题目大意
给出 (a_{1,2,...,n})，对于 (forall kin [1,n]) ，求出：
[sum_{i=1}^{n}a_i^k
]
(nle 2 imes 10^5)，答案对 (998244353) 取模 。
思路
我们考虑对答案构造生成函数：
[F(x)=sum_{k=0}^{infty} sum_{i=1}^{n}a_i^kx^k
]
[=sum_{i=1}^{n}frac{1}{1-a_ix}
]
[=sum_{i=1}^{n}(1+frac{a_ix}{1-a_ix})
]
[=n-x(sum_{i=1}^{n}frac{-a_i}{1-a_ix})
]
然后我们发现存在：
[(ln(1-a_ix))^{'}=frac{-a_i}{1-a_ix}
]
于是：
[F(x)=n-x(sum_{i=1}^{n}ln(1-a_ix))^{'}
]
[=n-x(ln prod_{i=1}^{n}(1-a_ix))^{'}
]
然后这个东西就可以直接分治解决了，时间复杂度 (Theta(nlog^2 n)) 。
( exttt{Code})
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline", "no-stack-protector", "unroll-loops")
#pragma GCC diagnostic error "-fwhole-program"
#pragma GCC diagnostic error "-fcse-skip-blocks"
#pragma GCC diagnostic error "-funsafe-loop-optimizations"
#include <bits/stdc++.h>
using namespace std;
#define SZ(x) ((int)x.size())
#define Int register int
#define mod 998244353
#define MAXN 1000005
int mul (int a,int b){return 1ll * a * b % mod;}
int dec (int a,int b){return a >= b ? a - b : a + mod - b;}
int add (int a,int b){return a + b >= mod ? a + b - mod : a + b;}
int qkpow (int a,int k){
int res = 1;for (;k;k >>= 1,a = 1ll * a * a % mod) if (k & 1) res = 1ll * res * a % mod;
return res;
}
int inv (int x){return qkpow (x,mod - 2);}
typedef vector <int> poly;
int w[MAXN],rev[MAXN];
void init_ntt (){
int lim = 1 << 19;
for (Int i = 0;i < lim;++ i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << 18);
int Wn = qkpow (3,(mod - 1) / lim);w[lim >> 1] = 1;
for (Int i = lim / 2 + 1;i < lim;++ i) w[i] = mul (w[i - 1],Wn);
for (Int i = lim / 2 - 1;i;-- i) w[i] = w[i << 1];
}
void ntt (poly &a,int lim,int type){
#define G 3
#define Gi 332748118
static unsigned long long d[MAXN];
for (Int i = 0,z = 19 - __builtin_ctz(lim);i < lim;++ i) d[rev[i] >> z] = a[i];
for (Int i = 1;i < lim;i <<= 1)
for (Int j = 0;j < lim;j += i << 1)
for (Int k = 0;k < i;++ k){
int x = mul (w[i + k],d[i + j + k]);
d[i + j + k] = dec (d[j + k],x),d[j + k] = add (d[j + k],x);
}
for (Int i = 0;i < lim;++ i) a[i] = d[i];
if (type == -1){
reverse (a.begin() + 1,a.begin() + lim);
for (Int i = 0,Inv = inv (lim);i < lim;++ i) a[i] = mul (a[i],Inv);
}
#undef G
#undef Gi
}
poly operator * (poly a,poly b){
int d = SZ (a) + SZ (b) - 1,lim = 1;while (lim < d) lim <<= 1;
a.resize (lim),b.resize (lim);
ntt (a,lim,1),ntt (b,lim,1);
for (Int i = 0;i < lim;++ i) a[i] = mul (a[i],b[i]);
ntt (a,lim,-1),a.resize (d);
return a;
}
poly inv (poly a,int n){
poly b(1,inv (a[0])),c;
for (Int l = 4;(l >> 2) < n;l <<= 1){
c.resize (l >> 1);
for (Int i = 0;i < (l >> 1);++ i) c[i] = i < n ? a[i] : 0;
c.resize (l),b.resize (l);
ntt (c,l,1),ntt (b,l,1);
for (Int i = 0;i < l;++ i) b[i] = mul (b[i],dec (2,mul (b[i],c[i])));
ntt (b,l,-1),b.resize (l >> 1);
}
b.resize (n);
return b;
}
poly inv (poly a){return inv (a,SZ (a));}
poly der (poly a){
for (Int i = 0;i < SZ (a) - 1;++ i) a[i] = mul (a[i + 1],i + 1);
a.pop_back ();return a;
}
poly ine (poly a){
a.push_back (0);
for (Int i = SZ (a) - 1;i;-- i) a[i] = mul (a[i - 1],inv (i));
a[0] = 0;return a;
}
poly ln (poly a,int n){
a = ine (der (a) * inv (a));
a.resize (n);
return a;
}
poly ln (poly a){return ln (a,SZ (a));}
template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
int n,a[MAXN];poly F;
poly divide (int l,int r){
if (l == r){
poly tmp;tmp.resize(2);
tmp[0] = 1,tmp[1] = mod - a[l];
return tmp;
}
int mid = (l + r) >> 1;
return divide (l,mid) * divide (mid + 1,r);
}
signed main(){
init_ntt ();
int T;read (T);
while (T --> 0){
read (n);
for (Int i = 1;i <= n;++ i) read (a[i]);
F = der (ln (divide (1,n))),F.resize (n + 1);
for (Int i = n;i;-- i) F[i] = mod - F[i - 1];F[0] = n;
int ans = 0;for (Int i = 1;i <= n;++ i) ans ^= (F[i] % mod + mod) % mod;
write (ans),putchar ('
');
}
return 0;
}