矩阵快速幂

可以找到规律。。答案是第2 * k + 3项减1，直接矩阵加速就好了

#include 
    
     #define INF 0x3f3f3f3f#define full(a, b) memset(a, b, sizeof a)#define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)using namespace std;typedef long long ll;inline int lowbit(int x){ return x & (-x); }inline int read(){    int ret = 0, w = 0; char ch = 0;    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }    while(isdigit(ch)) ret = (ret << 3) + (ret << 1) + (ch ^ 48), ch = getchar();    return w ? -ret : ret;}inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }inline int lcm(int a, int b){ return a / gcd(a, b) * b; }template 
     
      inline T max(T x, T y, T z){ return max(max(x, y), z); }template 
      
       inline T min(T x, T y, T z){ return min(min(x, y), z); }template 
       
        inline A fpow(A x, B p, C lyd){    A ans = 1;    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;    return ans;}const int MOD = 998244353;int n;struct Matrix{    ll m[2][2];    Matrix(){        full(m, 0);    }    Matrix operator * (Matrix &b){        Matrix ret;        for(int i = 0; i < 2; i ++){            for(int j = 0; j < 2; j ++){                for(int k = 0; k < 2; k ++){                    ret.m[i][j] += (m[i][k] * b.m[k][j]) % MOD;                    ret.m[i][j] %= MOD;                }            }        }        return ret;    }};Matrix e, ori, t;Matrix fpow(Matrix &a, int p){    Matrix res = e;    for(; p; p >>= 1, a = a * a) if(p & 1) res = a * res;    return res;}void init(){    full(e.m, 0), full(ori.m, 0), full(t.m, 0);    e.m[0][0] = 1, e.m[1][1] = 1;    ori.m[0][1] = 1, ori.m[1][1] = 0;    t.m[0][0] = t.m[0][1] = t.m[1][0] = 1;}int main(){    while(~scanf("%d", &n)){        init();        t = fpow(t, 2 * n + 2);        ori = t * ori;        printf("%lld\n", ((ori.m[0][1] - 1) % MOD + MOD) % MOD);    }    return 0;}