POJ - 3070 Fibonacci

题目已经告诉你斐波那契矩阵算法了...

所以需要用到矩阵快速幂，看起来名字很diao但其实和普通快速幂基本一毛一样...

 1 #include <iostream>
 2 #include <string.h>
 3 #include <cstdio>
 4 #include <queue>
 5 #include <set>
 6 #include <stack>
 7 #include <math.h>
 8 #include <string>
 9 #include <algorithm>
10 
11 #define SIGMA_SIZE 26
12 #define pii pair<int,int>
13 #define lson rt<<1
14 #define rson rt<<1|1
15 #define lowbit(x) (x&-x)
16 #define fode(i, a, b) for(int i=a; i>=b; i--)
17 #define foe(i, a, b) for(int i=a; i<=b; i++)
18 #define fod(i, a, b) for(int i=a; i>b; i--)
19 #define fo(i, a, b) for(int i=a; i<b; i++)
20 //#pragma warning ( disable : 4996 )
21 
22 using namespace std;
23 typedef long long LL;
24 inline LL LMax(LL a, LL b) { return a>b ? a : b; }
25 inline LL LMin(LL a, LL b) { return a>b ? b : a; }
26 inline LL lgcd(LL a, LL b) { return b == 0 ? a : lgcd(b, a%b); }
27 inline LL llcm(LL a, LL b) { return a / lgcd(a, b)*b; }  //a*b = gcd*lcm
28 inline int Max(int a, int b) { return a>b ? a : b; }
29 inline int Min(int a, int b) { return a>b ? b : a; }
30 inline int gcd(int a, int b) { return b == 0 ? a : gcd(b, a%b); }
31 inline int lcm(int a, int b) { return a / gcd(a, b)*b; }  //a*b = gcd*lcm
32 const LL INF = 0x3f3f3f3f3f3f3f3f;
33 const LL lmod = 1e9+7;
34 const int mod = 10000;
35 const double eps = 1e-8;
36 const int inf = 0x3f3f3f3f;
37 const int maxk = 1e5+5;
38 const int maxm = 510*510;
39 const int maxn = 2;
40 
41 struct Matrix {
42     int mm[maxn][maxn];
43     Matrix() {}
44     Matrix operator*(Matrix const &b) const
45     {
46         Matrix tmp; memset(tmp.mm, 0, sizeof(tmp.mm));
47         fo(i, 0, maxn)
48             fo(j, 0, maxn)
49                 fo(k, 0, maxn)
50                     tmp.mm[i][j] = (tmp.mm[i][j] + this->mm[i][k]*b.mm[k][j])%mod;
51         return tmp;
52     }
53 };
54 
55 Matrix qpow(Matrix base, int b)
56 {
57     Matrix res; 
58     memset(res.mm, 0, sizeof(res.mm));
59     fo(i, 0, maxn) res.mm[i][i] = 1;
60     while(b)
61     {
62         if (b&1) res = res*base;
63         base = base*base;
64         b >>= 1;
65     }
66     return res;
67 }
68 
69 
70 int main()
71 {
72 
73     #ifndef ONLINE_JUDGE
74         freopen("input.txt", "r", stdin);
75     #endif
76     
77     int x;
78     while(cin >> x)
79     {
80         if (x==-1) break;
81         //if (x==0) cout << 0 << endl;
82 
83         Matrix tmp, ans; memset(tmp.mm, 0, sizeof(tmp.mm));
84         tmp.mm[0][0] = tmp.mm[0][1] = tmp.mm[1][0] = 1;
85         tmp.mm[1][1] = 0;
86 
87         ans = qpow(tmp, x);
88         printf("%d\n", ans.mm[0][1]);
89     }
90 
91     return 0;
92 }
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posted @ 2018-10-12 20:28  LBNOQYX  阅读(197)  评论(0)  编辑  收藏  举报