洛谷 [P2480] 古代猪文

卢卡斯定理

注意特判底数和模数相等的情况

http://www.cnblogs.com/poorpool/p/8532809.html

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define ll long long
using namespace std;
const int MOD = 999911659;
ll n, g, p[5] = {0, 2, 3, 4679, 35617}, ans[5], fac[100000], ni[100000];
ll exgcd(ll a, ll b, ll & x, ll & y ){
	if(!b) {
		x = 1; y = 0;
		return a;
	}
	ll t = exgcd(b, a % b, y, x);
	y -= a / b * x;
	return t;
}
ll getni(ll x, ll p) {
	ll a, b;
	exgcd(x, p, a, b);
	(a += p) %= p;
	return a;
}
ll lucas(ll a, ll b, ll p) {
	if(a < b) return 0;
	if(a < p) return fac[a] * ni[b] * ni[a - b] % p;
	else return lucas(a % p, b % p, p) * lucas(a / p, b / p, p) % p;
}
void work(int x) {
	memset(fac, 0, sizeof(fac));
	memset(ni, 0, sizeof(ni));
	fac[1] = fac[0] = ni[0] = ni[1] = 1;
	for(int i = 2; i < p[x]; i++) fac[i] = fac[i - 1] * i % p[x];
	for(int i = 2; i < p[x]; i++) ni[i] = (p[x] - p[x] / i) * ni[p[x] % i] % p[x];
	for(int i = 2; i < p[x]; i++) (ni[i] *= ni[i - 1]) %= p[x];
	ll i = 1ll;
	for( ; i * i < n; i++) {
		if(n % i == 0) {
			ans[x] += lucas(n, i, p[x]);
			//cout << ans[x] << endl;
			ans[x] += lucas(n, n / i, p[x]);
			//cout << ans[x] << endl;
			ans[x] %= p[x];
			//cout << ans[x] << endl;
		}
	}
	if(i * i == n) ans[x] += lucas(n, i, p[x]);
	//cout << ans[x] << endl;
	//cout << endl;
	ans[x] %= p[x];
}
ll CRT() {
	ll M = MOD - 1;
	ll rt = 0ll;
	for(int i = 1; i <= 4; i++) {
		rt += ans[i] * getni(M / p[i], p[i]) * (M / p[i]) % M;
	}
	//cout << rt << endl;
	return rt % M;
}
ll quick_mod(ll a, ll k) {
	ll ans = 1ll;
	while(k) {
		if(k & 1ll) (ans *= a) %= MOD;
		(a *= a) %= MOD;
		k >>= 1;
	}
	return ans;
}
int main() {
	cin >> n >> g;
	if(g == MOD) {printf("0\n");return 0;}
	for(int i = 1; i <= 4; i++) {
		work(i);
		//cout << ans[i] << endl;
	}
	cout << quick_mod(g, CRT()) << endl;
	return 0;
}