多次剩余原根法

多次剩余：
求$x^a=b(modp)$，p是质数的所有解。

找到p的原根g，设$g^B=b$，这个用bsgs求。

$x^a=b(modp)$
等价于$g^{a\ast x}=g^B(modp)$
$a\ast x=B(mod(p-1))$

那么用exgcd解即可。

例题：
http://www.51nod.com/Challenge/AcceptedRank.html#!#problemId=1038

卡常：写bsgs时可以利用边集数组，会比哈希表快一些。

#include
#include
#include
#include
#define ll long long
#define pp printf
#define fo(i, x, y) for(int i = x; i <= y; i ++)
using namespace std;

int T;
ll a, b, mo, g;
ll u[20], v[20], u0;

ll ksm(ll x, ll y) {
	ll s = 1;
	for(; y; y /= 2, x = x * x % mo)
		if(y & 1) s = s * x % mo;
	return s;
}

void fen(ll x) {
	u0 = 0;
	for(int i = 2; i * i <= x; i ++) if(x % i == 0) {
		u[++ u0] = i; v[u0] = 0;
		while(x % i == 0) x /= i, v[u0] ++;
	}
	if(x > 1) u[++ u0] = x, v[u0] = 1;
}

int pdg(int x) {
	fo(i, 1, u0) if(ksm(x, (mo - 1) / u[i]) == 1) return 0;
	return 1;
}

void fg() {
	fo(i, 2, mo) if(pdg(i))
		{ g = i; return;}
}

const int M = 1e5 + 7;
int fi[M], to[M], nt[M], V[M], tot;

void link(int x, int y, int z) {
	nt[++ tot] = fi[x], to[tot] = y, V[tot] = z, fi[x] = tot;
}


int bsgs(int a, int b) {
	tot = 0; memset(fi, 0, sizeof fi);
	int m = ceil(sqrt(mo));
	ll x = b;
	fo(i, 1, m) {
		x = x * a % mo;
		link(x % M, x, i);
	}
	x = 1; ll am = ksm(a, m);
	fo(i, 1, m) {
		x = x * am % mo;
		for(int j = fi[x % M]; j; j = nt[j])
			if(to[j] == x) return i * m - V[j];
	}
	return -1;
}

int gcd(int x, int y) {
	return !y ? x : gcd(y, x % y);
}

void exgcd(int a, int b, int &x, int &y) {
	if(!b) {x = a; y = 0; return;}
	exgcd(b, a % b, y, x); y -= (a / b) * x;
}

int ans[100000];

void gg(int a, int b, int c) {
	int d = gcd(a, b);
	if(c % d) return;
	a /= d; b /= d; c /= d;
	int x, y; exgcd(a, b, x, y);
	x = (ll) (x % b + b) * c % b;
	fo(i, 1, d) ans[++ ans[0]] = (i - 1) * b + x;
}

int main() {
	for(scanf("%d", &T); T; T --) {
		scanf("%lld %lld %lld", &mo, &a, &b);
		if(mo == 2) {
			if(b == 0) pp("1\n0\n"); else
			pp("1\n1\n");
			continue;
		}
		fen(mo - 1); fg();
		int B = bsgs(g, b);
		ans[0] = 0;
		gg(a, mo - 1, B);
		fo(i, 1, ans[0]) ans[i] = ksm(g, ans[i]);
		sort(ans + 1, ans + ans[0] + 1);
		ans[0] = unique(ans + 1, ans + ans[0] + 1) - (ans + 1);
		if(ans[0] == 0) {
			pp("No Solution\n");
		} else {
			fo(i, 1, ans[0]) pp("%d ", ans[i]);
			pp("\n");
		}
	}
}