哎呀大水题。。我写了一个多小时。。好没救啊。。

数论板子X合一？

注意: 本文中变量名称区分大小写。

题意: 给一个\(n\)阶递推序列\(f_k=\prod^{n}_{i=1} f_{k-i}b_i\mod P\)其中\(P=998244353\), 输入\(b_1,b_2,...,b_n\)以及已知\(f_1,f_2,...,f_{n-1}=1\), 再给定一个数\(m\)和第\(m\)项的值\(f_m\), 求出一个合法的\(f_n\)值使得按照这个值递推出来的序列满足第\(m\)项的值为给定的\(f_m\).

题解: 首先一个显然的结论是\(f_m\)可以表示成\(\prod^{n}_{i=1} f_i^{a_i}\), 而且由于\(i=1,2,...,n-1\)时\(f_i\)的任何次幂都为\(1\), 因此就是\(f_m=f_n^{a}\). 令\(A(m)\)为\(f_m\)内\(f_n\)的次数，则有\(A[1..n]=[0,0,0,0,...,0,1]\), \(A_m=\sum^{n-1}_{i=1} A(m-i)b_i (m>n)\), 即\(A\)数组满足一个常系数线性递推序列。因此可以用矩阵乘法在\(O(n^3\log m)\)的时间内求出\(A(m)\). 注意因为是指数的运算(\((a^n)^m=a^{nm}\)), 根据费马小定理，这个指数应该模\(\phi(P)=P-1\)而不是\(P\) (\((a^n)^m\mod P=a^{nm\mod (P-1)}\mod P\))

求出来\(a=A(m)\)之后这题就变成了，\(f_m=f_n^a\mod P\), 已知\(f_m, a\), 求出一组合法的\(f_n\).

根据常识，\(998244353\)有原根\(3\), 我们下文令\(G=3\) (实际上任何一个原根均可). 设\(f_m=G^p, f_n=G^q\), 则有\(G^p\equiv (G^q)^a (\mod P)\), \(p\equiv qa(\mod P-1)\), 然后用BSGS求离散对数\(p\), exgcd解出\(q\)就可以了啊……

时间复杂度\(O(\sqrt P\log P+n^3\log P)\)

坑: 注意解同余方程的时候那个\(P\)的系数不要设成负的。

代码

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<map>
#define llong long long
using namespace std;const int N = 100;
const int G = 3;
const int P = 998244353;
llong quickpow(llong x,llong y)
{llong cur = x,ret = 1ll;for(int i=0; y; i++){if(y&(1ll<<i)) {y-=(1ll<<i); ret = ret*cur%P;}cur = cur*cur%P;}return ret;
}
struct Matrix
{llong a[N+3][N+3]; int sz1,sz2,sz;void init() {for(int i=1; i<=sz1; i++) for(int j=1; j<=sz2; j++) a[i][j] = 0ll;}Matrix() {}Matrix(int _sz) {sz = sz1 = sz2 = _sz; init();}Matrix(int _sz1,int _sz2) {sz1 = _sz1,sz2 = _sz2; init();}void uinit(int _sz) {sz = sz1 = sz2 = _sz; for(int i=1; i<=sz; i++) for(int j=1; j<=sz; j++) a[i][j] = (i==j)?1:0;}void output() {for(int i=1; i<=sz1; i++) {for(int j=1; j<=sz2; j++) printf("%lld ",a[i][j]); puts("");}}
};
Matrix operator *(Matrix x,Matrix y)
{Matrix ret = Matrix(x.sz1,y.sz2);for(int i=1; i<=x.sz1; i++){for(int j=1; j<=x.sz2; j++){for(int k=1; k<=y.sz2; k++){ret.a[i][j] = (ret.a[i][j]+x.a[i][k]*y.a[k][j])%(P-1ll);}}}return ret;
}
Matrix mquickpow(Matrix x,llong y)
{Matrix cur = x,ret; ret.uinit(x.sz);for(int i=0; y; i++){if(y&(1ll<<i)) {y-=(1ll<<i); ret = ret*cur;}cur = cur*cur;}return ret;
}
namespace BSGS
{const int B = 31595;map<llong,int> mp;void init(){llong bs = quickpow(G,B); llong j = 1ll;for(int i=0; i<=P; i+=B,j=(j*bs)%P){mp[j] = i;}}llong Logarithm(llong x){llong j = 1ll;for(int i=0; i<=B; i++,j=(j*G)%P){llong tmp = x*j%P;if(mp.count(tmp)){llong ret = (mp[tmp]-i+(P-1))%(P-1);return ret;}}return P-1;}
}
Matrix mA,mB,mC;
llong a[N+3],b[N+3];
int n; llong m,p,q,lq,lx;llong gcd(llong x,llong y) {return y==0 ? x : gcd(y,x%y);}
void exgcd(llong _a,llong _b,llong &_x,llong &_y)
{if(_b==0ll) {_x = 1ll,_y = 0ll; return;}exgcd(_b,_a%_b,_x,_y);llong tmp = _x; _x = _y; _y = tmp-(_a/_b)*_y;
}
llong CongruenceEquation(llong _a,llong _b,llong mod)
{llong g = gcd(_a,mod),x,y;exgcd(_a/g,mod/g,x,y);return (x*(_b/g)%mod+mod)%mod;
}int main()
{BSGS::init();scanf("%d",&n);for(int i=1; i<=n; i++) scanf("%I64d",&b[i]);scanf("%I64d",&m);mA = Matrix(1,n); mA.a[1][1] = 1ll;mB = Matrix(n); for(int i=1; i<n; i++) mB.a[i][i+1] = 1ll; for(int i=1; i<=n; i++) mB.a[i][1] = b[i];mC = mA*mquickpow(mB,m-n); p = mC.a[1][1]; scanf("%I64d",&q);lq = BSGS::Logarithm(q);if(lq%gcd(P-1,p)!=0) {printf("-1\n"); return 0;}lx = CongruenceEquation(p,lq,P-1);llong ans = quickpow(G,lx);printf("%I64d\n",ans);return 0;
}