Problem Description
We define a sequence F:
⋅ F0=0,F1=1;
⋅ Fn=Fn−1+Fn−2 (n≥2).
Give you an integer k, if a positive number n can be expressed by
n=Fa1+Fa2+...+Fak where 0≤a1≤a2≤⋯≤ak, this positive number is mjf−good. Otherwise, this positive number is mjf−bad.
Now, give you an integer k, you task is to find the minimal positive mjf−bad number.
The answer may be too large. Please print the answer modulo 998244353.
⋅ F0=0,F1=1;
⋅ Fn=Fn−1+Fn−2 (n≥2).
Give you an integer k, if a positive number n can be expressed by
n=Fa1+Fa2+...+Fak where 0≤a1≤a2≤⋯≤ak, this positive number is mjf−good. Otherwise, this positive number is mjf−bad.
Now, give you an integer k, you task is to find the minimal positive mjf−bad number.
The answer may be too large. Please print the answer modulo 998244353.
Input
There are about 500 test cases, end up with EOF.
Each test case includes an integer k which is described above. (1≤k≤109)
Each test case includes an integer k which is described above. (1≤k≤109)
Output
For each case, output the minimal mjf−bad number mod 998244353.
Sample Input
1
Sample Output
4
思路：找规律，当k=1时，n=F5-1=4。k=2，n=F7-1=12。k=3，n=F9-1=33。所以大胆推测n=F(2*k+3)-1；再用矩阵快速幂输出F(2n+3)-1。
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <vector> 5 using namespace std; 6 const int mod = 998244353; 7 typedef long long LL; 8 LL n; 9 typedef vector<LL>vec; 10 typedef vector<vec>mat; 11 mat mul(mat &A, mat &B) 12 { 13 mat C(A.size(), vec(B[0].size()));///分配大小，A的行，B的列 14 for (int i = 0; i<A.size(); i++) ///矩阵A的行 15 for (int k = 0; k<B.size(); k++) ///矩阵B的行 16 for (int j = 0; j<B[0].size(); j++) ///矩阵B的列 17 C[i][j] = (C[i][j] + A[i][k] * B[k][j] % mod + mod) % mod; 18 return C; 19 } 20 ///计算A^n 21 mat pow(mat A, LL n) 22 { 23 mat B(A.size(), vec(A.size()));///和矩阵A的大小相同 24 for (int i = 0; i<A.size(); i++) 25 B[i][i] = 1; 26 while (n>0) 27 { 28 if (n & 1) B = mul(B, A); 29 A = mul(A, A); 30 n >>= 1; 31 } 32 return B; 33 } 34 void solve() 35 { 36 mat A(2, vec(2));///2*2的矩阵 37 A[0][0] = 1; 38 A[0][1] = 1; 39 A[1][0] = 1; 40 A[1][1] = 0; 41 A = pow(A, n); 42 printf("%d ", (A[1][0] % mod - 1 + mod) % mod); 43 } 44 int main() 45 { 46 while (~scanf("%lld", &n)) 47 { 48 n = 2 * n + 3; 49 solve(); 50 } 51 }