【E：Faulty Factorial】(数论，循环节)

**链接**:http://sua2018.contest.codeforces.com/group/J1GbXiu37w/contest/227105

**题意**:在1*2*3*......*n中改变一个数为比它小的数，使得%p等于r.输出任意一个解

2<=n&&n<=1e18

2<=p&&p<1e7

**分析**：n很大，从p入手，考虑循环节。但是真的很暴力

看代码：

#pragma warning(disable:4996)
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n;
ll p, r;

ll qpow(ll a, ll b) {
	ll res = 1;
	while (b) {
		if (b&1) {
			res = res * a%p;
		}
		a = a * a%p;
		b >>= 1;
	}
	return res;
}

int main() {
	scanf("%lld%lld%lld", &n, &p, &r);
	if (n == p) {
		if (r == 0) {
			if (n == 2)printf("-1 -1\n");
			else printf("2 1\n");
		}
		else {
			int f = 0;
			printf("%lld %lld\n", p, p - r);
		}
	}
	else if (n >= 2 * p) {
		if (r == 0) printf("%lld 1\n", p + 1);
		else printf("-1 -1\n");
	}
	else if (n > p) {
		if (r == 0)printf("%lld 1\n", p + 1);
		else {
			int f = 0;
			ll t = 1;
			for (ll i = 2; i <= n; i++) 
				if (i != p) t = t * i%p;
			for (ll i = 1; i < p; i++) {
				if (t*i%p == r) {
					printf("%lld %lld\n", p, i);
					f = 1;
					break;
				}
			}
			if (!f)printf("-1 -1\n");
		}
	}
	else if(n<p) {
		if (r == 0) cout << -1 << " " << -1 << endl;
		else {
			ll t = 1;
			for (ll i = 2; i <= n; i++) t = t * i%p;
			t = qpow(t, p - 2);
			for (ll i = 2; i <= n; i++) {
				ll tmp = r * i%p*t%p;
				if (tmp >= 1 && tmp<i) { cout << i << " " << tmp << endl; return 0; }
			}
			cout << -1 << " " << -1 << endl;
		}
	}
}