Fibonacci(矩阵快速幂)
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 24331 | Accepted: 16339 |
Description
In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …
An alternative formula for the Fibonacci sequence is
.
Given an integer n, your goal is to compute the last 4 digits of Fn.
Input
The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.
Output
For each test case, print the last four digits of Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).
Sample Input
0 9 999999999 1000000000 -1
Sample Output
0 34 626 6875
解题思路:因为n很大所以不能用递推来算;
a[2] a[1] a[0] ------> a[3] a[2] a[1]
1 1 0 2 1 1
构造一个矩阵 1行3列的矩阵*(莫一个矩阵)=1行3列的矩阵 -----> 矩阵大小为3行3列
递推式= a[n]=a[n-1]]+a[n-2]; a[n+1]=a[n]+a[n-1];
后一个由第一个怎么变换到达 -----> 构造的矩阵为
1 1 0
1 0 1
0 0 0
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <cstring> 3 #include <algorithm> 4 #define ll long long 5 #define mod(x) ((x)%MOD) 6 using namespace std; 7 8 ll n; 9 const int maxn=3; 10 const int MOD=10000; 11 12 struct mat{ 13 int m[maxn][maxn]; 14 }unit; 15 16 mat operator *(mat a,mat b){ 17 mat ret; 18 ll x; 19 for(int i=0;i<maxn;i++) 20 for(int j=0;j<maxn;j++){ 21 x=0; 22 for(int k=0;k<maxn;k++){ 23 x+=mod(1LL*a.m[i][k]*b.m[k][j]); 24 } 25 ret.m[i][j]=mod(x); 26 } 27 return ret; 28 } 29 30 void init(){ 31 for(int i=0;i<maxn;i++) 32 unit.m[i][i]=1; 33 return; 34 } 35 36 mat pow_mat(mat a,ll m){ 37 mat ret=unit; 38 while(m){ 39 if(m&1) ret=ret*a; 40 a=a*a; 41 m=m/2; 42 } 43 return ret; 44 } 45 46 int main(){ 47 ios::sync_with_stdio(false); 48 init(); 49 while(cin>>n&&n!=-1){ 50 if(n==0) cout << 0 << endl; 51 else if(n==1) cout << 1 << endl; 52 else if(n==2) cout << 1 << endl; 53 else{ 54 mat a,b; 55 a.m[0][0]=1,a.m[0][1]=1,a.m[0][2]=0; 56 57 b.m[0][0]=1,b.m[0][1]=1,b.m[0][2]=0; 58 b.m[1][0]=1,b.m[1][1]=0,b.m[1][2]=1; 59 b.m[2][0]=0,b.m[2][1]=0,b.m[2][2]=0; 60 61 b=pow_mat(b,n-2); 62 a=a*b; 63 cout << mod(a.m[0][0]) << endl; 64 } 65 } 66 return 0; 67 }