[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。
  1     /*
  2 ━━━━━┒ギリギリ♂ eye！
  3 ┓┏┓┏┓┃キリキリ♂ mind！
  4 ┛┗┛┗┛┃＼○／
  5 ┓┏┓┏┓┃ /
  6 ┛┗┛┗┛┃ノ)
  7 ┓┏┓┏┓┃
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  9 ┓┏┓┏┓┃
 10 ┛┗┛┗┛┃
 11 ┓┏┓┏┓┃
 12 ┛┗┛┗┛┃
 13 ┓┏┓┏┓┃
 14 ┃┃┃┃┃┃
 15 ┻┻┻┻┻┻
 16 */
 17 #include <algorithm>
 18 #include <iostream>
 19 #include <iomanip>
 20 #include <cstring>
 21 #include <climits>
 22 #include <complex>
 23 #include <fstream>
 24 #include <cassert>
 25 #include <cstdio>
 26 #include <bitset>
 27 #include <vector>
 28 #include <deque>
 29 #include <queue>
 30 #include <stack>
 31 #include <ctime>
 32 #include <set>
 33 #include <map>
 34 #include <cmath>
 35 using namespace std;
 36 #define fr first
 37 #define sc second
 38 #define cl clear
 39 #define BUG puts("here!!!")
 40 #define W(a) while(a--)
 41 #define pb(a) push_back(a)
 42 #define Rint(a) scanf("%d", &a)
 43 #define Rll(a) scanf("%lld", &a)
 44 #define Rs(a) scanf("%s", a)
 45 #define Cin(a) cin >> a
 46 #define FRead() freopen("in", "r", stdin)
 47 #define FWrite() freopen("out", "w", stdout)
 48 #define Rep(i, len) for(int i = 0; i < (len); i++)
 49 #define For(i, a, len) for(int i = (a); i < (len); i++)
 50 #define Cls(a) memset((a), 0, sizeof(a))
 51 #define Clr(a, x) memset((a), (x), sizeof(a))
 52 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
 53 #define lrt rt << 1
 54 #define rrt rt << 1 | 1
 55 #define pi 3.14159265359
 56 #define RT return
 57 #define lowbit(x) x & (-x)
 58 #define onenum(x) __builtin_popcount(x)
 59 typedef long long LL;
 60 typedef long double LD;
 61 typedef unsigned long long ULL;
 62 typedef pair<int, int> pii;
 63 typedef pair<string, int> psi;
 64 typedef pair<LL, LL> pll;
 65 typedef map<string, int> msi;
 66 typedef vector<int> vi;
 67 typedef vector<LL> vl;
 68 typedef vector<vl> vvl;
 69 typedef vector<bool> vb;
 70 
 71 const int mod = 19999997;
 72 const int maxn = 5;
 73 LL n;
 74 
 75 typedef struct Matrix {
 76     LL m[maxn][maxn];
 77     int r;
 78     int c;
 79     Matrix(){
 80         r = c = 0;
 81         memset(m, 0, sizeof(m));
 82     } 
 83 } Matrix;
 84 
 85 Matrix mul(Matrix m1, Matrix m2, int mod) {
 86     Matrix ans = Matrix();
 87     ans.r = m1.r;
 88     ans.c = m2.c;
 89     for(int i = 1; i <= m1.r; i++) {
 90         for(int j = 1; j <= m2.r; j++) {
 91                for(int k = 1; k <= m2.c; k++) {
 92                 if(m2.m[j][k] == 0) continue;
 93                 ans.m[i][k] = ((ans.m[i][k] + m1.m[i][j] * m2.m[j][k] % mod) % mod) % mod;
 94             }
 95         }
 96     }
 97     return ans;
 98 }
 99 
100 Matrix quickmul(Matrix m, int n, int mod) {
101     Matrix ans = Matrix();
102     for(int i = 1; i <= m.r; i++) {
103         ans.m[i][i]  = 1;
104     }
105     ans.r = m.r;
106     ans.c = m.c;
107     while(n) {
108         if(n & 1) {
109             ans = mul(m, ans, mod);
110         }
111         m = mul(m, m, mod);
112         n >>= 1;
113     }
114     return ans;
115 }
116 
117 int main() {
118     // FRead();
119     while(cin >> n) {
120         Matrix p, q;
121         p.r = p.c = 2;
122         p.m[1][1] = 1; p.m[1][2] = 1;
123         p.m[2][1] = 1; p.m[2][2] = 0;
124         q.r = 2; q.c = 1;
125         if(n <= 2) {
126             cout << n << endl;
127             continue;
128         }
129         q = quickmul(p, n-1, mod);
130         cout << (q.m[1][1] + q.m[1][2]) % mod << endl;
131     }
132     RT 0;
133 }