hdu 4291 A Short problem (矩阵快速幂+广义斐波那契循环节||暴力找循环节)

Posted by 111qqz on Monday, October 31, 2016

TOC

题目链接

题意:

Given n (1 <= n <= 1018), You should solve for

g(g(g(n))) mod 109 + 7
where

g(n) = 3g(n - 1) + g(n - 2)

g(1) = 1

g(0) = 0思路:找循环节。首先由于模数固定,可以暴力一下找到循环节。

得到1E9+7的循环节是222222224,222222224的循环节是183120.

然后三次矩阵快速幂就行了。

需要注意每次都要判断那一层的n是否为0和1。

暴力解法:

/* ***********************************************
Author :111qqz
Created Time :Mon 31 Oct 2016 02:47:01 PM CST
File Name :code/hdu/4291.cpp
************************************************ */

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair

using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const LL loop0 = 1E9+7;
const LL loop1=222222224,loop2=183120; //暴力得到每一层的循环节。
/*
LL find_loop(LL mod)
{
    LL a,b,c;
    a = 0 ;
    b = 1 ;
    for ( int i = 2 ; ; i++)
    {
    c = (a+3*b%mod)% mod;
    a = b;
    b = c; 
//	cout<<"a:"<<a<<" b:"<<b<<endl;
    if (a==0&&b==1)
    {
        return i-1; // i的时候得到第i项,判断循环的时候是判断g[i-1]==g[0]&&g[i]==g[1],所以循环节长度是i-1;
    }
    }
}*/
LL n;
struct Mat
{
    LL mat[2][2];
    void clear()
    {
    ms(mat,0);
    }
    void out()
    {
    cout<<mat[0][0]<<" "<<mat[0][1]<<endl;
    cout<<mat[1][0]<<" "<<mat[1][1]<<endl;
    cout<<endl;
    }
}M,M1;
Mat mul(Mat a,Mat b,LL MOD)
{
    Mat c;
    c.clear();
    for ( int i = 0 ; i < 2 ; i++)
    for  ( int j = 0 ; j < 2 ; j++)
        for ( int k = 0 ; k < 2 ; k++)
        c.mat[i][j] = (c.mat[i][j] + a.mat[i][k]*b.mat[k][j]%MOD)%MOD;
    return c;
}
Mat power(Mat a,LL b,LL MOD)
{
    Mat res;
    res.clear();
    for ( int i = 0 ;i < 2 ; i++) res.mat[i][i] = 1;
    while (b>0)
    {
    if (b&1) res = mul(res,a,MOD);
    b = b>>1LL;
    a = mul(a,a,MOD);
    }
    return res;
}
void Mat_init()
{
    M.clear();
    M.mat[0][1] = 1;
    M.mat[1][0] = 1;
    M.mat[1][1] = 3;
    M1.clear();
    M1.mat[1][0] = 1;
}
int main()
{
    #ifndef  ONLINE_JUDGE 
    freopen("code/in.txt","r",stdin);
  #endif
    /*
    loop1 = find_loop(loop0);
    printf("loop1:%lld\n",loop1);
    loop2 = find_loop(loop1);
    printf("loop2:%lld\n",loop2);
    */
    Mat_init();
    while (scanf("%lld\n",&n)!=EOF)
    {
        if (n==0)
        {
        puts("0");
        continue;
        }
        if (n==1)
        {
        puts("1");
        continue;
        }

        LL cur = n;
        Mat ans;
        ans = power(M,cur-1,loop2);
        ans = mul(ans,M1,loop2);
        cur = ans.mat[1][0];
        if (cur!=0&&cur!=1) //三次快速幂,每次都要记得判断初始项。
        {

        ans = power(M,cur-1,loop1);
        ans = mul(ans,M1,loop1);
        cur = ans.mat[1][0];
        }
        if (cur!=0&&cur!=1)
        {
        ans = power(M,cur-1,loop0);
        ans = mul(ans,M1,loop0);
        cur = ans.mat[1][0];
        }
        printf("%lld\n",cur);
    }


#ifndef ONLINE_JUDGE  
    fclose(stdin);
#endif
    return 0;
}

再来一个比较一般的做法:

参考递推式循环节

Acdreamer的博客_广义Fibonacci数列找循环节

摘重点:

想了想,5对于斐波那契数列来讲,不就是x^2=x+1的delta么,那么这题的递推式是x^2=3x+1,delta=33+4=13,然后我就把勒让德符号判断二次剩余那里改成13,然后对应的暴力出13及13以内的素数对应的循环节,交了一发,AC了


是模
的二次剩余时,枚举
的因子


是模
的二次非剩余时,枚举
的因子

对于第二种非剩余的情况,理论上是枚举(p+1)(p-1)的因子,实际上常常只是枚举2(p+1)的因子*

对于c=5的情况,是有定理:

screenshot-from-2016-10-31-21-23-45

不过对于c不等于5的情况。。。该结论是否一定成立呢…感觉不是很好证,求指教。

/* ***********************************************
Author :111qqz
Created Time :Mon 31 Oct 2016 08:22:17 PM CST
File Name :code/hdu/4291_2.cpp
************************************************ */

