[SDOI2011]計算器

YM zky跑得飛快的BSGS

數論模板題,複習下數論

過不了樣例都能A是鬧哪樣啊233

Code:

#include <cstdio>
#include <map>
#include <cmath>
using namespace std;
map<long long, long long> s;
long long y, z, p, T, k;
inline long long exgcd(long long a, long long b, long long &x, long long &y)
{
	if(!b)
	{
		x = 1, y = 0;
		return a;
	}
	else
	{
		long long d = exgcd(b, a % b, x, y), t = x;
		x = y, y = t - a / b * y;
		return d;
	}
}
inline long long quick_pow(long long n, long long m, long long p)
{
	long long res = 1;
	while (m)
	{
		if (m & 1)
			res = res * n % p;
		m >>= 1;
		n = n * n % p;
	}
	return res;
}
inline void INV(long long a, long long b, long long p)
{
	a = quick_pow(a, p-2, p);
	b = b * a % p;
	if (!b)
		puts("Orz, I cannot find x!");
	else
		printf("%lld\n", b);
}
inline long long gcd(long long a, long long b)
{
	long long t;
	while(b)
	{
		t = a % b;
		a = b;
		b = t;
	}
	return a;
}
inline long long inv(long long a)
{
	if (gcd(a, p) == 1)
		return quick_pow(a, p-2, p);
	long long d, x, y;
	d = exgcd(a, p, x, y);
	return d == 1 ? (x + p) % p : -1;

}
inline void BSGS(long long a, long long b, long long p)
{
	int i;
	long long m = sqrt(p) + .5, v = inv(quick_pow(a,m,p)), e = 1;
	s.clear();
	s[1] = 0;
	for(i = 1;i < m; ++i)
		e = e * a % p, s[e] = i;
	for(i = 0;i <= m; ++i)
	{
		if(s.count(b))
		{
			printf("%lld\n", i * m + s[b]);
			return;
		}
		b = b * v % p;
	}
	puts("Orz, I cannot find x!");
}
inline void solve(long long type)
{
	switch (type)
	{
		case (1) :{
			printf("%d\n", quick_pow(y, z, p));
			break;
		}
		case (2) :{
			INV(y, z, p);
			break;
		}
		case (3) :{
			BSGS(y, z, p);
			break;
		}
	}
}
char c;
inline void read(long long &x)
{
	for (c = getchar();c > '9' || c < '0';c = getchar());
	for (x = 0;c >= '0' && c <= '9';c = getchar())
		x = (x << 3) + (x << 1) + c - '0';
}
int main()
{
	read(T), read(k);
	while (T--)
	{
		read(y), read(z), read(p);
		solve(k);
	}
}