题意

给定 $f_i,p$ ，求 $x$ 使得

$$\left\{\begin{matrix} x^i & \equiv & f_{a_i} \pmod{p}\\ & \vdots & \end{matrix}\right.$$

其中 $a_i$ 是一个位置的排列。

$n\leq2\times10^5,x< p$ 。

题解

注意到解一定在 $f_i$ 中。就用等比数列求和一个一个判。

代码

const int N = 2e5;

inline ll Read() {
	ll x = 0, f = 1;
	char c = getchar();
	while (c != '-' && (c < '0' || c > '9')) c = getchar();
	if (c == '-') f = -f, c = getchar();
	while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
	return x * f;
}

namespace Main {
	int n, mod;
	ll a[N], sum;
	ll qpow (ll a, ll b) {
		ll ans = 1;
		for (; b; b >>= 1, a = a * a % mod)
			if (b & 1) ans = ans * a % mod;
		return ans;
	}
	int main () {
		n = Read(), mod = Read();
		for (int i = 1; i <= n; i++) (sum += (a[i] = Read())) %= mod;
		for (int i = 1; i <= n; i++) {
			if (a[i] == 1) continue;
			ll fsum = (qpow (a[i], n) - 1) * qpow (a[i] - 1, mod - 2) % mod;
			if (fsum == sum) {
				printf ("%lld\n", a[i]);
				return 0;
			}
		}
		return 0;
	}
}

int main () {
//	freopen(".in", "r", stdin);
//	freopen(".out", "w", stdout);
	Main::main();
	return 0;
}