acdream oj 1124 喵喵的遗憾 (斐波那契数列循环节)

Posted by 111qqz on Thursday, November 3, 2016

TOC

题目链接

题意:

F0 = 1 , F1 = 1 , F2 = 2 , Fn = Fn-1+Fn-2

求:

FFFn Mod P

( 也就是 F[ F[ F[n] ] ]  % P )

思路:原来这是适牛出的题2333.

需要注意的是p可能为1,因此n==0或者1的时候,特判要输出1%p而不是1.

其他的没了。就是求斐波那契数列循环节的经典做法。

/* ***********************************************
Author :111qqz
Created Time :Thu 03 Nov 2016 08:09:26 AM CST
File Name :1124.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;
        }
    }
}
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]==2) rec = 3;
        else if (pri[i]==3) rec = 8;
        else if (pri[i]==5) rec = 20;
        else
        {
            if (legendre(5,pri[i])==1)
                get_fac(pri[i]-1);
            else get_fac(2*pri[i]+2);
            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]==1&&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] = M.mat[1][1] = 1;
    M1.clear();
    M1.mat[0][0] = M1.mat[1][0] = 1;
}
LL get_fib(LL n,LL mod)
{
//    cout<<"n:"<<n<<" mod:"<<mod<<endl;
    if (mod==1) return 0;
    Mat ret;
    ret = mat_ksm(M,n-1,mod);
    ret = mul(ret,M1,mod);
    return (ret.mat[1][0])%mod;
}
LL n,P;
LL loop[N];
int main()
{
    #ifndef  ONLINE_JUDGE 
    freopen("code/in.txt","r",stdin);
  #endif
    int T;
    isprime();
    init();
    cin>>T;
    while (T--)
    {
        scanf("%lld%lld",&n,&P);
        if (n==0||n==1)
        {
        printf("%lld\n",1%P);//p可以等于1....
        continue;
        }
        loop[0]=P;

        LL cur = n ;
        for (int i = 1; i <= 2 ; i++) loop[i] = find_loop(loop[i-1]);
        //check
//	    for ( int i = 0  ;i  <= 2 ; i++) printf("loop:%lld\n",loop[i]);
        for ( int i  = 2 ; i >= 0 ; i--) {
        cur = get_fib(cur,loop[i]);
//		printf("cur:%lld\n",cur);
        }
        printf("%lld\n",cur);
    }
  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}

「真诚赞赏,手留余香」

111qqz的小窝

真诚赞赏,手留余香

使用微信扫描二维码完成支付