2017 ACM/ICPC Asia Regional Shenyang Online 1005 number number number (HDU6198)

題目鏈接：number number number
思路：找規律，當k=1時，n=F5-1=4。k=2，n=F7-1=12。k=3，n=F9-1=33。所以大膽推測n=F(2*k+3)-1，得到序列4、12、33、88、…
遞推公式：f(n) = 3*f(n-1) - f(n-2) + 1 用矩陣快速冪求解
構造矩陣：
| 3 1 0|
|-1 0 0| * [f(n) f(n-1) 1] = [f(n+1) f(n) 1]
| 1 0 1|

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#define Max(a,b) a>b?a:b
#define Min(a,b) a>b?b:a
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
int dir[4][2]= {{1,0},{-1,0},{0,1},{0,-1}};
const double eps = 1e-6;
const double Pi = acos(-1.0);
const int INF=0x3f3f3f3f;

const int dim = 3;
const ll MOD = 998244353;
#define mod(x) ((x)%MOD)

ll n;

struct mat{
    ll m[dim][dim];
}unit;

mat operator * (mat a,mat b){
    mat ret;
    ll x;
    for(ll i = 0;i < dim;i++){
        for(ll j = 0;j < dim;j++){
            x = 0;
            for(ll k = 0;k < dim;k++)
                x += mod((ll)a.m[i][k]*b.m[k][j]);
            ret.m[i][j] = mod(x);
        }
    }
    return ret;
}

void init_unit(){
    for(ll i = 0;i < dim;i++)
        unit.m[i][i] = 1;
    return ;
}

mat pow_mat(mat a,ll n){
    mat ret = unit;
    while(n){
        if(n&1) ret = ret*a;
        a = a*a;
        n >>= 1;
    }
    return ret;
}

int main(){
    ll n;
    init_unit();
    while(~scanf("%lld",&n)){
        if(n == 1) printf("4\n");
        else if(n == 2) printf("12\n");
        else{
            mat a,b;
            b.m[0][0]=3,b.m[0][1]=1,b.m[0][2]=0;
            b.m[1][0]=-1,b.m[1][1]=0,b.m[1][2]=0;
            b.m[2][0]=1,b.m[2][1]=0,b.m[2][2]=1;

            a.m[0][0]=12,a.m[0][1]=4,a.m[0][2]=1;

            b = pow_mat(b,n-2);
            a = a*b;
            printf("%lld\n",(a.m[0][0]+MOD)%MOD);
        }
    }
    return 0;
}