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:
.
题意:题意:求第n个Fibonacci数mod(m)的结果,当n=-1时,break。其中n(where 0 ≤ n ≤ 1,000,000,000) ,m=10000;
思路:常规方法肯定超时,这道题学会了用矩阵快速幂求斐波那契。如下图:
A = F(n - 1), B = F(N - 2),这样使构造矩阵的n次幂乘以初始矩阵得到的结果就是。
#include <stdio.h> #include <math.h> #include <string.h> #include <stdlib.h> #include <iostream> #include <algorithm> #include <set> #include <map> #include <queue> using namespace std; const int inf=0x3f3f3f3f; const int mod=10000; struct node { int mp[3][3]; }init,res; struct node Mult(struct node x,struct node y) { struct node tmp; int i,j,k; for(i=0;i<2;i++) for(j=0;j<2;j++) { tmp.mp[i][j]=0; for(k=0;k<2;k++) { tmp.mp[i][j]=(tmp.mp[i][j]+x.mp[i][k]*y.mp[k][j])%mod; } } return tmp; } struct node expo(struct node x, int k) { int i,j; struct node tmp; for(i=0;i<2;i++) for(j=0;j<2;j++) { if(i==j) tmp.mp[i][j]=1; else tmp.mp[i][j]=0; } while(k) { if(k&1) tmp=Mult(tmp,x); x=Mult(x,x); k>>=1; } return tmp; } int main() { int k; while(~scanf("%d",&k)) { if(k==-1) break; init.mp[0][0]=1; init.mp[0][1]=1; init.mp[1][0]=1; init.mp[1][1]=0; res=expo(init,k); printf("%d\n",res.mp[0][1]); } return 0; }
POJ 3070-Fibonacci(矩阵快速幂求斐波那契数列)
原文:http://blog.csdn.net/u013486414/article/details/44064467