Luogu5110 块速递推

题面

题解

线性常系数齐次递推sb板子题

$a_n=233a_{n-1}+666a_{n-2}$的特征方程为

$$ x^2=233x+666 \ x^2-233x+666=0 \ x_1=frac{233+sqrt{56953}}2,x_2=frac{233-sqrt{56953}}2 \ herefore a_n=alpha x_1^n+eta x_2^n \ ecause a_0=0,a_1=1 \ herefore egin{cases} alpha+eta=0 \ alpha x_1+eta x_2=1 end{cases} \ herefore egin{cases} alpha=frac1{sqrt{56953}} \ eta=-frac1{sqrt{56953}} end{cases} \ herefore a_n=frac1{sqrt{56953}}left(left(frac{233+sqrt{56953}}2 ight)^n-left(frac{233-sqrt{56953}}2 ight)^n ight) \ ecause 188305837^2 equiv 56953 ; ( ext{mod};10^9+7) \ herefore a_n equiv 233230706 imesleft(94153035^n-905847205^n ight) $$

求里面两个底数的$n$次方如何$O(1)$求？分段打表

设$f_1(n)=x^{65536n},f_2(n)=x^n$

则：

$$ x^n=f_1(n/65536) imes f_2(n\%65536) $$

就可以了。

复杂度$ ext{O}(T)$，$T$是询问次数

代码

#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#define RG register

namespace Maker
{
    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;
    }
}

const int Mod(1e9 + 7), alpha(233230706), x_1(94153035),
	  x_2(905847205), x_3(64353223), x_4(847809841);
const int maxn(65536 + 5);
int f_1[maxn], f_2[maxn], f_3[maxn], f_4[maxn], T, ans;

inline int Pow_1(int x) { return 1ll * f_3[x >> 16] * f_1[x & 65535] % Mod; }
inline int Pow_2(int x) { return 1ll * f_4[x >> 16] * f_2[x & 65535] % Mod; }

int main()
{
	f_1[0] = f_2[0] = f_3[0] = f_4[0] = 1;
	for(RG int i = 1; i < 65536; i++) f_1[i] = 1ll * f_1[i - 1] * x_1 % Mod;
	for(RG int i = 1; i < 65536; i++) f_2[i] = 1ll * f_2[i - 1] * x_2 % Mod;
	for(RG int i = 1; i < 65536; i++) f_3[i] = 1ll * f_3[i - 1] * x_3 % Mod;
	for(RG int i = 1; i < 65536; i++) f_4[i] = 1ll * f_4[i - 1] * x_4 % Mod;
	scanf("%d", &T); Maker::init(); unsigned long long n;
	while(T--) n = Maker::rand() % (Mod - 1),
		ans ^= 1ll * alpha * (Pow_1(n) - Pow_2(n) + Mod) % Mod;
	printf("%d
", ans);
	return 0;
}