【51NOD 1195】斐波那契数列的循环节





  1. 将模数质因数分解
  2. 对于每一个$p_i^m$我们计算其循环节$l_i$
  3. 将所有$l_i$取$lcm$

那么$\bmod p^m$的最短循环节为$G(p)\cdot p^{m-1}$

然后我们就可以通过从小到大枚举因数+$O(\log n)$判断的方法得到答案了

另外的话,这个算法似乎没有比较准确的复杂度。不过我们可以假装他是$O(\sqrt{n} \log n)$的
再另外的话,这份代码会$T$,但我又不想优化了,反正这个东西也不可能考 QwQ


#define ll long long
using namespace std;

inline int read() {
	char c=getchar(); int ret=0,f=1;
	while (c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
	while (c<='9'&&c>='0') {ret=ret*10+c-'0';c=getchar();}
	return ret*f;

class Fibonacci{
	int ans[4],tra[4],MOD;
		inline bool cal(int t, int mod) {
			ans[1] = tra[1] = tra[2] = tra[3] = 1;
			ans[0] = ans[2] = ans[3] = tra[0] = 0; MOD = mod;
			Pow(tra, tra, t - 2); Mul(ans, ans, tra);
			return ans[1] == 1 && !ans[0];
		inline void Pow(int *ans, int *a, int t) {
			static int ret[4],cur[4]; 
			ret[0]=ret[3]=1; ret[1]=ret[2]=0;
			for (;t;t>>=1,Mul(cur,cur,cur))
				if (t&1) Mul(ret,cur,ret);
			memcpy(ans, ret, sizeof(ret));
		inline void Mul(int *ans, int *a, int *b) {
			static int ret[4];
			ret[0] = ((ll)a[0] * b[0] + (ll)a[1] * b[2]) % MOD;
			ret[1] = ((ll)a[0] * b[1] + (ll)a[1] * b[3]) % MOD;
			ret[2] = ((ll)a[2] * b[0] + (ll)a[3] * b[2]) % MOD;
			ret[3] = ((ll)a[2] * b[1] + (ll)a[3] * b[3]) % MOD;
			memcpy(ans, ret, sizeof(ret));

ll GCD(ll a, ll b) {
	return b? GCD(b, a%b): a;

int Pow(int w, int t, int mod) {
	int ret = 1;
	for (;t;t>>=1,w=(ll)w*w%mod)
		if (t&1) ret=(ll)ret*w%mod;
	return ret;

inline ll G(ll p) {
	if (p == 2) return 3;
	else if (p == 3) return 8;
	else if (p == 5) return 20;
	static ll LIM = 0; stack<int> stk;
	if (Pow(5, (p-1)/2, p) == 1) LIM = p - 1;
	else LIM = p + 1 << 1; 
	for (int i=1;i*i<=LIM;i++) {
		if (LIM % i == 0) {
			if (fib.cal(i+2, p)) return i;
			else stk.push(LIM / i);
	for (int ret;!stk.empty();) {
		ret = stk.top(); stk.pop();
		if (fib.cal(ret+2, p)) return ret;

inline ll cal(int a, int b) {
	static ll INF = 4e18, ret;
	for (ret=G(a);b>1;b--) ret = ret * a;
	return ret;

inline ll solve(int mod) {
	static const int N = 1e5;
	static int pri[N],cnt[N],tot;
	if (mod == 1) return 1; tot = 0; 
	for (int i=2;i*i<=mod;i++) {
		if (mod % i == 0) {
			pri[++tot] = i; cnt[tot] = 0;
			for (;mod%i==0;mod/=i) ++cnt[tot];
	} if (mod>1) pri[++tot] = mod, cnt[tot]=1;
	ll ret = 1;
	for (int i=1,tmp;i<=tot;i++) {
		tmp = cal(pri[i], cnt[i]);
		ret = ret / GCD(ret, tmp) * tmp;
	} return ret;

int main() {
	for (int T=read();T;T--) 
	return 0;

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