POJ - 3070 Fibonacci (矩阵快速幂运算)

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:

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题解如下：
#include
#include
using namespace std;
typedef long long ll;
const ll mod = 10000;
ll n;
ll e[2][2]={{1,1},{1,0}};
ll a[2][2];
void mi(ll x1)
{
ll ans[2][2];
a[0][0]=1;
a[0][1]=0;
a[1][0]=0;
a[1][1]=1;

    e[0][0]=1;
    e[0][1]=1;
    e[1][0]=1;
    e[1][1]=0;

    while(x1)
    {
        if(x1&1)
        {
            for(int i=0;i<2;i++){
                for(int j=0;j<2;j++){
                    ans[i][j] = a[i][0]*e[0][j]+a[i][1]*e[1][j];
                }
            }

            a[0][0] = ans[0][0]%mod;
            a[0][1] = ans[0][1]%mod;
            a[1][0] = ans[1][0]%mod;
            a[1][1] = ans[1][1]%mod;
        }

        x1>>=1;

        for(int i=0;i<2;i++){
            for(int j=0;j<2;j++){
                ans[i][j] = e[i][0]*e[0][j]+e[i][1]*e[1][j];
            }
        }
        e[0][0] = ans[0][0]%mod;
        e[0][1] = ans[0][1]%mod;
        e[1][0] = ans[1][0]%mod;
        e[1][1] = ans[1][1]%mod;
    }
}
int main()
{
    while(scanf("%lld",&n) && n!=-1)
    {
        mi(n);
        printf("%lld\n",a[1][0]);
    }
    return 0;
}