题目：
https://ac.nowcoder.com/acm/problem/15254?&headNav=acm
给你 n n n， k k k，求 ∀ 0 ≤ t ≤ k − 1 \forall 0\le t\le k-1 ∀0≤t≤k−1
s t = [ ∑ k ∣ i , 0 ≤ i + t ≤ n ( n i + t ) ] m o d 998244353 s_t=\Big[\sum_{k|i,0\le i+t\le n}{n\choose i+t}\Big]\;mod\;998244353 st=[k∣i,0≤i+t≤n∑(i+tn)]mod998244353
最后输出 s t s_t st的异或和
1 ≤ n ≤ 1 0 1 0 6 1\le n\le 10^{10^6} 1≤n≤10106， 1 ≤ k ≤ 2 20 1\le k\le 2^{20} 1≤k≤220， k k k为 2 2 2的幂
思路：
单位根反演
[ n ∣ k ] = 1 n ∑ i = 0 n − 1 ω n k i [n|k]=\frac{1}{n}\sum_{i=0}^{n-1}\omega^{ki}_n [n∣k]=n1i=0∑n−1ωnki
s t = ∑ k ∣ i , 0 ≤ i + t ≤ n ( n i + t ) = ∑ i = 0 n [ k ∣ i − t ] ( n i ) = ∑ i = 0 n ∑ j = 0 k − 1 1 k ω k j ( i − t ) ( n i ) = 1 k ∑ j = 0 k − 1 ω k − j t ∑ i = 0 n ω k j i ( n i ) = 1 k ∑ j = 0 k − 1 ω k − j t ( ω k j + 1 ) n \begin{aligned} s_t&=\sum_{k|i,0\le i+t\le n}{n\choose i+t}\\ &=\sum_{i=0}^{n}[k|i-t]{n\choose i}\\ &=\sum_{i=0}^{n}\sum_{j=0}^{k-1}\frac{1}{k}\omega^{j(i-t)}_k{n\choose i}\\ &=\frac{1}{k}\sum_{j=0}^{k-1}\omega^{-jt}_k\sum_{i=0}^{n}\omega^{ji}_k{n\choose i}\\ &=\frac{1}{k}\sum_{j=0}^{k-1}\omega^{-jt}_k(\omega^j_k+1)^n \end{aligned} st=k∣i,0≤i+t≤n∑(i+tn)=i=0∑n[k∣i−t](in)=i=0∑nj=0∑k−1k1ωkj(i−t)(in)=k1j=0∑k−1ωk−jti=0∑nωkji(in)=k1j=0∑k−1ωk−jt(ωkj+1)n
注意到这个形式就是 I D F T IDFT IDFT， ( ω k j + 1 ) n (\omega^j_k+1)^n (ωkj+1)n是点值，算的 s t s_t st就是系数。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<utility>
#include<iostream>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
void open(const char *s)
{
#ifndef ONLINE_JUDGE
char str[100];
sprintf(str,"%s.in",s);
freopen(str,"r",stdin);
sprintf(str,"%s.out",s);
freopen(str,"w",stdout);
#endif
}
const ll p=998244353;
const ll p2=998244352;
const ll g=3;
ll fp(ll a,ll b)
{
ll s=1;
for(;b;b>>=1,a=a*a%p)
if(b&1)
s=s*a%p;
return s;
}
char s[1000010];
ll a[2000010];
int rev[2000010];
void ntt(ll *a,int n,int t)
{
for(int i=1;i<n;i++)
{
rev[i]=(rev[i>>1]>>1)|(i&1?n>>1:0);
if(i>rev[i])
swap(a[i],a[rev[i]]);
}
for(int i=2;i<=n;i<<=1)
{
int wn=fp(g,(p-1)/i);
if(t==-1)
wn=fp(wn,p-2);
for(int j=0;j<n;j+=i)
{
int w=1;
for(int k=j;k<j+i/2;k++)
{
int u=a[k];
int v=(ll)a[k+i/2]*w%p;
a[k]=(u+v)%p;
a[k+i/2]=(u-v+p)%p;
w=(ll)w*wn%p;
}
}
}
if(t==-1)
{
ll inv=fp(n,p-2);
for(int i=0;i<n;i++)
a[i]=a[i]*inv%p;
}
}
int main()
{
// freopen("e.in","r",stdin);
scanf("%s",s+1);
int n=0,k;
scanf("%d",&k);
int len=strlen(s+1);
for(int i=1;i<=len;i++)
n=((ll)n*10+s[i]-'0')%p2;
ll w=fp(g,(p-1)/k);
ll now=1;
for(int i=0;i<k;i++)
{
a[i]=fp(now+1,n)%p;
now=now*w%p;
}
ntt(a,k,-1);
ll ans=0;
for(int i=0;i<k;i++)
ans=ans^a[i];
printf("%lld\n",ans);
return 0;
}