NYOJ148 fibonacci数列(二)

fibonacci数列(二)

时间限制：1000 ms  |  内存限制：65535 KB

难度：3

 

描述

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.

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:

.

 

输入

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.

输出

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).

样例输入

0
9
1000000000
-1

样例输出

0
34
6875

/*
//代码一：找取模后的周期---因为是加法，所以完全可以根据取模后的周期推倒正确的结果
#include <iostream>
#include <cstdio>
using namespace std;

const int MAX = 15000 + 1;
int fib[MAX];

int fun(int *fib, int mod)
{
    fib[0] = 0;
    fib[1] = 1;
    int i = 2;
    do
    {
        fib[i] = (fib[i - 1] + fib[i - 2]) % mod;
        ++i;
    }while(fib[i - 1] != 1 || fib[i - 2] != 0);
    return i - 2;
}

int main()
{
    int T = fun(fib, 10000);
    //cout << T;     T = 15000;
    int n;
    while(scanf("%d", &n) && n != -1)
    {
        printf("%d\n",fib[n % T]);
    }
    return 0;
}


*/

//代码二：二分幂
#include <cstdio>
#include <iostream>

using namespace std;

void fun(int a1[][2], int a2[][2])     // 相当于 a1 = a1 * a2
{
    int c[2][2];
    for(int i=0; i < 2; i++)
        for(int j = 0; j < 2; j++)
        {
            c[i][j] = 0;
            for(int k = 0; k < 2; k++)
                c[i][j] = (c[i][j] + a1[i][k] * a2[k][j])%10000;
        }
    for(int i = 0; i < 2; i++)
        for(int j = 0; j < 2; j++)
            a1[i][j] = c[i][j];
}

int main()
{
    int n;
    while(scanf("%d", &n) && n != -1)
    {
        int a[2][2] = {{1,1}, {1,0}};
        int b[2][2] = {{1,0}, {0,1}};     //初始化为单位矩阵，保存fib的值
        if(n == -1) break;
        while(n)                         //求 a 的 n 次幂
        {
            if(n & 1)
                fun(b, a);
            fun(a, a);
            n >>= 1;
        }
        printf("%d\n", b[1][0]); //最后n必然从1变为0，所以最后一次总要执行 fun(b, a);
    }
    return 0;
}