[HIHO1143]骨牌覆盖问题·一（矩阵快速幂，递推）

题目链接：http://hihocoder.com/problemset/problem/1143

这个递推还是很经典的，结果是斐波那契数列。f(i) = f(i-1) + f(i-2)。数据范围太大了，应该用快速幂加速下。

     /*
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 ┓┏┓┏┓┃キリキリ♂ mind！
 ┛┗┛┗┛┃＼○／
 ┓┏┓┏┓┃ /
 ┛┗┛┗┛┃ノ)
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 */
 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <fstream>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>
 using namespace std;
 #define fr first
 #define sc second
 #define cl clear
 #define BUG puts("here!!!")
 #define W(a) while(a--)
 #define pb(a) push_back(a)
 #define Rint(a) scanf("%d", &a)
 #define Rll(a) scanf("%lld", &a)
 #define Rs(a) scanf("%s", a)
 #define Cin(a) cin >> a
 #define FRead() freopen("in", "r", stdin)
 #define FWrite() freopen("out", "w", stdout)
 #define Rep(i, len) for(int i = 0; i < (len); i++)
 #define For(i, a, len) for(int i = (a); i < (len); i++)
 #define Cls(a) memset((a), 0, sizeof(a))
 #define Clr(a, x) memset((a), (x), sizeof(a))
 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
 #define lrt rt << 1
 #define rrt rt << 1 | 1
 #define pi 3.14159265359
 #define RT return
 #define lowbit(x) x & (-x)
 #define onenum(x) __builtin_popcount(x)
 typedef long long LL;
 typedef long double LD;
 typedef unsigned long long ULL;
 typedef pair<int, int> pii;
 typedef pair<string, int> psi;
 typedef pair<LL, LL> pll;
 typedef map<string, int> msi;
 typedef vector<int> vi;
 typedef vector<LL> vl;
 typedef vector<vl> vvl;
 typedef vector<bool> vb;

 const int mod = ;
 const int maxn = ;
 LL n;

 typedef struct Matrix {
     LL m[maxn][maxn];
     int r;
     int c;
     Matrix(){
         r = c = ;
         memset(m, , sizeof(m));
     }
 } Matrix;

 Matrix mul(Matrix m1, Matrix m2, int mod) {
     Matrix ans = Matrix();
     ans.r = m1.r;
     ans.c = m2.c;
     for(int i = ; i <= m1.r; i++) {
         for(int j = ; j <= m2.r; j++) {
                for(int k = ; k <= m2.c; k++) {
                 if(m2.m[j][k] == ) continue;
                 ans.m[i][k] = ((ans.m[i][k] + m1.m[i][j] * m2.m[j][k] % mod) % mod) % mod;
             }
         }
     }
     return ans;
 }

 Matrix quickmul(Matrix m, int n, int mod) {
     Matrix ans = Matrix();
     for(int i = ; i <= m.r; i++) {
         ans.m[i][i]  = ;
     }
     ans.r = m.r;
     ans.c = m.c;
     while(n) {
         if(n & ) {
             ans = mul(m, ans, mod);
         }
         m = mul(m, m, mod);
         n >>= ;
     }
     return ans;
 }

 int main() {
     // FRead();
     while(cin >> n) {
         Matrix p, q;
         p.r = p.c = ;
         p.m[][] = ; p.m[][] = ;
         p.m[][] = ; p.m[][] = ;
         q.r = ; q.c = ;
         if(n <= ) {
             cout << n << endl;
             continue;
         }
         q = quickmul(p, n-, mod);
         cout << (q.m[][] + q.m[][]) % mod << endl;
     }
     RT ;
 }