[SDOI2010]古代猪文

Description

[SDOI2010]古代猪文

Solution

给(n)，(G)，求(G^{sum_{d|n}C^{d}_{n}}:mod:999911659)
先套一下欧拉定理，因为(999911659)是质数，所以，
(G^{sum_{d|n}C^{d}_{n}}:mod:999911659)=(G^{sum_{d|n}C^{d}_{n}:mod:999911658}:mod:999911659)
我们先求(sum_{d|n}C^{d}_{n}:mod:999911658)
我们发现，(999911658=2*3*4679*35617)，四个质数相乘，而且质数的次数都为(1)
用(Lucas)定理求出(sum_{d|n}C^{d}_{n})分别在(2)，(3)，(4679)，(35617)下的余数(a1)，(a2)，(a3)，(a4)
设答案为(x)，得到方程
(left{ egin{aligned} x&equiv a1&(mod&:2)\ x&equiv a2&(mod&:3)\ x&equiv a3&(mod&:4679)\ x&equiv a4&(mod&:35617)\ end{aligned} ight.)
再套一下中国剩余定理的板子，求出(x)，再用一个快速幂，就可以求出答案了

Code

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define MOD 999911659
#define MoD 999911658
#define MOD1 2
#define MOD2 3
#define MOD3 4679
#define MOD4 35617
ll M1, M2, M3, M4, n, G, a1, a2, a3, a4, x, sum;
ll h1[MOD1 + 10], h2[MOD2 + 10], h3[MOD3 + 10], h4[MOD4 + 10];
inline ll read() {
	ll s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}
inline ll Pow(ll a, ll b, ll p) {
	ll ans = 1;
	while (b) {
		if (b & 1) (ans *= a) %= p;
		(a *= a) %= p;
		b >>= 1;
	}
	return ans;
}
inline ll Inv(ll a, ll p) {
	return Pow(a, p - 2, p);
}
inline ll C(ll n, ll m, ll p) {
    if (m > n) return 0;
    if (p == MOD1) return h1[n] * Inv(h1[m], p) % p * Inv(h1[n - m], p) % p;
    if (p == MOD2) return h2[n] * Inv(h2[m], p) % p * Inv(h2[n - m], p) % p;
    if (p == MOD3) return h3[n] * Inv(h3[m], p) % p * Inv(h3[n - m], p) % p;
    if (p == MOD4) return h4[n] * Inv(h4[m], p) % p * Inv(h4[n - m], p) % p;
}

ll Lucas(ll n, ll m, ll p) {
    if (!m) return 1;
    return Lucas(n / p, m / p, p) * C(n % p, m % p, p) % p;
}
inline void Init() {
	h1[0] = 1; for (register ll i = 1; i <= MOD1; i++) h1[i] = (h1[i - 1] * i) % MOD1;
	h2[0] = 1; for (register ll i = 1; i <= MOD2; i++) h2[i] = (h2[i - 1] * i) % MOD2;
	h3[0] = 1; for (register ll i = 1; i <= MOD3; i++) h3[i] = (h3[i - 1] * i) % MOD3;
	h4[0] = 1; for (register ll i = 1; i <= MOD4; i++) h4[i] = (h4[i - 1] * i) % MOD4;	
}
int main() {
	M1 = MoD / MOD1;
	M2 = MoD / MOD2;
	M3 = MoD / MOD3;
	M4 = MoD / MOD4;
	Init();
	n = read(), G = read();
	if (G % MOD == 0) {
		printf("0");
		return 0;
	}
	for (register ll i = 1; i <= sqrt(n); i++)
		if (n % i == 0) {
			a1 = Lucas(n, i, MOD1), a2 = Lucas(n, i, MOD2), a3 = Lucas(n, i, MOD3), a4 = Lucas(n, i, MOD4);
			x = ((a1 % MoD * M1 % MoD * Inv(M1, MOD1) % MoD) +
				 (a2 % MoD * M2 % MoD * Inv(M2, MOD2) % MoD) +
				 (a3 % MoD * M3 % MoD * Inv(M3, MOD3) % MoD) +
				 (a4 % MoD * M4 % MoD * Inv(M4, MOD4) % MoD)) % MoD;
			(sum += x) %= MoD;
			if (i * i == n) continue;
			a1 = Lucas(n, n / i, MOD1), a2 = Lucas(n, n / i, MOD2), a3 = Lucas(n, n / i, MOD3), a4 = Lucas(n, n / i, MOD4);
			x = ((a1 % MoD * M1 % MoD * Inv(M1, MOD1) % MoD) +
				 (a2 % MoD * M2 % MoD * Inv(M2, MOD2) % MoD) +
				 (a3 % MoD * M3 % MoD * Inv(M3, MOD3) % MoD) +
				 (a4 % MoD * M4 % MoD * Inv(M4, MOD4) % MoD)) % MoD;
			(sum += x) %= MoD;
		}
	printf("%lld", Pow(G, sum, MOD));
	return 0;
}