bzoj 4555: [Tjoi2016&Heoi2016]求和 （NTT）

题目描述

传送门

题解

这道题解决的关键是知道第二类斯特林数的通项公式

S(n,m)=1m!∑k=0m(−1)kC(m,k)(m−k)n

然后将通项公式带入题目中的原式化简

f(n)=∑i=0n∑j=0iS(i,j)∗2j∗j!

∑i=0n∑j=0i1j!∑k=0j(−1)kC(j,k)(j−k)i∗2j∗j!

∑i=0n∑j=0i∑k=0j(−1)k∗(j−k)ik!(j−k)!∗2j∗j!

∑j=0n2j∗j!∑k=0j(−1)k∗∑ni=0(j−k)ik!(j−k)!

然后就可以用一遍NTT求解了

代码

#include
#include
#include
#include
#include
#define N 500003
#define LL long long 
#define p 998244353
using namespace std;
LL f[N],g[N],inv[N],jc[N],mi[N];
int n,n1,L,R[N];
LL quickpow(LL num,LL x)
{
    LL base=(num%p+p)%p; LL ans=1;
    while (x) {
        if (x&1) ans=ans*base%p;
        x>>=1;
        base=base*base%p;
    }
    return ans;
}
void NTT(LL a[N],int n,int opt)
{
    for (int i=0;iif (ifor (int i=1;i1){
        LL wn=quickpow(3,(p-1)/(i<<1));
        for (int p1=i<<1,j=0;j1;
            for (int k=0;k*wn)%p) {
                LL x=a[j+k],y=w*a[j+k+i]%p;
                a[j+k]=(x+y)%p; a[j+k+i]=(x-y+p)%p;
            }
        }
     } 
     if (opt==-1) reverse(a+1,a+n);
}
int main()
{
    freopen("a.in","r",stdin);
    scanf("%d",&n); 
    inv[0]=inv[1]=1; jc[0]=1; mi[0]=1;
    for (int i=1;i<=n;i++) jc[i]=jc[i-1]*i%p;
    for (int i=1;i<=n;i++) mi[i]=mi[i-1]*2%p;
    for (int i=2;i<=n;i++) inv[i]=(p-p/i)*inv[p%i]%p;
    for (int i=1;i<=n;i++) inv[i]=inv[i-1]*inv[i]%p;
    for (int i=0;i<=n;i++) 
     if (i&1) f[i]=-inv[i];
     else f[i]=inv[i];
    g[0]=1; g[1]=n+1;
    for (int i=2;i<=n;i++) {
        LL t=(quickpow(i,n+1)-1+p)%p;
        t=t*quickpow(i-1,p-2)%p; 
        g[i]=t*inv[i]%p;
    }
    for (n1=1;n1<=n*2+1;n1<<=1) L++;
    for (int i=0;i<=n1;i++) R[i]=(R[i>>1]>>1)|((i&1)<<(L-1));
    NTT(f,n1,1); NTT(g,n1,1);
    for (int i=0;i<=n1;i++) f[i]=f[i]*g[i]%p;
    NTT(f,n1,-1);
    LL t=quickpow(n1,p-2);
    for (int i=0;i<=n1;i++) f[i]=f[i]*t%p;
    LL ans=0;
    for (int i=0;i<=n;i++) ans=(ans+f[i]*jc[i]%p*mi[i]%p)%p;
    printf("%lld\n",(ans%p+p)%p); 
}