【蓝桥杯】入门训练 Fibonacci数列

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 using namespace std;
 5 const int maxn=1000000+10;
 6 int f[maxn];
 7 int main()
 8 {
 9     int n;
10     scanf("%d",&n);
11     f[1]=1;
12     f[2]=1;
13     for(int i=2;i<=n;i++)
14         f[i]=(f[i-1]+f[i-2])%10007;
15     printf("%d\n",f[n]);
16 
17 }

 1 /*
 2 ID: sdj22251
 3 PROG: subset
 4 LANG: C++
 5 */
 6 #include <iostream>
 7 #include <vector>
 8 #include <list>
 9 #include <map>
10 #include <set>
11 #include <deque>
12 #include <queue>
13 #include <stack>
14 #include <bitset>
15 #include <algorithm>
16 #include <functional>
17 #include <numeric>
18 #include <utility>
19 #include <sstream>
20 #include <iomanip>
21 #include <cstdio>
22 #include <cmath>
23 #include <cstdlib>
24 #include <cctype>
25 #include <string>
26 #include <cstring>
27 #include <cmath>
28 #include <ctime>
29 #define MAXN 305
30 #define INF 100000000
31 #define eps 1e-7
32 #define PI 3.1415926535898
33 using namespace std;
34 int n = 2, m;
35 int tt[2][2]={{1, 1}, {1, 0}};
36 struct wwj
37 {
38     int r, c;
39     int mat[3][3];
40 } need, ready;
41 void init()
42 {
43     memset(need.mat, 0, sizeof(need.mat));
44     need.r = n;
45     need.c = n;
46     for(int i = 1; i <= n; i++)
47     {
48         need.mat[i][i] = 1;
49     }
50     ready.c = n;
51     ready.r = n;
52     for(int i = 1; i <= n; i++)
53     {
54         for(int j = 1; j <= n; j++)
55         ready.mat[i][j] = tt[i - 1][j - 1];
56     }
57 }
58 wwj multi(wwj x, wwj y)
59 {
60     wwj t;
61     int i, j, k;
62     memset(t.mat, 0, sizeof(t.mat));
63     t.r = x.r;
64     t.c = y.c;
65     for(i = 1; i <= x.r; i++)
66     {
67         for(k = 1; k <= x.c; k++)
68             if(x.mat[i][k])
69             {
70                 for(j = 1; j <= y.c; j++)
71                 {
72                     t.mat[i][j] += (x.mat[i][k] * y.mat[k][j]) % 10007;
73                     t.mat[i][j] %= 10007;
74                 }
75             }
76     }
77     return t;
78 }
79 int main()
80 {
81     int m;
82     scanf("%d", &m);
83         init();
84         while(m)
85         {
86             if(m & 1)
87             {
88                 need = multi(ready, need);
89             }
90             ready = multi(ready, ready);
91             m = m >> 1;
92         }
93         printf("%d\n", need.mat[1][2] % 10007);
94     return 0;
95 }