2627: JZPKIL
题意：求
\[ \sum_{i=1}^n (n,i)^x [i,n]^y,\ [i,n] = lcm(i,n) \]
\(n \le 10^{18},\ x,y\le 3000\)
本题带来了一种新技巧，n太大，转化成一个积性函数然后求这个积性函数，质因子分解利用积性，这养只与质因子的数量和指数有关
官方题解清橙上有
首先套路推♂倒
\[ n^y \sum_{d \mid n} d^x \sum_{e \mid \frac{n}{d}} \mu(e) e^y \sum_{i=1}^{\frac{n}{de}} i^{y+1-k} \\ \]
这里使用\(D=de\)然后整除分块并不好做
代入伯努利数，交换求和顺序
\[ \frac{n^y}{y+1} \sum_{k=0}^y \binom{y+1}{k} B_k^+ \sum_{d \mid n} d^x \sum_{e \mid \frac{n}{d}} \mu(e) e^y \sum_{i=1}^{\frac{n}{de}} i^{y+1-k} \\ \]
后面那块就是\((id^x *( (\mu \cdot id^y) * id^z)(n) = ( (\mu \cdot id^y) * id^x * id^z)(n)\)
考虑求这个积性函数
用Pollard-rho质因子分解n，然后求
\(( (\mu \cdot id^y) * id^x * id^z)(p^c)\)
这个很好求，\(\mu\)那里只能是\(1,p\)，剩下的直接\(O(c)\)计算就行了
质因子个数和指数都是log级的，所以复杂度大约是\(O(y^2 + n^{\frac{1}{4}}logn+ylog^2n)\)
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int N = 3005, P = 1e9+7, mo = P, inv2 = (P+1)/2;
inline int read(){
char c=getchar(); int x=0,f=1;
while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
return x*f;
}
namespace qwq {
ll mul(ll a, ll b, ll P) {
ll t = (a * b - (ll) ((long double) a / P * b + 0.5) * P);
return t<0 ? t+P : t;
}
ll Pow(ll a, ll b, ll P) {
ll ans = 1; a %= P;
for(; b; b>>=1, a = mul(a, a, P))
if(b&1) ans = mul(ans, a, P);
return ans;
}
bool witness(ll a, ll n, ll u, int t) {
ll x = Pow(a, u, n), y = x;
while(t--) {
x = mul(x, x, n);
if(x == 1 && y != 1 && y != n-1) return true;
y = x;
}
return x != 1;
}
int p[10] = {2, 3, 5, 7, 11, 13, 17, 19, 23};
bool m_r(ll n) {
if(n == 2) return true;
if(n <= 1 || ~n&1) return false;
ll u = n-1, t = 0;
while(~u&1) u >>= 1, t++;
for(int i=0; i<9; i++) {
if(p[i] == n) return true;
if(witness(p[i], n, u, t)) return false;
}
return true;
}
ll gcd(ll a, ll b) {return !b ? a : gcd(b, a%b);}
ll _rho(ll n, ll c) {
int k = 2; ll x = rand()%n, y = x, d = 1;
for(int i=1; d==1; i++) {
x = (mul(x, x, n) + c) %n;
d = gcd(n, x>y ? x-y : y-x);
if(i == k) y = x, k <<= 1;
}
return d;
}
ll d[N];
void p_r(ll n) {
if(n == 1) return;
if(m_r(n)) {d[++d[0]] = n; return;}
ll t = n;
while(t == n) t = _rho(n, rand() % (n-1) + 1);
p_r(t); p_r(n/t);
}
} using qwq::d;
ll Pow(ll a, ll b) {
ll ans = 1; a %= P;
for(; b; b>>=1, a=a*a%P)
if(b&1) ans=ans*a%P;
return ans;
}
int b[N];
ll inv[N], fac[N], facInv[N];
inline ll C(int n, int m) {return fac[n] * facInv[m] %P * facInv[n-m] %P;}
void init(int n) {
inv[1] = fac[0] = facInv[0] = 1;
for(int i=1; i<=n+1; i++) {
if(i != 1) inv[i] = (P - P/i) * inv[P%i] %P;
fac[i] = fac[i-1] * i %P;
facInv[i] = facInv[i-1] * inv[i] %P;
}
b[0] = 1; b[1] = P-inv2;
for(int m=2; m<=n; m++) if(!(m&1)) {
for(int k=0; k<=m-1; k++) b[m] = (b[m] - C(m+1, k) * b[k]) %P;
b[m] = b[m] * Pow(C(m+1, m), P-2) %P;
if(b[m] < 0) b[m] += P;
}
b[1] = inv2;
}
struct meow{ll p; int k;} a[N]; int m;
ll p1[N], p2[N];
void solve(ll n, int x, int y) {
d[0] = 0; qwq::p_r(n);
sort(d+1, d+d[0]+1);
m = 0; a[++m] = (meow){d[1], 1};
for(int i=2; i<=d[0]; i++) {
if(d[i] == d[i-1]) a[m].k++;
else a[++m] = (meow){d[i], 1};
}
//for(int i=1; i<=m; i++) printf("factor %d %lld %d\n", i, a[i].p, a[i].k);
ll ans = 0;
for(int k=0; k<=y; k++) {
ll t = C(y+1, k) * b[k] %mo;
int z = y + 1 - k;
for(int i=1; i<=m; i++) {
ll p = a[i].p, t1 = 0, t2 = 0;
int c = a[i].k;
p1[0] = 1; ll u = p1[1] = Pow(p, x);
p2[0] = 1; ll v = p2[1] = Pow(p, z);
for(int j=2; j<=c; j++) p1[j] = p1[j-1] * u %mo, p2[j] = p2[j-1] * v %mo;
for(int j=0; j<=c; j++) t1 = (t1 + p1[j] * p2[c-j]) %mo;
for(int j=0; j