acdream oj 1124 喵喵的遗憾 (斐波那契数列循环节)
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;
}