Fibonacci矩阵乘法(POJ 3070)
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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
Hint
As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by
.
Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:
.
矩乘讲解:
直接递推计算时间复杂度显然为O(n),而本题 n<=1e9,直接递推显然超时
我们知道,要求出 Fib(n) ,我们只需要知道Fib(n-1) 和 Fib(n-2),我们在递推时只需要保存最近的两个斐波那契数即可
设 F(n) 表示一个1*2 的矩阵,F(n) = { F(n) , F(n+1) }
我们希望根据 F(n-1) = { F(n-1) , F(n) } 计算出 F(n) 。我们设一个矩阵 A {0,1},{1,1},那么:F(n) =F(n-1)*A=F(1)*An-1
A的次方,不就可以用快速幂优化吗?硬是把O(n) 优化为O( 23log(n) )
code
#include<stdio.h> #include<string.h> #include<algorithm> #define ll long long using namespace std; const int mod=10000; int k; void mul(ll f[2],ll ju[2][2]) { ll c[2];memset(c,0,sizeof(c)); for(int i=0;i<2;++i) for(int j=0;j<2;++j) c[i]=(c[i]+f[j]*ju[j][i])%mod; memcpy(f,c,sizeof(c)); } void mulself(ll ju[2][2]) { ll c[2][2];memset(c,0,sizeof(c)); for(int i=0;i<2;++i) // i 行 for(int j=0;j<2;++j) //j 列 for(int k=0;k<2;++k) c[i][j]=(c[i][j]+ju[i][k]*ju[k][j])%mod; memcpy(ju,c,sizeof(c)); } int main() { while(scanf("%d",&k)!=EOF) { if(k==-1) break; ll f[2]={0,1}; ll ju[2][2]={{0,1},{1,1}}; while(k) { if(k&1) mul(f,ju); mulself(ju); k>>=1; } printf("%lld ",f[0]); } return 0; }
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