广义Fibonacci数列找循环节 (二次剩余)
**问题:**给定
,满足
,求
的循
环节长度。
这里只说做法
我们先写出递推式的特征式子 x^2 =ax + b,整理得到 x^2-ax-b=0求出 delta = a^2+4b
对于质因数小于等于delta的部分,我们选择暴力求循环节。
暴力求循环节的代码如下:
1int get_loop( LL p) //暴力得到不大于13的素数的循环节。
2{
3 LL a,b,c;
4 a = 0 ;
5 b = 1 ;
6 for ( int i = 2; ; i++)
7 {
8 c = (a+3*b%p)%p; //此处为递推式
9 a = b;
10 b = c;
11 if (a==0&&b==1) return i-1;
12 }
13}
通常做法是先暴力求出来之后,写进一个表里。
通常不会有很多项。
对于质因数大于delta的部分,我们判断每个质因数是否是delta的二次剩余,如果是,枚举(prime-1)的因子,否则枚举2*(prime+1)的因子。
贴一个板子,需要注意如果是多个函数嵌套的形式,我们只需要从外向里,依次求循环节即可。
/* ***********************************************
Author :111qqz
Created Time :Mon 31 Oct 2016 08:22:17 PM CST
File Name :code/hdu/4291_2.cpp
************************************************ */
1#include <cstdio>
2#include <cstring>
3#include <iostream>
4#include <algorithm>
5#include <vector>
6#include <queue>
7#include <set>
8#include <map>
9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6struct Mat
7{
8 LL mat[2][2];
9 void clear()
10 {
11 ms(mat,0);
12 }
13}M,M1;
14Mat mul (Mat a,Mat b,LL mod)
15{
16 Mat c;
17 c.clear();
18 for ( int i = 0 ; i < 2 ; i++)
19 for ( int j = 0 ; j < 2 ; j++)
20 for ( int k = 0 ; k < 2 ; k++)
21 c.mat[i][j] = (c.mat[i][j] + a.mat[i][k] * b.mat[k][j]%mod)%mod;
22 return c;
23}
24Mat mat_ksm(Mat a,LL b,LL mod)
25{
26 Mat res;
27 res.clear();
28 for ( int i = 0 ; i < 2 ; i++) res.mat[i][i] = 1;
29 while (b>0)
30 {
31 if (b&1) res = mul(res,a,mod);
32 b = b >> 1LL;
33 a = mul(a,a,mod);
34 }
35 return res;
36}
37LL gcd(LL a,LL b)
38{
39 return b?gcd(b,a%b):a;
40}
41const int N = 1E6+7;
42bool prime[N];
43int p[N];
44void isprime() //一个普通的筛
45{
46 int cnt = 0 ;
47 ms(prime,true);
48 for ( int i = 2 ; i < N ; i++)
49 {
50 if (prime[i])
51 {
52 p[cnt++] = i ;
53 for ( int j = i+i ; j < N ; j+=i) prime[j] = false;
54 }
55 }
56}
57LL ksm( LL a,LL b,LL mod)
58{
59 LL res = 1;
60 while (b>0)
61 {
62 if (b&1) res = (res * a) % mod;
63 b = b >> 1LL;
64 a = a * a % mod;
65 }
66 return res;
67}
68LL legendre(LL a,LL p) //勒让德符号,判断二次剩余
69{
70 if (ksm(a,(p-1)>>1,p)==1) return 1;
71 return -1;
72}
73LL pri[N],num[N];//分解质因数的底数和指数。
74int cnt; //质因子的个数
75void solve(LL n,LL pri[],LL num[])
76{
77 cnt = 0 ;
78 for ( int i = 0 ; p[i] * p[i] <= n ; i++)
79 {
80 if (n%p[i]==0)
81 {
82 int Num = 0 ;
83 pri[cnt] = p[i];
84 while (n%p[i]==0)
85 {
86 Num++;
87 n/=p[i];
88 }
89 num[cnt] = Num;
90 cnt++;
91 }
92 }
93 if (n>1)
94 {
95 pri[cnt] = n;
96 num[cnt] = 1;
97 cnt++;
98 }
99}
100LL fac[N];
101int cnt2; //n的因子的个数
102void get_fac(LL n)//得到n的所有因子
103{
104 cnt2 = 0 ;
105 for (int i = 1 ; i*i <= n ; i++)
106 {
107 if (n%i==0)
108 {
109 if (i*i!=n)
110 {
111 fac[cnt2++] = i ;
112 fac[cnt2++] = n/i;
113 }
114 else fac[cnt2++] = i;
115 }
116 }
117}
118int get_loop( LL p) //暴力得到不大于13的素数的循环节。
119{
120 LL a,b,c;
121 a = 0 ;
122 b = 1 ;
123 for ( int i = 2; ; i++)
124 {
125 c = (a+3*b%p)%p;
126 a = b;
127 b = c;
128 if (a==0&&b==1) return i-1;
129 }
130}
131/*
132 2->3
133 3->2
134 5->12
135 7->16
136 11->8
137 13->52
138 */
139const LL LOOP[10]={3,2,12,16,8,52};
140LL ask_loop(int id)
141{
142 return LOOP[id];
143}
144LL find_loop(LL n)
145{
146 solve(n,pri,num);
147 LL ans = 1;
148 for ( int i = 0 ; i < cnt ; i++)
149 {
150 LL rec = 1;
151 if (pri[i]<=13) rec = ask_loop(i);
152 else
153 {
154 if (legendre(13,pri[i])==1) //13就是delta值
155 get_fac(pri[i]-1);
156 else get_fac((pri[i]+1)*(3-1)); //为什么可以假设循环节不大于2*(p+1)???
