hdu 3221 Brute-force Algorithm

題目：hdu 3221 Brute-force Algorithm

題意：很簡單，枚舉幾項就知道了

狀態不好，sb了，錯了好多遍

#include <cstdio>
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
typedef __int64 LL;
LL mod,p,fib[60];
struct Matrix
{
	LL m[3][3];
}E,D;
void init()
{
	for (int i = 1; i <= 2; i++)
	{
		for (int j = 1; j <= 2; j++)
		{
			E.m[i][j] = (i == j);
			D.m[i][j] = 1;
		}
	}
	D.m[2][2] = 0;
}
Matrix Multi(Matrix A, Matrix B)
{
	Matrix ans;
	for (int i = 1; i <= 2; i++)
	{
		for (int j = 1; j <= 2; j++)
		{
			ans.m[i][j] = 0;
			for (int k = 1; k <= 2; k++)
				ans.m[i][j] = (ans.m[i][j] + A.m[i][k] * B.m[k][j])%mod;
		}
	}
	return ans;
}
Matrix Pow(Matrix A, LL k)
{
	Matrix ans = E;
	while (k)
	{
		if (k & 1)
		{
			k--;
			ans = Multi(ans, A);
		}
		else
		{
			k /= 2;
			A = Multi(A, A);
		}
	}
	return ans;
}
LL euler(LL x)
{
	LL i, res = x;
	for (i = 2; i*i <= x; i++)
	{
		if (x%i == 0)
		{
			res = res / i*(i - 1);
			while (x%i == 0)
				x /= i;
		}
	}
	if (x > 1)
		res = res / x*(x - 1);
	return res;
}
LL Pow(LL a, LL b)
{
	LL ans = 1;
	while (b)
	{
		if (b & 1)
		{
			b--;
			ans = (ans*a)%p;
		}
		else
		{
			b /= 2;
			a = (a*a)%p;
		}
	}
	return ans;
}
LL Fib(int n)
{
	if (n == 0)
		return 0;
	if (n == 1)
		return 1;
	fib[0] = 0;
	fib[1] = 1;
	int i;
	for (i = 2; i <= n; i++)
	{
		fib[i] = fib[i - 1] + fib[i - 2];
		if (fib[i] >= mod)
			break;
	}
	if (i > n)
		return fib[n];
	return Pow(D, n).m[2][1] + mod;
}
LL solve(LL a,LL b,LL n)
{
	if (n == 1)
		return a%p;
	if (n == 2)
		return b%p;
	return Pow(a, Fib(n - 2))*Pow(b, Fib(n - 1))%p;
}
int main()
{
	int T;
	scanf("%d", &T);
	for (int cas = 1; cas <= T; cas++)
	{
		LL a, b, n;
		scanf("%I64d%I64d%I64d%I64d", &a, &b, &p, &n);
		mod = euler(p);
		init();
		printf("Case #%d: %I64d\n", cas, solve(a, b, n));
	}
	return 0;
}