i]%mod;
        ntt(a,0),ntt(b,0);
        a.resize(n);
        return b*2-a;
    }
    il poly getln(poly a,int n=-1){
        if(n==-1)n=a.size();
        a.resize(n);
        return jifen(mul(qiudao(a),getinv(a),n));
    }
    il poly getexp(poly a){
        if(a.size()==1)return a[0]=1,a;
        int n=a.size(),m=a.size()+1>>1;
        poly _a(m);
        for(int i=0;i<m;++i)_a[i]=a[i];
        poly b=getexp(_a);
        return mul(b,poly(1,1)-getln(b,a.size())+a,a.size());
    }
    il poly operator^(poly a,int b){
        int n=a.size();
        a=getexp(getln(a)*b);a.resize(n);
        return a;
    }
    il poly sqrt(poly a){
        if(a.size()==1)return a;
        int n=a.size(),m=a.size()+1>>1;
        poly _a(m);
        for(int i=0;i<m;++i)_a[i]=a[i];
        poly b=sqrt(_a);b.resize(n);
        return (b+mul(a,getinv(b),n))*(mod+1>>1);
    }
    il vd poly_init(){
        int G=3,iG=332748118;
        for(int i=1;i<maxn;i<<=1)P[i]=pow(G,(mod-1)/(i<<1)),iP[i]=pow(iG,(mod-1)/(i<<1));
    }
    il vd add(int&a,int b){a=a+b>=mod?a+b-mod:a+b;}
    int fact[20000010],ifact[20000010];
    int pr[4000010],Pr,PL[20000010];
    bool yes[20000010];
    int main(){
    #ifdef XZZSB
        freopen("in.in","r",stdin);
        freopen("out.out","w",stdout);
    #endif
        poly_init();
        int N=gi(),M=gi(),S=gi(),L=gi(),o=std::max(N,L);
        fact[0]=1;for(int i=1;i<=o;++i)fact[i]=1ll*i*fact[i-1]%mod;
        ifact[o]=pow(fact[o],mod-2);for(int i=o-1;~i;--i)ifact[i]=1ll*ifact[i+1]*(i+1)%mod;
        PL[0]=0,PL[1]=1;
        for(int i=2;i<=L;++i){
            if(!yes[i])pr[++Pr]=i,PL[i]=pow(i,L);
            for(int j=1;j<=Pr&&i*pr[j]<=L;++j){
                yes[i*pr[j]]=1;
                PL[i*pr[j]]=1ll*PL[pr[j]]*PL[i]%mod;
                if(i%pr[j]==0)break;
            }
        }
        poly f(L+1),g(L+1);
        for(int i=0,mul=1;i<=L;++i,mul=mod-mul)f[i]=1ll*mul*ifact[i]%mod;
        for(int i=0;i<=L;++i)g[i]=1ll*PL[i]*ifact[i]%mod;
        f=mul(f,g,L+1);
        while(S--){
            int n=gi(),m=gi(),k=gi(),ans=0,o=std::min(k,std::min(m,L));
            for(int i=0;i<=o;++i)
                ans=(ans+1ll*f[i]*ifact[m-i]%mod*fact[n-i]%mod*ifact[k-i])%mod;
            printf("%d\n",1ll*ans*fact[m]%mod*fact[k]%mod*ifact[n]%mod);
        }
        return 0;
    }
    
    • 1

    信息

    ID
    4338
    时间
    2000ms
    内存
    512MiB
    难度
    7
    标签
    递交数
    0
    已通过
    0
    上传者