洛谷P5110 块速递推 [分块]

传送门

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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。

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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;

}

洛谷P5110 块速递推 [分块]的相关教程结束。