Fibonacci
| Time limit:1000 ms |
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Memory limit:65536 K |
| Total submissions:9650 |
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Accepted:6856 |
Description
In the Fibonacci integer sequence,F0 = 0,F1 = 1, andFN=FN? 1 +FN? 2N≥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 integerN, Your goal is to compute the last 4 digitsFN.
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 digitsFN. If the last four digitsFNAre all zeros, print '0'; otherwise, omit any leading zeros (I. e., printFNMoD 10000 ).
Sample Input
099999999991000000000-1
Sample output
0346266875
Hint
As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given
.
Also, note that raising any 2 × 2 matrix to The 0th power gives the identity matrix:
.
Source
Stanford local 2006 is written in the same way as a normal fast power. The difference is that you need to calculate the Matrix Multiplication. As long as you write the multiplication of the matrix, it is no longer difficult.
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define LL long longstruct node{ LL s11 , s12 , s21 , s22 ;};node f(node a,node b){ node p ; p.s11 = (a.s11*b.s11 + a.s12*b.s21)%10000 ; p.s12 = (a.s11*b.s12 + a.s12*b.s22)%10000 ; p.s21 = (a.s21*b.s11 + a.s22*b.s21)%10000 ; p.s22 = (a.s21*b.s12 + a.s22*b.s22)%10000 ; return p ;}node pow(node p,int n){ node q ; q.s11 = q.s22 = 1 ; q.s12 = q.s21 = 0 ; if(n == 0) return q ; q = pow(p,n/2); q = f(q,q); if( n%2 ) q = f(q,p); return q ;}int main(){ int n ; node p ; while(scanf("%d", &n) && n != -1) { p.s11 = p.s12 = p.s21 = 1 ; p.s22 = 0 ; p = pow(p,n); printf("%d\n", p.s12); } return 0;}
Poj3070 -- Fibonacci (Fast Power of matrix)