poj 3070 Fibonacci 【矩阵快速幂 求第N个斐波那契数%1000】

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11123		Accepted: 7913

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:

.

题意：给你一个求解第N个斐波那契数的公式。 让你求出Fn % 10000。

构造单位矩阵，然后就是矩阵快速幂了。

AC代码：

#include 
#include 
#include 
#define MAXN 100
#define LL long long
#define MOD 10000
using namespace std;
struct Matrix
{
    LL a[MAXN][MAXN];
    int r, c;//行数 列数
};
Matrix ori, res;//初始矩阵 和 结果矩阵
void init()//初始化矩阵
{
    memset(res.a, 0, sizeof(res.a));
    res.r = 2; res.c = 2;
    for(int i = 1; i <= 2; i++)//构造单位矩阵
        res.a[i][i] = 1;
    ori.r = 2; ori.c = 2;
    ori.a[1][1] = ori.a[1][2] = ori.a[2][1] = 1;
    ori.a[2][2] = 0;
}
Matrix multi(Matrix x, Matrix y)
{
    Matrix z;
    memset(z.a, 0, sizeof(z.a));
    z.r = x.r, z.c = y.c;//新矩阵行数等于x矩阵的行数 列数等于y矩阵的列数
    for(int i = 1; i <= x.r; i++)//x矩阵的行数
    {
        for(int k = 1; k <= x.c; k++)//矩阵x的列数等于矩阵y的行数 即x.c = y.r
        {
            if(x.a[i][k] == 0) continue;//优化
            for(int j = 1; j<= y.c; j++)//y矩阵的列数
                z.a[i][j] = (z.a[i][j] + (x.a[i][k] * y.a[k][j]) % MOD) % MOD;
        }
    }
    return z;
}
void Matrix_mod(int n)
{
    while(n)//N次幂
    {
        if(n & 1)
            res = multi(ori, res);
        ori = multi(ori, ori);
        n >>= 1;
    }
    printf("%lld\n", res.a[1][2] % MOD);
}
int main()
{
    int N;
    while(scanf("%d", &N), N!=-1)
    {
        init();//初始化单位矩阵
        Matrix_mod(N);//矩阵快速幂
    }
    return 0;
}