矩阵快速幂
能够找到规律。。答案是第2 * k + 3项减1,直接矩阵加速就行了html
#include <bits/stdc++.h> #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 <typename T> inline T max(T x, T y, T z){ return max(max(x, y), z); } template <typename T> inline T min(T x, T y, T z){ return min(min(x, y), z); } template <typename A, typename B, typename C> 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; }