【NTT】loj#6261. 一个人的高三楼

去年看过t老师写这题博客；以为是道神仙题

题目大意

求一个数列的$k$次前缀和。$nle 10^5$.

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题目分析

【计数】cf223C. Partial Sums 加强版。注意到最后的式子是$f_i=sumlimits_{j+k=i}pre_j a_k$的样子，因此在预处理$pre_j$之后就是卷积的板子了。

#include<bits/stdc++.h>
#define MO 998244353
const int maxn = 400035;
const int inv3 = 332748118;

int n,len,dt;
int cov[maxn],a[maxn],f[maxn],pre[maxn],fac[maxn],inv[maxn];
long long k;

int qmi(int a, int b)
{
    int ret = 1;
    for (; b; b>>=1, a=1ll*a*a%MO)
        if (b&1) ret = 1ll*ret*a%MO;
    return ret;
}
void NTT(int *a, int opt)
{
    for (int i=0; i<len; i++)
        if (i < cov[i]) std::swap(a[i], a[cov[i]]);
    for (int i=1; i<len; i<<=1)
    {
        int Wn = qmi(3, (MO-1)/(i<<1));
        if (opt==-1) Wn = qmi(inv3, (MO-1)/(i<<1));
        for (int j=0, p=i<<1; j<len; j+=p)
        {
            int w = 1;
            for (int k=0; k<i; k++, w=1ll*w*Wn%MO)
            {
                int valx = a[j+k], valy = 1ll*w*a[i+j+k]%MO;
                a[j+k] = (valx+valy)%MO, a[i+j+k] = (valx-valy+MO)%MO;
            }
        }
    }
    if (opt==-1){
        int inv = qmi(len, MO-2);
        for (int i=0; i<len; i++) a[i] = 1ll*a[i]*inv%MO;
    }
}
void init()
{
    pre[0] = fac[0] = fac[1] = inv[0] = inv[1] = 1;
    for (int i=2; i<=n; i++)
        fac[i] = 1ll*fac[i-1]*i%MO, 
        inv[i] = MO-1ll*(MO/i)*inv[MO%i]%MO;
    for (int i=1; i<=n; i++)
        pre[i] = 1ll*pre[i-1]*(k%MO+i-1)%MO*inv[i]%MO;
}
int main()
{
    freopen("loj6261.in","r",stdin);
    freopen("loj6261.out","w",stdout);
    scanf("%d%lld",&n,&k);
    for (int i=0; i<n; i++) scanf("%d",&a[i]);
    if (k){
        init();
        for (len=1; len<=n+n; len<<=1) ++dt;
        for (int i=0; i<len; i++)
            cov[i] = (cov[i>>1]>>1)|((i&1)<<(dt-1));
        NTT(a, 1), NTT(pre, 1);
        for (int i=0; i<len; i++) f[i] = 1ll*a[i]*pre[i]%MO;
        NTT(f, -1);
    }else for (int i=0; i<n; i++) f[i] = a[i];
    for (int i=0; i<n; i++) printf("%d
",f[i]);
    return 0;
}

END