POJ3070 Fibonacci

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12833		Accepted: 9124

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

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:

.

Source

Stanford Local 2006

一个矩阵乘法就砸了上去

代码是偷来的2333 

from：http://blog.csdn.net/orion_rigel/article/details/51926671

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
int n;
void cheng (int a[2][2],int b[2][2])
{
    int c[2][2];
    memset(c,0,sizeof(c));
    for (int i=0;i<2;i++)
        for (int j=0;j<2;j++)
          for (int k=0;k<2;k++)
            c[i][j]=(c[i][j]+a[i][k]*b[k][j])%10000;
    memcpy(a,c,sizeof(c));
}
int main()
{
    while (cin>>n && n!=-1)
    {
        int f[2][2]={{0,1},{0,0}};
        int a[2][2]={{0,1},{1,1}};
        while (n>0)
     {
        if(n&1) cheng(f,a);
        cheng (a,a);
        n>>=1;
     }
      printf("%d
",f[0][0]);
    }
}