斐波那契数列矩阵快速幂

题意

  用矩阵乘法求fibonacci数列的第n项。

Solution

  矩乘入门题啊,题目把题解已经说的很清楚里= =。

  矩乘其实很简单,通过自己YY或者是搜索对于一个递推公式求出它所对应的矩阵,然后套个快速幂就可以迅速求解第n项。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<vector>
 5 #include<queue>
 6 #include<cstring>
 7 #define mp make_pair
 8 #define pb push_back
 9 #define first fi
10 #define second se
11 #define pw(x) (1ll << (x))
12 #define sz(x) ((int)(x).size())
13 #define all(x) (x).begin(),(x).end()
14 #define rep(i,l,r) for(int i=(l);i<(r);i++)
15 #define per(i,r,l) for(int i=(r);i>=(l);i--)
16 #define FOR(i,l,r) for(int i=(l);i<=(r);i++)
17 #define eps 1e-9
18 #define PIE acos(-1)
19 #define cl(a,b) memset(a,b,sizeof(a))
20 #define fastio ios::sync_with_stdio(false);cin.tie(0);
21 #define lson l , mid , ls
22 #define rson mid + 1 , r , rs
23 #define ls (rt<<1)
24 #define rs (ls|1)
25 #define INF 0x3f3f3f3f
26 #define LINF 0x3f3f3f3f3f3f3f3f
27 #define freopen freopen("in.txt","r",stdin);
28 #define cfin ifstream cin("in.txt");
29 #define lowbit(x) (x&(-x))
30 #define sqr(a) a*a
31 #define ll long long
32 #define ull unsigned long long
33 #define vi vector<int>
34 #define pii pair<int, int>
35 #define dd(x) cout << #x << " = " << (x) << ", "
36 #define de(x) cout << #x << " = " << (x) << "
"
37 #define endl "
"
38 using namespace std;
39 //**********************************
40 const int mod=1e4;
41 typedef vector<vi> mat;
42 //**********************************
43 mat mul(mat& a,mat& b)
44 {
45     mat c(2,vi(2));
46     rep(i,0,2)rep(j,0,2)rep(k,0,2)
47     c[i][j]+=(a[i][k]*b[k][j]),c[i][j]%=mod;
48     return c;
49 }
50 mat qpow(mat a,int b)
51 {
52     mat c(2,vi(2));
53     c[0][0]=c[1][1]=1;c[1][0]=c[0][1]=0;    
54     while(b){
55         if(b&1)c=mul(c,a);
56         a=mul(a,a);
57         b>>=1;
58     }
59     return c;
60 }
61 //**********************************
62 int main()
63 {
64     int n;
65     while(cin>>n,~n){
66         mat v(2,vi(2));v[0][0]=1;v[0][1]=1;v[1][0]=1;v[1][1]=0;
67         v=qpow(v,n);
68         cout<<v[1][0]<<endl;
69     }
70     return 0;
71 }
View Code
原文地址:https://www.cnblogs.com/klaycf/p/9682980.html