nyist 148 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

解题方法：
使用矩阵的来解决
1,1
1,0
的n次幂就能得到我们要的结果了
证明和具体的思路课参照网上的资料

下面是实现代码

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

#define N 10000

typedef struct matrix
{
    long long a[2][2];
    matrix()
    {
        memset(a,0,sizeof(a));
    }
}matrix;

void  chenfa(matrix x,matrix y,matrix &z)
{

        z.a[0][0] = (x.a[0][0]*y.a[0][0] + x.a[0][1]*y.a[1][0]) % N;
        z.a[0][1] = (x.a[0][0]*y.a[0][1] + x.a[0][1]*y.a[1][1]) % N;
        z.a[1][0] = (x.a[1][0]*y.a[0][0] + x.a[1][1]*y.a[1][0]) % N;
        z.a[1][1] = (x.a[1][0]*y.a[0][1] + x.a[1][1]*y.a[1][1]) % N;
}

void kuaishu(matrix x,matrix &z,long long n)
{
    z.a[0][0] = 1;
    z.a[0][1] = 0;
    z.a[1][0] = 0;
    z.a[1][1] = 1;

    while(n)
    {
        if(n % 2 == 1)
            chenfa(z,x,z);
        n /= 2;
        chenfa(x,x,x);
    }

}

/*void print(matrix z)
{
    int i = 0;
    int j = 0;
    for(i = 0; i< 2; i++)
        {
            for(j = 0; j< 2;j++)
                printf("%d ",z.a[i][j]);
                printf("\n");
        }
}
*/
int main()
{
    long long n;
    while(1)
    {
        scanf("%lld",&n);
        if(n == -1)
            break;

        if(n < 3)
        {
            if(n == 0)
                printf("0\n");
            else
            printf("1\n");
        }
        else
        {
            matrix z;
            matrix x;
            x.a[0][0] = 1;
            x.a[0][1] = 1;
            x.a[1][0] = 1;
            x.a[1][1] = 0;
            kuaishu(x,z,n);
            //print(z);
            printf("%lld\n",z.a[0][1]);
        }
    }
    return 0;
}