# 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

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。

``````/* ***********************************************
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;
}
``````

Acdreamer的博客_广义Fibonacci数列找循环节

``````/* ***********************************************
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};
{
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;
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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