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;
}