157 sort(fac,fac+cnt2);
158 for ( int j = 0 ; j < cnt2 ; j++) //挨个验证因子
159 {
160 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
161 tmp = mul(tmp,M1,pri[i]);
162 if (tmp.mat[0][0]==0&&tmp.mat[1][0]==1)
163 {
164 rec = fac[j];
165 break;
166 }
167 }
1 }
2 for ( int j = 1 ; j < num[i] ; j++)
3 rec *=pri[i];
4 ans = ans / gcd(ans,rec) * rec;
5 }
6 return ans;
7}
8void init()
9{
10 M.clear();
11 M.mat[0][1] = M.mat[1][0] = 1;
12 M.mat[1][1] = 3;
13 M1.clear();
14 M1.mat[1][0] = 1;
15}
16LL n;
17LL loop0 = 1E9+7;
18LL loop1,loop2;
19int main()
20{
21 #ifndef ONLINE_JUDGE
22 freopen("code/in.txt","r",stdin);
23 #endif
24 /*
25 printf("%d\n",get_loop(2));
26 printf("%d\n",get_loop(3));
27 printf("%d\n",get_loop(5));
28 printf("%d\n",get_loop(7));
29 printf("%d\n",get_loop(11));
30 printf("%d\n",get_loop(13));
31 */
32 init();
33 isprime();
34 while (~scanf("%lld\n",&n))
35 {
36 if (n==0||n==1)
37 {
38 printf("%lld\n",n);
39 continue;
40 }
41 LL loop1 = find_loop(loop0);
42 LL loop2 = find_loop(loop1);
43// printf("loop1:%lld loop2:%lld\n",loop1,loop2);
44 LL cur = n;
45 Mat ans = mat_ksm(M,cur-1,loop2);
46 ans = mul(ans,M1,loop2);
47 cur = ans.mat[1][0];
48 if (cur!=0&&cur!=1)
49 {
50 Mat ans = mat_ksm(M,cur-1,loop1);
51 ans = mul(ans,M1,loop1);
52 cur = ans.mat[1][0];
53 }
54 if (cur!=0&&cur!=1)
55 {
56 Mat ans = mat_ksm(M,cur-1,loop0);
57 ans = mul(ans,M1,loop0);
58 cur = ans.mat[1][0];
59 }
60 printf("%lld\n",cur);
}
1#ifndef ONLINE_JUDGE
2 fclose(stdin);
3#endif
4 return 0;
5}
再补一个,更为常用的,求斐波那契循环节的板子:
/* ***********************************************
Author :111qqz
Created Time :Mon 31 Oct 2016 03:54:15 AM CST
File Name :code/hdu/3977.cpp
************************************************ */
1#include <cstdio>
2#include <cstring>
3#include <iostream>
4#include <algorithm>
5#include <vector>
6#include <queue>
7#include <set>
8#include <map>
9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6LL P;
7struct Mat
8{
9 LL mat[2][2];
10 void clear()
11 {
12 ms(mat,0);
13 }
14}M,M1;
15Mat mul (Mat a,Mat b,LL mod)
16{
17 Mat c;
18 c.clear();
19 for ( int i = 0 ; i < 2 ; i++)
20 for ( int j = 0 ; j < 2 ; j++)
21 for ( int k = 0 ; k < 2 ; k++)
22 c.mat[i][j] = (c.mat[i][j] + a.mat[i][k] * b.mat[k][j]%mod)%mod;
23 return c;
24}
25Mat mat_ksm(Mat a,LL b,LL mod)
26{
27 Mat res;
28 res.clear();
29 for ( int i = 0 ; i < 2 ; i++) res.mat[i][i] = 1;
30 while (b>0)
31 {
32 if (b&1) res = mul(res,a,mod);
33 b = b >> 1LL;