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair

using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
struct Mat
{
    LL mat[2][2];
    void clear()
    {
    ms(mat,0);
    }
}M,M1;
Mat mul (Mat a,Mat b,LL mod)
{
    Mat c;
    c.clear();
    for ( int i = 0 ; i < 2 ; i++)
    for ( int j = 0 ; j < 2 ; j++)
        for ( int k  = 0 ; k < 2 ; k++)
        c.mat[i][j] = (c.mat[i][j] + a.mat[i][k] * b.mat[k][j]%mod)%mod;
    return c;
}
Mat mat_ksm(Mat a,LL b,LL mod)
{
    Mat res;
    res.clear();
    for ( int i = 0 ; i < 2 ; i++) res.mat[i][i] = 1;
    while (b>0)
    {
    if (b&1) res = mul(res,a,mod);
    b = b >> 1LL;
    a = mul(a,a,mod);
    }
    return res;
}
LL gcd(LL a,LL b)
{
    return b?gcd(b,a%b):a;
}
const int N = 1E6+7;
bool prime[N];
int p[N];
void isprime() //一个普通的筛
{
    int cnt = 0 ;
    ms(prime,true);
    for ( int i = 2 ; i < N ; i++)
    {
    if (prime[i])
    {
        p[cnt++] = i ;
        for ( int j = i+i ; j < N ; j+=i) prime[j] = false;
    }
    }
}
LL ksm( LL a,LL b,LL mod)
{
   LL res = 1;
   while (b>0)
   {
       if (b&1) res = (res * a) % mod;
       b = b >> 1LL;
       a = a * a % mod;
   }
   return res;
}
LL legendre(LL a,LL p) //勒让德符号,判断二次剩余
{
    if (ksm(a,(p-1)>>1,p)==1) return 1;
    return -1;
}
LL pri[N],num[N];//分解质因数的底数和指数。
int cnt; //质因子的个数
void solve(LL n,LL pri[],LL num[])
{
    cnt = 0 ;
    for ( int  i = 0 ; p[i] * p[i] <= n ; i++)
    {
    if (n%p[i]==0)
    {
        int Num = 0 ;
        pri[cnt] = p[i];
        while (n%p[i]==0)
        {
        Num++;
        n/=p[i];
        }
        num[cnt] = Num;
        cnt++;
    }
    }
    if (n>1)
    {
    pri[cnt] = n;
    num[cnt] = 1;
    cnt++;
    }
}
LL fac[N];
int cnt2; //n的因子的个数
void get_fac(LL n)//得到n的所有因子
{
    cnt2 = 0 ;
    for (int i =  1 ; i*i <= n ; i++)
    {
    if (n%i==0)
    {
        if (i*i!=n)
        {
        fac[cnt2++] = i ;
        fac[cnt2++] = n/i;
        }
        else fac[cnt2++] = i;
    }
    }
}
int get_loop( LL p) //暴力得到不大于13的素数的循环节。
{		    
    LL a,b,c;
    a = 0 ;
    b = 1 ;
    for ( int i = 2; ; i++)
    {
    c = (a+3*b%p)%p;
    a = b;
    b = c;
    if (a==0&&b==1) return i-1;
    }
}
/*
    2->3
    3->2
    5->12
    7->16
    11->8
    13->52
    */
const LL LOOP[10]={3,2,12,16,8,52};
LL ask_loop(int id)
{
    return LOOP[id];
}
LL find_loop(LL n)
{
    solve(n,pri,num);
    LL ans = 1;
    for ( int i = 0 ; i < cnt ; i++)
    {
    LL rec = 1;
    if (pri[i]<=13) rec = ask_loop(i);
    else
    {
        if (legendre(13,pri[i])==1)
        get_fac(pri[i]-1);
        else get_fac((pri[i]+1)*(3-1)); //为什么可以假设循环节不大于2*(p+1)???
        sort(fac,fac+cnt2);
        for ( int j = 0 ; j < cnt2 ; j++) //挨个验证因子
        {
        Mat tmp = mat_ksm(M,fac[j],pri[i]); //下标从0开始,验证fac[j]为循环节,应该看fib[0]==fib[fac[j]]和fib[1]==fib[fac[j]+1]是否成立
        tmp = mul(tmp,M1,pri[i]);
        if (tmp.mat[0][0]==0&&tmp.mat[1][0]==1)
        {
            rec = fac[j];
            break;
        }
        }

    }
    for ( int j = 1 ; j < num[i] ; j++)
        rec *=pri[i];
    ans = ans / gcd(ans,rec) * rec;
    }
    return ans;
}
void init()
{
    M.clear();
    M.mat[0][1] = M.mat[1][0] = 1;
    M.mat[1][1] = 3;
    M1.clear();
    M1.mat[1][0] = 1;
}
LL n;
LL loop0 = 1E9+7;
LL loop1,loop2;
int main()
{
    #ifndef  ONLINE_JUDGE 
    freopen("code/in.txt","r",stdin);
  #endif
    /*
    printf("%d\n",get_loop(2));
    printf("%d\n",get_loop(3));
    printf("%d\n",get_loop(5));
    printf("%d\n",get_loop(7));
    printf("%d\n",get_loop(11));
    printf("%d\n",get_loop(13));
    */
    init();
    isprime();
    while (~scanf("%lld\n",&n))
    {
        if (n==0||n==1)
        {
        printf("%lld\n",n);
        continue;
        }
        LL loop1 = find_loop(loop0);
        LL loop2 = find_loop(loop1);
//	    printf("loop1:%lld loop2:%lld\n",loop1,loop2);
        LL cur = n;
        Mat ans = mat_ksm(M,cur-1,loop2);
        ans = mul(ans,M1,loop2);
        cur = ans.mat[1][0];
        if (cur!=0&&cur!=1)
        {
        Mat ans = mat_ksm(M,cur-1,loop1);
        ans = mul(ans,M1,loop1);
        cur = ans.mat[1][0];
        }
        if (cur!=0&&cur!=1)
        {
        Mat ans = mat_ksm(M,cur-1,loop0);
        ans = mul(ans,M1,loop0);
        cur = ans.mat[1][0];
        }
        printf("%lld\n",cur);

    }

#ifndef ONLINE_JUDGE  
    fclose(stdin);
#endif
    return 0;
}

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