HDU 3977 斐波那契循环节

这类型的题目其实没什么意思..知道怎么做后，就有固定套路了..而且感觉这东西要出的很难的话，有这种方法解常数会比较大吧..所以一般最多套一些比较简单的直接可以暴力求循环节的题目了..

/** @Date    : 2017-09-26 16:37:05
  * @FileName: HDU 3977 斐波那契循环节.cpp
  * @Platform: Windows
  * @Author  : Lweleth (SoungEarlf@gmail.com)
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <bits/stdc++.h>
#define LL long long
#define PII pair<int ,int>
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const double eps = 1e-8;

/////////
LL mul(LL x, LL y, LL mod)
{
    return (x * y - (LL)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

struct Matrix
{
    LL m[2][2];
};

Matrix A;
Matrix I = {1, 0, 0, 1};

Matrix multi(Matrix a, Matrix b, LL MOD)
{
    Matrix c;
    for(int i = 0; i < 2; i++)
    {
        for(int j = 0; j < 2; j++)
        {
            c.m[i][j] = 0;
            for(int k = 0; k < 2; k++)
                c.m[i][j] = (c.m[i][j] % MOD + (a.m[i][k] % MOD) * (b.m[k][j] % MOD) % MOD) % MOD;
            c.m[i][j] %= MOD;
        }
    }
    return c;
}

Matrix power(Matrix a, LL k, LL MOD)
{
    Matrix ans = I, p = a;
    while(k)
    {
        if(k & 1)
        {
            ans = multi(ans, p, MOD);
            k--;
        }
        k >>= 1;
        p = multi(p, p, MOD);
    }
    return ans;
}

LL gcd(LL a, LL b)
{
    return b ? gcd(b, a % b) : a;
}

const int N = 400005;
const int NN = 5005;

LL num[NN], pri[NN];
LL fac[NN];
int cnt, c;

bool prime[N];
int p[N];
int k;

void isprime()
{
    k = 0;
    memset(prime, true, sizeof(prime));
    for(int i = 2; i < N; i++)
    {
        if(prime[i])
        {
            p[k++] = i;
            for(int j = i + i; j < N; j += i)
                prime[j] = false;
        }
    }
}

LL fpow(LL a, LL b, LL m)
{
    LL ans = 1;
    a %= m;
    while(b)
    {
        if(b & 1)
        {
            ans = mul(ans,  a , m);
            b--;
        }
        b >>= 1;
        a = mul(a , a , m);
    }
    return ans;
}

LL legendre(LL a, LL p)
{
    if(fpow(a, (p - 1) >> 1, p) == 1)
        return 1;
    else return -1;
}

void Solve(LL n, LL pri[], LL num[])
{
    cnt = 0;
    LL t = (LL)sqrt(1.0 * n);
    for(int i = 0; p[i] <= t; i++)
    {
        if(n % p[i] == 0)
        {
            int a = 0;
            pri[cnt] = p[i];
            while(n % p[i] == 0)
            {
                a++;
                n /= p[i];
            }
            num[cnt] = a;
            cnt++;
        }
    }
    if(n > 1)
    {
        pri[cnt] = n;
        num[cnt] = 1;
        cnt++;
    }
}

void Work(LL n)
{
    c = 0;
    LL t = (LL)sqrt(1.0 * n);
    for(int i = 1; i <= t; i++)
    {
        if(n % i == 0)
        {
            if(i * i == n) fac[c++] = i;
            else
            {
                fac[c++] = i;
                fac[c++] = n / i;
            }
        }
    }
}

LL get_loop(LL n)
{
    Solve(n, pri, num);
    LL ans = 1;
    for(int i = 0; i < cnt; i++)
    {
        LL record = 1;
        if(pri[i] == 2)
            record = 3;
        else if(pri[i] == 3)
            record = 8;
        else if(pri[i] == 5)
            record = 20;
        else
        {
            if(legendre(5, pri[i]) == 1)
                Work(pri[i] - 1);
            else
                Work(2 * (pri[i] + 1));
            sort(fac, fac + c);
            for(int k = 0; k < c; k++)
            {
                Matrix a = power(A, fac[k] - 1, pri[i]);
                LL x = (a.m[0][0] % pri[i] + a.m[0][1] % pri[i]) % pri[i];
                LL y = (a.m[1][0] % pri[i] + a.m[1][1] % pri[i]) % pri[i];
                if(x == 1 && y == 0)
                {
                    record = fac[k];
                    break;
                }
            }
        }
        for(int k = 1; k < num[i]; k++)
            record *= pri[i];
        ans = ans / gcd(ans, record) * record;
    }
    return ans;
}

LL fib[5005];

void Init()
{
    A.m[0][0] = 1;
    A.m[0][1] = 1;
    A.m[1][0] = 1;
    A.m[1][1] = 0;
    fib[0] = 0;
    fib[1] = 1;
    for(int i = 2; i < 5005; i++)
        fib[i] = fib[i - 1] + fib[i - 2];
}
//////////
int main()
{
	Init();
	isprime();
	int T;
	cin >> T;
	int icas = 0;
	while(T--)
	{
		LL n;
		cin >> n;
		LL ans = get_loop(n);
		printf("Case #%d: %lld
", ++icas, ans);
	}
    return 0;
}