HDU - 6198 number number number（规律+矩阵快速幂）

题意：已知F₀ = 0，F₁ = 1，F_n = F_{n - 1} + F_{n - 2}(n >= 2)， 且若n=Fa1+Fa2+...+Fak where 0≤a1≤a2≤⋯≤a，n为正数，则n为mjf-good，否则为mjf-bad，给定k，求最小的mjf-bad。

分析：找规律可得，F_2*k+3 - 1，矩阵快速幂即可。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const LL MOD = 998244353;
const double pi = acos(-1.0);
const int MAXN = 100000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
struct Matrix{
    LL r, c;
    LL matrix[2][2];
    Matrix(LL rr, LL cc):r(rr), c(cc){
        memset(matrix, 0, sizeof matrix);
    }
};
Matrix mul(Matrix a, Matrix b){
    Matrix ans(a.r, b.c);
    for(int i = 0; i < a.r; ++i){
        for(int j = 0; j < b.c; ++j){
            for(int k = 0; k < a.c; ++k){
                (ans.matrix[i][j] += (a.matrix[i][k] * b.matrix[k][j]) % MOD) %= MOD;
            }
        }
    }
    return ans;
}
int Q_POW(Matrix a, int n){
    Matrix ans(a.r, a.c);
    ans.matrix[0][0] = ans.matrix[1][1] = 1;
    while(n){
        if(n & 1) ans = mul(ans, a);
        a = mul(a, a);
        n >>= 1;
    }
    return ans.matrix[0][0] % MOD;
}
int main(){
    int k;
    while(scanf("%d", &k) == 1){
        int n = 5 + (k - 1) * 2;
        Matrix a(2, 2);
        a.matrix[0][0] = a.matrix[0][1] = a.matrix[1][0] = 1;
        printf("%lld\n", (Q_POW(a, n - 1) - 1 + MOD) % MOD);
    }
    return 0;
}