34 a = mul(a,a,mod);
35 }
36 return res;
37}
38LL gcd(LL a,LL b)
39{
40 return b?gcd(b,a%b):a;
41}
42const int N = 1E6+7;
43bool prime[N];
44int p[N];
45void isprime() //一个普通的筛
46{
47 int cnt = 0 ;
48 ms(prime,true);
49 for ( int i = 2 ; i < N ; i++)
50 {
51 if (prime[i])
52 {
53 p[cnt++] = i ;
54 for ( int j = i+i ; j < N ; j+=i) prime[j] = false;
55 }
56 }
57}
58LL ksm( LL a,LL b,LL mod)
59{
60 LL res = 1;
61 while (b>0)
62 {
63 if (b&1) res = (res * a) % mod;
64 b = b >> 1LL;
65 a = a * a % mod;
66 }
67 return res;
68}
69LL legendre(LL a,LL p) //勒让德符号,判断二次剩余
70{
71 if (ksm(a,(p-1)>>1,p)==1) return 1;
72 return -1;
73}
74LL pri[N],num[N];//分解质因数的底数和指数。
75int cnt; //质因子的个数
76void solve(LL n,LL pri[],LL num[])
77{
78 cnt = 0 ;
79 for ( int i = 0 ; p[i] * p[i] <= n ; i++)
80 {
81 if (n%p[i]==0)
82 {
83 int Num = 0 ;
84 pri[cnt] = p[i];
85 while (n%p[i]==0)
86 {
87 Num++;
88 n/=p[i];
89 }
90 num[cnt] = Num;
91 cnt++;
92 }
93 }
94 if (n>1)
95 {
96 pri[cnt] = n;
97 num[cnt] = 1;
98 cnt++;
99 }
100}
101LL fac[N];
102int cnt2; //n的因子的个数
103void get_fac(LL n)//得到n的所有因子
104{
105 cnt2 = 0 ;
106 for (int i = 1 ; i*i <= n ; i++)
107 {
108 if (n%i==0)
109 {
110 if (i*i!=n)
111 {
112 fac[cnt2++] = i ;
113 fac[cnt2++] = n/i;
114 }
115 else fac[cnt2++] = i;
116 }
117 }
118}
119LL find_loop(LL n)
120{
121 solve(n,pri,num);
122 LL ans = 1;
123 for ( int i = 0 ; i < cnt ; i++)
124 {
125 LL rec = 1;
126 if (pri[i]==2) rec = 3;
127 else if (pri[i]==3) rec = 8;
128 else if (pri[i]==5) rec = 20;
129 else
130 {
131 if (legendre(5,pri[i])==1)
132 get_fac(pri[i]-1);
133 else get_fac(2*pri[i]+2);
134 sort(fac,fac+cnt2);
135 for ( int j = 0 ; j < cnt2 ; j++) //挨个验证因子
136 {
137 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
138 tmp = mul(tmp,M1,pri[i]);
139 if (tmp.mat[0][0]==1&&tmp.mat[1][0]==1)
140 {
141 rec = fac[j];
142 break;
143 }
144 }
1 }
2 for ( int j = 1 ; j < num[i] ; j++)
3 rec *=pri[i];
4 ans = ans / gcd(ans,rec) * rec;
5 }
6 return ans;
7}
8void init()
9{
10 M.clear();
11 M.mat[0][1] = M.mat[1][0] = M.mat[1][1] = 1;
12 M1.clear();
13 M1.mat[0][0] = M1.mat[1][0] = 1;
14}
15int main()
16{
17 #ifndef ONLINE_JUDGE
18 freopen("code/in.txt","r",stdin);
19 #endif
20 int T;
21 int cas = 0 ;
22 isprime();
23 scanf("%d",&T);
24 while (T--)
25 {
26 init();
27 scanf("%lld",&P);
28 printf("Case #%d: ",++cas);
29 LL ret = find_loop(P);
30 printf("%lld\n",ret);
31 }
1 #ifndef ONLINE_JUDGE
2 fclose(stdin);
3 #endif
4 return 0;
5}