[CQOI2016]密钥破解

嘟嘟嘟


这题我读了两遍才懂，然后感觉要解什么高次同余方程……然后我又仔细的看了看题，发现只要求得\(p\)和\(q\)就能求出\(r\)，继而用exgcd求出\(d\)，最后用快速幂求出\(n\)。
再看看这个数据范围，用Pollard-Rho最适合不过了。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
#include<ctime>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
//const int maxn = ;
inline ll read()
{
	ll ans = 0;
	char ch = getchar(), last = ' ';
	while(!isdigit(ch)) {last = ch; ch = getchar();}
	while(isdigit(ch)) {ans = (ans << 1) + (ans << 3) + ch - '0'; ch = getchar();}
	if(last == '-') ans = -ans;
	return ans;
}
inline void write(ll x)
{
	if(x < 0) x = -x, putchar('-');
	if(x >= 10) write(x / 10);
	putchar(x % 10 + '0');
}

ll e, N, c, r;

In ll mul(ll a, ll b, ll mod)
{
	ll d = ((long double)a / mod * b + 1e-8);
	ll r = a * b - d * mod;
	return r < 0 ? r + mod : r;
}
In ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
In ll f(ll x, ll a, ll mod) {return (mul(x, x, mod) + a) % mod;}

int a[] = {2, 3, 5, 7, 11};
const int M = (1 << 7) - 1;
In ll find(ll n)
{
	for(int i = 0; i < 5; ++i) if(n % a[i] == 0) return a[i];
	ll x = rand(), y = x, a = rand() % (n - 2) + 2, t = 1;
	for(int k = 2;; k <<= 1, y = x)
	{
		ll p = 1;
		for(int i = 1; i <= k; ++i)
		{
			x = f(x, a, n);
			p = mul(p, abs(x - y), n);
			if(!(i & M))
			{
				t = gcd(p, n);
				if(t > 1) break;
			}
		}
		if(t > 1 || (t = gcd(p, n)) > 1) break;
	}
	return t;
}
In ll pollard_rho(ll x)
{
	ll p = x;
	while(p == x) p = find(x);
	return p;
}

In void exgcd(ll a, ll b, ll& x, ll& y, ll& t)
{
	if(!b) t = a, x = 1, y = 0;
	else exgcd(b, a % b, y, x, t), y -= a / b * x;
}

In ll quickpow(ll a, ll b, ll mod)
{
	a %= mod;
	ll ret = 1;
	for(; b; b >>= 1, a = mul(a, a, mod))	//别忘了这里也会爆long long 
		if(b & 1) ret = mul(ret, a, mod);
	return ret;
}

int main()
{
	srand(time(0));
	e = read(), N = read(), c = read();
	ll p = pollard_rho(N), q = N / p; ll r = (p - 1) * (q - 1);
	ll d, y, t;
	exgcd(e, r, d, y, t);
	t = r / t;
	d = (d % t + t) % t;
	ll n = quickpow(c, d, N);
	write(d), space, write(n), enter;
	return 0;
}