gym100299J Captain Obvious and the Rabbit-Man

题意

定义\(f(n)=\sum_{i=1}^k fib_i^n a_i\)，其中\(fib_1=1,fib_2=2,fib_n=fib_{n-1}+fib_{n-2}\)

已知\(f(1)\dots f(k)\)，求\(f(k+1)\)。

所有运算对质数\(M\)取模，保证\(fib_i\)模\(m\)后两两不同，保证存在唯一解。

需要使用\(O(k^2)\)的做法。 ### 题解 由于可以解出\(a_i\)，显然\(f(k+1)=\sum_{i=1}^k f(i)b_i\)，其中\(b_i\)和\(f(i)\)无关。

那么\(b_i\)就满足\(fib_n^{k+1}=\sum_{i=1}^k b_i fib_n^i\)。

设\(F(x)=x^{k+1}-\sum_{i=1}^k b_ix^i\)，则\(fib_1\dots fib_k\)是\(F(x)\)的\(k\)个零点。

则\(F(x)=G(x)\prod (x-fib_i)\)，由于最高次项为\(x^{k+1}\)，则\(G(x)=x\)。

时间复杂度\(O(k^2)\)。

/*
Author: QAQAutoMaton
Lang: C++
Code: J.cpp
Mail: lk@qaq-am.com
Blog: https://www.qaq-am.com/
*/
#include<bits/stdc++.h>
#define int long long
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
using namespace std;
typedef unsigned uint;
typedef long long ll;
typedef unsigned long long ull;
typedef complex<double> cp;
typedef pair<int,int> pii;
int inf;
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T,class T2>int chkmin(T &a,T2 b){return a>b?a=b,1:0;}
template<class T,class T2>int chkmax(T &a,T2 b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T,class T2>T mmin(T a,T2 b){return a<b?a:b;}
template<class T,class T2>T mmax(T a,T2 b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>int dcmp(T a,T b){return a>b;}
template<int *a>int cmp_a(int x,int y){return a[x]<a[y];}
template<class T>bool sort2(T &a,T &b){return a>b?swap(a,b),1:0;}
#define min mmin
#define max mmax
#define abs aabs
struct __INIT__{
	__INIT__(){
		fill((unsigned char*)&inf,(unsigned char*)&inf+sizeof(inf),0x3f);
	}
}__INIT___;
namespace io {
	const int SIZE = (1 << 21) + 1;
	char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
	// getchar
	#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
	// print the remaining part
	inline void flush () {
		fwrite (obuf, 1, oS - obuf, stdout);
		oS = obuf;
	}
	// putchar
	inline void putc (char x) {
		*oS ++ = x;
		if (oS == oT) flush ();
	}
	template<typename A>
	inline bool read (A &x) {
		for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
		for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
		return 1;
	}
	inline bool read (char &x) {
		while((x=gc())==' '||x=='\n' || x=='\r');
		return x!=EOF;
	}
	inline bool read(char *x){
		while((*x=gc())=='\n' || *x==' '||*x=='\r');
		if(*x==EOF)return 0;
		while(!(*x=='\n'||*x==' '||*x=='\r'||*x==EOF))*(++x)=gc();
		*x=0;
		return 1;
	}
	template<typename A,typename ...B>
	inline bool read(A &x,B &...y){
		return read(x)&&read(y...);
	}
	template<typename A>
	inline bool write (A x) {
		if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
		while (x) qu[++ qr] = x % 10 + '0',  x /= 10;
		while (qr) putc (qu[qr --]);
		return 0;
	}
	inline bool write (char x) {
		putc(x);
		return 0;
	}
	inline bool write(const char *x){
		while(*x){putc(*x);++x;}
		return 0;
	}
	inline bool write(char *x){
		while(*x){putc(*x);++x;}
		return 0;
	}
	template<typename A,typename ...B>
	inline bool write(A x,B ...y){
		return write(x)||write(y...);
	}
	//no need to call flush at the end manually!
	struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;

int p;
struct Z{
	uint x;
	Z(){}
	Z(uint a){
		x=a;
	}
};
inline uint modp(const uint x){
	return x<p?x:x-p;
}
inline Z operator+(const Z x1, const Z x2) { return modp(x1.x+x2.x);}
inline Z operator-(const Z x1, const Z x2) { return modp(x1.x+p-x2.x);}
inline Z operator-(const Z x) {return x.x?p-x.x:0;}
inline Z operator*(const Z x1, const Z x2) { return static_cast<ull>(x1.x)*x2.x%p;}
void exgcd(int a,int b,int &x,int &y){
	if(!b){x=1;y=0;return;}
	exgcd(b,a%b,y,x);
	y-=(a/b)*x;
}
inline Z Inv(const Z a){
	int x,y;
	exgcd(p,a.x,x,y);
	return y<0?y+=p:y;
}
inline Z operator/(const Z x1, const Z x2) { return x1*Inv(x2);}

inline Z &operator++(Z &x1){x1.x==p-1?x1.x=0:++x1.x;return x1;} 
inline Z &operator--(Z &x1){x1.x?--x1.x:x1.x=p-1;return x1;}
inline Z &operator+=(Z &x1, const Z x2) { return x1 = x1 + x2; }
inline Z &operator-=(Z &x1, const Z x2) { return x1 = x1 - x2; }
inline Z &operator*=(Z &x1, const Z x2) { return x1 = x1 * x2; }
inline Z &operator/=(Z &x1, const Z x2) { return x1 = x1 / x2; }
inline Z fpm(Z a,int b){Z c(1);for(;b;b>>=1,a*=a)if(b&1)c*=a;return c;}
Z f[4005];
Z a[4005];
Z w[4005];
signed main(){
#ifdef QAQAutoMaton 
	freopen("J.in","r",stdin);
	freopen("J.out","w",stdout);
#endif
	int t;
	read(t);
	int n;
	for(;t;--t){
		read(n,p);
		a[1]=1;
		a[2]=2;
		for(int i=3;i<=n;++i)a[i]=a[i-1]+a[i-2];
		for(int i=0;i<n;++i)read(w[i]);
		f[0]=1;
		for(int i=1;i<=n;++i){
			f[i]=0;
			for(int j=i-1;~j;--j){
				f[j+1]+=f[j];
				f[j]*=-a[i];
			}
		}
		Z ans(0);
		for(int i=0;i<n;++i)ans-=w[i]*f[i];
		write(ans.x,'\n');
	}
	return 0;
}