The Nth Item
思路：
先用特征根法求出通向公式，然后通向公式中出现了(sqrt{17})，这个可以用二次剩余求出来，然后可以O((log(n)))求出。
但是还不够，我们先对(n)欧拉降幂，然后求base为(sqrt{1e9})的快速幂，预处理一些东西，就可以类似O(1)求出了。
代码：
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<double, double>
#define mem(a, b) memset(a, b, sizeof(a))
#define debug(x) cerr << #x << " = " << x << "
";
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head
const int MOD = 998244353;
const LL fsqrt17 = 473844410;
const LL sqrt17 = MOD-fsqrt17;
const LL inv2 = (MOD+1)/2;
const LL invsqrt17 = 438914993;
const int N = 4e4 + 5;
LL p[N], pp[N], P[N], PP[N], q, n;
LL f(LL n) {
return (p[n%(N-1)]*P[n/(N-1)])%MOD;
}
LL F(LL n) {
return (pp[n%(N-1)]*PP[n/(N-1)])%MOD;
}
int main() {
p[0] = pp[0] = 1;
for (int i = 1; i < N; ++i) p[i] = p[i-1]*(3+sqrt17)%MOD*inv2%MOD, pp[i] = pp[i-1]*(3+fsqrt17)%MOD*inv2%MOD;
P[0] = PP[0] = 1;
for (int i = 1; i < N; ++i) {
P[i] = P[i-1]*p[N-1]%MOD;
PP[i] = PP[i-1]*pp[N-1]%MOD;
}
scanf("%lld %lld", &q, &n);
LL res = 0;
for (int i = 1; i <= q; ++i) {
LL ans = -(f(n%(MOD-1))-F(n%(MOD-1)))*invsqrt17%MOD;
ans = (ans + MOD) % MOD;
res ^= ans;
n ^= ans*ans;
}
printf("%lld
", res);
return 0;
}