题目链接
题意:求一个长度为(n)的数列的(k)阶前缀和对998244353取模的结果.((nle10^{5} kle2^{60}))
设$$f_k(x)=sum_{i=0}{n-1}S_{i+1}{(k)}x^i$$
[g(x)=sum_{i=0}^{n-1}x^i
]
于是有$$f_k(x)g(x)equiv f_{k+1}(x)pmod{x^n}$$
所求即$$g(x)^kf_0(x)$$
直接快速幂 复杂度(O(n log n log k)) 无法通过此题
考虑(g(x)^k)各项的意义，设$$g(x)k=sum_{i=0}{n-1}a_ix^i$$
(a_i)即为有序地选k个([0,n-1])之间的整数，和为i的方案数，用插板法可得(a_i=inom{k-1+i}{i})，从而可以(O(n))或(O(n log n))得到(g(x)^k)
最后用(NTT)将(f_0(x))和(g(x)^k)合并
总复杂度(O(n log n))
#include<cstdio>
#include<algorithm>
#include<ctype.h>
#include<string.h>
#include<math.h>
using namespace std;
#define ll long long
#define rep(i,x,y) for(int i=(x);i<=(y);++i)
#define travel(i,x) for(int i=h[x];i;i=pre[i])
inline char read() {
static const int IN_LEN = 1000000;
static char buf[IN_LEN], *s, *t;
return (s == t ? t = (s = buf) + fread(buf, 1, IN_LEN, stdin), (s == t ? -1 : *s++) : *s++);
}
template<class T>
inline void read(T &x) {
static bool iosig;
static char c;
for (iosig = false, c = read(); !isdigit(c); c = read()) {
if (c == '-') iosig = true;
if (c == -1) return;
}
for (x = 0; isdigit(c); c = read()) x = ((x + (x << 2)) << 1) + (c ^ '0');
if (iosig) x = -x;
}
const int OUT_LEN = 10000000;
char obuf[OUT_LEN], *ooh = obuf;
inline void print(char c) {
if (ooh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), ooh = obuf;
*ooh++ = c;
}
template<class T>
inline void print(T x) {
static int buf[30], cnt;
if (x == 0) print('0');
else {
if (x < 0) print('-'), x = -x;
for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;
while (cnt) print((char)buf[cnt--]);
}
}
inline void flush() { fwrite(obuf, 1, ooh - obuf, stdout); }
const int N = 1<<18, P = 998244353;
ll k;
int n, p, a[N], b[N];
inline int Pow(ll x, int y=P-2){
ll ass=1;
for(; y; y>>=1, x=x*x%P) if(y&1) ass=ass*x%P;
return ass;
}
inline void NTT(int *f, int g){
for(int i=0, j=0; i<p; ++i){
if(i>j) swap(f[i], f[j]);
for(int k=p>>1; (j^=k)<k; k>>=1);
}
for(int i=1; i<p; i<<=1){
int w0=(g==1?Pow(3, (P-1)/i/2):Pow(Pow(3, (P-1)/i/2)));
for(int j=0; j<p; j+=i<<1){
int w=1;
for(int k=j; k<j+i; ++k){
int t=(ll)w*f[k+i]%P;
f[k+i]=(P+f[k]-t)%P;
f[k]=(f[k]+t)%P;
w=(ll)w*w0%P;
}
}
}
if(g==-1){
int I=Pow(p);
rep(i, 0, p-1) f[i]=(ll)f[i]*I%P;
}
}
int main() {
read(n), read(k);
rep(i, 0, n-1) read(a[i]);
b[0]=1;
rep(i, 1, n-1) b[i]=(k-1+i)%P*b[i-1]%P*Pow(i)%P;
for(p=1; p<n*2-1; p<<=1);
NTT(a, 1), NTT(b, 1);
rep(i, 0, p-1) a[i]=(ll)a[i]*b[i]%P;
NTT(a, -1);
rep(i, 0, n-1) print(a[i]), print('
');
return flush(), 0;
}