POJ 3070-Fibonacci(矩阵快速幂求斐波那契数列) - c++编程基础

Fibonacci Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 3070 Appoint description:

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_n_{? 1} + F_n_{? 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

.<??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vcD4KPHA+R2l2ZW4gYW4gaW50ZWdlciA8ZW0+bjwvZW0+LCB5b3VyIGdvYWwgaXMgdG8gY29tcHV0ZSB0aGUgbGFzdCA0IGRpZ2l0cyBvZiA8ZW0+RjxzdWI+bjwvc3ViPjwvZW0+LjwvcD4KCgoKCjxwIGNsYXNzPQ=="pst"> 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 F_n. If the last four digits of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n mod 10000).

Sample Input

Sample Output

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 
  
   
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            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; }