2019南昌网络赛H The Nth Item(二阶线性数列递推 + 广义斐波那契循环节 + 分段打表)题解
题意:
传送门
已知\(F(n)=3F(n-1)+2F(n-2) \mod 998244353,F(0)=0,F(1)=1\),给出初始的\(n_1\)和询问次数\(q\),设每一次的答案\(a_i=F(n_i)\),而\(n_{i+1}=n_i\oplus(a_i*a_i)\),求\(a_1\oplus a_2\dots\oplus a_q\)。
思路:
原式是一个二次常数递归式,我们可以求得它的通项为:
\[F(n)=\frac{1}{\sqrt{17}}[(\frac{3+\sqrt{17}}{2})^n-(\frac{3-\sqrt{17}}{2})^n]
\]
经过二次剩余等乱七八糟的操作,我们能直接\(O(qlogn)\)得到答案,但是时间还是太多。
1:
然后有一个广义斐波那契数列的循环节,那么用\(unordered\_map\)记忆化一下。
2:
因为直接\(logn\)求一个\(F(n)\)会超时,那么使用分段打表把复杂度降到\(O(1)\)。由扩展欧拉定理可得,\(a^n\equiv a^{n\%\varphi(p)+\varphi(p)} \mod p\),那么公式中的指数上限降到\(2(mod -1)\)。然后我们打表出\(\sqrt{2(mod -1)}<5e4\)次幂的结果,那么只要小于\(5e4\)的幂次直接可得。然后打表打出\(0*5e4,1*5e4\dots5e4*5e4\)次幂的结果,那么对于大于\(5e4\)的幂次可以转化为\(a^{k*5e4}*a^c\)前后结论都已打表,那么也是\(O(1)\)即可求解。
代码:
//1
#include<map>
#include<set>
#include<queue>
#include<cmath>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
#include<unordered_map>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int maxn = 1e6 + 5;
const int MAXM = 3e6;
const ll MOD = 998244353;
const ull seed = 131;
const int INF = 0x3f3f3f3f;
const ll r1 = 262199973, r2 = 736044383, invs17 = 559329360;
unordered_map<ll, ll> st;
ll ppow(ll a, ll b){
ll ret = 1;
while(b){
if(b & 1) ret = ret * a % MOD;
a = a * a % MOD;
b >>= 1;
}
return ret;
}
ll solve(ll n){
if(st.count(n)) return st[n];
return st[n] = (ppow(r1, n) - ppow(r2, n) + MOD) % MOD * invs17 % MOD;
}
int main(){
st.clear();
int q;
ll n;
scanf("%d%lld", &q, &n);
ll ans = 0, tmp;
for(int i = 1; i <= q; i++){
tmp = solve(n);
ans ^= tmp;
n = n ^ (tmp * tmp);
}
printf("%lld\n", ans);
return 0;
}
//2
#include<map>
#include<set>
#include<queue>
#include<cmath>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
#include<unordered_map>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int maxn = 1e6 + 5;
const int MAXM = 3e6;
const ll MOD = 998244353;
const ull seed = 131;
const int INF = 0x3f3f3f3f;
const ll r1 = 262199973, r2 = 736044383, invs17 = 559329360;
int N = 5e4;
ll p1[50005], p2[50005], pp1[50005], pp2[50005];
ll ppow1(int b){
if(b <= N) return p1[b];
else return pp1[b / N] * p1[b % N] % MOD;
}
ll ppow2(int b){
if(b <= N) return p2[b];
else return pp2[b / N] * p2[b % N] % MOD;
}
ll solve(ll n){
n = n % (MOD - 1) + MOD - 1;
return (ppow1(n) - ppow2(n) + MOD) % MOD * invs17 % MOD;
}
void init(){
p1[0] = p2[0] = 1;
for(int i = 1; i <= N; i++){
p1[i] = p1[i - 1] * r1 % MOD;
p2[i] = p2[i - 1] * r2 % MOD;
}
pp1[0] = pp2[0] = 1;
for(int i = 1; i <= N; i++){
pp1[i] = pp1[i - 1] * p1[N] % MOD;
pp2[i] = pp2[i - 1] * p2[N] % MOD;
}
}
int main(){
int q;
ll n;
init();
scanf("%d%lld", &q, &n);
ll ans = 0, tmp;
for(int i = 1; i <= q; i++){
tmp = solve(n);
ans ^= tmp;
n = n ^ (tmp * tmp);
}
printf("%lld\n", ans);
return 0;
}