POJ 3070 Fibonacci

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11996		Accepted: 8516

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:

.

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int maxn = 5;
const int MOD = 10000;

struct Matrix
{
    int mat[maxn][maxn];
    Matrix()        //构造函数
    {
        memset(mat, 0, sizeof(mat));
    }

    Matrix operator * (Matrix A)    //重载运算符*
    {
        Matrix res;
        for (int i = 0; i < 2; i++) //矩阵乘法
        {
            for (int j = 0; j < 2; j++)
            {
                for (int k = 0; k < 2; k++)
                    res.mat[i][j] = (res.mat[i][j] + (mat[i][k] * A.mat[k][j]) % MOD) % MOD;
            }
        }
        return res;
    }
};
//快速求矩阵A^n
Matrix pow_mul(Matrix A, int n)
{
    Matrix res;
    //一个矩阵的0次方为主对角线全为1，其他全为0的矩阵
    for (int i = 0; i < maxn; i++) res.mat[i][i] = 1;
    while (n)
    {
        if (n & 1) res = res * A;   //若n为奇数
        A = A * A;
        n >>= 1;                    //n = n / 2
    }
    return res;
}

int main()
{
    int n;
    while (~scanf("%d", &n) && n != -1)
    {
        Matrix A;
        A.mat[0][0] = A.mat[0][1] = A.mat[1][0] = 1;
        A = pow_mul(A, n);
        cout << A.mat[0][1] << endl;
    }
    return 0;
}