OJ-Problems-Source/HDOJ/4794_autoAC.cpp

#include <iostream>
#include <vector>
#include <algorithm>
#include <string.h>
#include <cstring>
#include <stdio.h>
#include <cmath>
#include <math.h>
#define rep(i,a,b) for(int i=(a);i<(b);++i)
#define rrep(i,b,a) for(int i = (b); i >= (a); --i)
#define clr(a,x) memset(a,(x),sizeof(a))
#define LL unsigned long long
#define eps 1e-8
using namespace std;
LL n;
LL gcd(LL a, LL b)
{
    while (a && b) {
        if (a > b) a %= b;
        else b %= a;
    }
    return a + b;
}
LL lcm(LL a,LL b)
{
    return a * b / gcd(a,b);
}
void mul(LL A[2][2],LL B[2][2],LL ret[2][2],LL mod)
{
    LL C[2][2] = { 0 };
    rep(i,0,2) rep(j,0,2) rep(k,0,2)
        C[i][j] = (C[i][j] + A[i][k] * B[k][j] % mod) % mod;
    rep(i,0,2) rep(j,0,2)
        ret[i][j] = C[i][j];
}
void qpow(LL base[2][2],LL p,LL dest[2][2],LL mod)
{
    LL ret[2][2] = { 0 };
    ret[0][0] = ret[1][1] = 1;
    while (p > 0) {
        if (p & 1) mul(ret,base,ret,mod);
        mul(base,base,base,mod);
        p >>= 1;
    }
    rep(i,0,2) rep(j,0,2) dest[i][j] = ret[i][j];
}
LL qpow(LL base,LL p,LL mod)
{
    LL ret = 1;
    while (p) {
        if (p & 1) ret = ret * base % mod;
        base = base * base % mod;
        p >>= 1;
    }
    return ret;
}
LL S[10000],c;
LL f1,f2;
void F(LL p,LL mod)
{
    LL A[2][2] = { 0 };
    A[0][0] = 1; A[0][1] = 1;
    A[1][0] = 1; A[1][1] = 0;
    qpow(A,p-1,A,mod);
    f1 = (A[1][0]+A[1][1]) % mod; f2 = (A[0][0] + A[0][1]) % mod;
}
LL loop(LL mod)
{
    if (mod == 2) return 3;
    else if (mod == 3) return 8;
    else if (mod == 5) return 20;
    LL p;
    if (qpow(5,(mod-1)>>1,mod) == 1) p = mod - 1;
    else p = 2*(mod + 1);
    c = 0;
    for(LL i = 1; i * i <= p; ++i) if (p % i == 0) {
        LL x = i , y = p / i;
        F(x,mod);
        if (f1 == 0 && f2 == 1) return x;
        if (y != x) S[c++] = y;
    }
    while (c > 0) {
        F(S[--c],mod);
        if (f1 == 0 && f2 == 1) return S[c];
    }
    return 0;
}
int main()
{
    while (cin >> n) {
        LL x = n;
        LL ans = 1;
        for(LL i = 2; i * i <= x; ++i) if (x % i == 0) {
            LL len = loop(i);
            LL S = 1;
            do x /= i, S = S * i;
            while (x % i == 0);
            S /= i; S = S * len;
            ans = lcm(ans,S);
        }
        if (x > 1) {
            LL len = loop(x);
            ans = lcm(ans,len);
        }
        if (ans % 2 == 0) ans /= 2;
        printf("%I64u\n",ans);
    }
}