洛谷P5110 块速递推 [分块]

传送门

---

思路

显然可以特征根方程搞一波（生成函数太累），得到结果：

\[a_n=\frac 1 {13\sqrt{337}} [(\frac{233+13\sqrt{337}}{2})^n-(\frac{233+13\sqrt{337}}{2})^n] \]

（其实我也不知道是不是，网上抄的，懒得算了）

放在模意义下，得到

\[a_n= 233230706\times (94153035^n-905847205^n) \pmod {1e9+7} \]

后面两个可以分块，预处理出\(x^{[1,\sqrt{{mod}}]}\)，再处理出\(x^{\sqrt{mod}\times[1,\sqrt{mod}]}\)，就可以\(O(1)\)得到\(x^n\)了。

---

代码

#include<bits/stdc++.h>
clock_t t=clock();
namespace my_std{
	using namespace std;
	#define pii pair<int,int>
	#define fir first
	#define sec second
	#define MP make_pair
	#define rep(i,x,y) for (int i=(x);i<=(y);i++)
	#define drep(i,x,y) for (int i=(x);i>=(y);i--)
	#define go(x) for (int i=head[x];i;i=edge[i].nxt)
	#define BASE 32768
	#define mod 1000000007
	#define templ template<typename T>
	typedef long long ll;
	typedef unsigned long long ull; 
	typedef double db;
	mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}
	templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}
	templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}
	templ inline void read(T& t)
	{
		t=0;char f=0,ch=getchar();double d=0.1;
		while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();
		while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();
		if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}
		t=(f?-t:t);
	}
	template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}
	char sr[1<<21],z[20];int C=-1,Z=0;
    inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
    inline void print(register int x)
    {
    	if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;
    	while(z[++Z]=x%10+48,x/=10);
    	while(sr[++C]=z[Z],--Z);sr[++C]='\n';
    }
	void file()
	{
		#ifndef ONLINE_JUDGE
		freopen("a.in","r",stdin);
		#endif
	}
	inline void chktime()
	{
		#ifndef ONLINE_JUDGE
		cout<<(clock()-t)/1000.0<<'\n';
		#endif
	}
	#ifdef mod
	ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}
	ll inv(ll x){return ksm(x,mod-2);}
	#else
	ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}
	#endif
//	inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;

namespace Mker
{
    unsigned long long SA,SB,SC;
    void init(){scanf("%llu%llu%llu",&SA,&SB,&SC);}
    inline unsigned long long rand()
    {
        SA^=SA<<32,SA^=SA>>13,SA^=SA<<1;
        unsigned long long t=SA;
        SA=SB,SB=SC,SC^=t^SA;return SC;
    }
}

struct POW
{
	ll a;
	ll pow1[BASE+2],pow2[BASE+2];
	void init(int aa)
	{
		a=aa;
		pow1[0]=1;rep(i,1,BASE) pow1[i]=pow1[i-1]*a%mod;
		pow2[0]=1;rep(i,1,BASE) pow2[i]=pow2[i-1]*pow1[BASE]%mod;
	}
	inline ll query(register int n){return pow1[n&32767]*pow2[n>>15]%mod;}
}a,b;


int main()
{
	file();
	a.init(94153035),b.init(905847205);
	int T;read(T);Mker::init();
	ll ans=0;
	rep(i,1,T)
	{
		ll n=Mker::rand()%(mod-1);
		ll cur=233230706ll*(a.query(n)-b.query(n)+mod)%mod;
		ans^=cur;
	}
	cout<<ans;
	return 0;
}