集训第六周矩阵快速幂 K题

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

Sample Output

1.使用一个结构体存下矩阵，再写一个二维矩阵乘法函数

2.然后求[1 1 1 0]的n次方？当然不是。

注意：0 ≤ n ≤ 1,000,000,000

如果这样直接乘以n次肯定会超时

可以使用二进制求快速幂

利用二进制求指数幂

举例：

3 ^ 999 = 3 * 3 * 3 * … * 3

直接乘要做998次乘法。但事实上可以这样做，先求出2^k次幂：

3 ^ 2 = 3 * 3

3 ^ 4 = (3 ^ 2) * (3 ^ 2)

3 ^ 8 = (3 ^ 4) * (3 ^ 4)

3 ^ 16 = (3 ^ 8) * (3 ^ 8)

3 ^ 32 = (3 ^ 16) * (3 ^ 16)

3 ^ 64 = (3 ^ 32) * (3 ^ 32)

3 ^ 128 = (3 ^ 64) * (3 ^ 64)

3 ^ 256 = (3 ^ 128) * (3 ^ 128)

3 ^ 512 = (3 ^ 256) * (3 ^ 256)

再相乘：

3 ^ 999

= 3 ^ (512 + 256 + 128 + 64 + 32 + 4 + 2 + 1)

= (3 ^ 512) * (3 ^ 256) * (3 ^ 128) * (3 ^ 64) * (3 ^ 32) * (3 ^ 4) * (3 ^ 2) * 3

把999转为2进制数：1111100111，其个位就是要乘的数。

1 pow ← 1

2 while (n > 0)

3 do if (n mod 2 = 1)

4 then pow ← pow * x

5 x ← x * x

6 n ← n / 2

7 return pow

#include"iostream"

#include"cstdio"

using namespace std;

typedef struct

{

    int m[][];

}node;

node work(node a,node b)

{

    node c;

    c.m[][]=(a.m[][]*b.m[][]+a.m[][]*b.m[][])%;

    c.m[][]=(a.m[][]*b.m[][]+a.m[][]*b.m[][])%;

    c.m[][]=(a.m[][]*b.m[][]+a.m[][]*b.m[][])%;

    c.m[][]=(a.m[][]*b.m[][]+a.m[][]*b.m[][])%;

    return c;

}

void caculate(int c)

{

    node ans,base;

    base.m[][]=base.m[][]=base.m[][]=;

    base.m[][]=;

    ans.m[][]=ans.m[][]=;

    ans.m[][]=ans.m[][]=;

    while(c)

    {

        if(c&) ans=work(ans,base);

        base=work(base,base);

        c>>=;

    }

    cout<<ans.m[][]<<endl;

}

int main()

{

    int n;

    while(cin>>n&&n>=)

    {

        caculate(n);

    }

}

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集训第六周矩阵快速幂 K题

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