poj 3070 矩阵快速幂求斐波拉契数列

Fibonacci

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12324 Accepted: 8746

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

.

Given an integer n, your goal is to compute the last 4 digits of F_n.

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

099999999991000000000-1

Sample Output

0346266875

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：给出一个求斐波拉契数列的矩阵，Fn即为进行n次矩阵乘法后得到矩阵的a[0][1]，让你用矩阵乘法来求斐波拉契数列

思路：将快速幂运用到矩阵上，非常通俗的方法。其实这道题拓神在去年的讲课上就已经说过了，当时回来就手敲了一个，一直不知道竟然在POJ上有题……今天改了一下交了下。虽然真的是手敲……

代码

#include<stdio.h>int ans[2][2];void kuaisumi(long long b){  long long temp[2][2];  long long templ[2][2]; // for(int i=0;i<2;i++)   // for(int j=0;j<2;j++) // temp[i][j]=a[i][j];   ans[0][0]=1;  ans[0][1]=1;  ans[1][0]=1;  ans[1][1]=0;   temp[0][0]=1;  temp[0][1]=1;  temp[1][0]=1;  temp[1][1]=0;  while(b>0)     {if(b%2==1)   {templ[0][0]=ans[0][0]%10000*temp[0][0]%10000+ans[0][1]%10000*temp[1][0]%10000;    templ[0][1]=ans[0][0]%10000*temp[1][0]%10000+ans[0][1]%10000*temp[1][1]%10000;    templ[1][0]=ans[1][0]%10000*temp[0][0]%10000+ans[1][1]%10000*temp[1][0]%10000;    templ[1][1]=ans[1][0]%10000*temp[0][1]%10000+ans[1][1]%10000*temp[1][1]%10000;    for(int i=0;i<2;i++)          for(int j=0;j<2;j++)          ans[i][j]=templ[i][j]%10000;      }      b=b/2;    templ[0][0]=temp[0][0]%10000*temp[0][0]%10000+temp[0][1]%10000*temp[1][0]%10000;    templ[0][1]=temp[0][0]%10000*temp[1][0]%10000+temp[0][1]%10000*temp[1][1]%10000;    templ[1][0]=temp[1][0]%10000*temp[0][0]%10000+temp[1][1]%10000*temp[1][0]%10000;    templ[1][1]=temp[1][0]%10000*temp[0][1]%10000+temp[1][1]%10000*temp[1][1]%10000;    for(int i=0;i<2;i++)          for(int j=0;j<2;j++)          temp[i][j]=templ[i][j]%10000; }    }int main(){int juzheng[3][3];long long i,j,k,m,n;int fm,fn;int f1,f2; f1=1; f2=1; while(scanf("%lld",&n)!=EOF)  {if(n==-1)break;   if(n==0)printf("0\n");   else    if(n==1 || n==2)printf("1\n");   else   { kuaisumi(n-2);     printf("%lld\n",(ans[0][0])%10000);     }    }    return 0; }