HDU1005（矩阵快速幂）

最新推荐文章于 2022-07-29 11:41:21 发布

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最新推荐文章于 2022-07-29 11:41:21 发布

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分类专栏：矩阵快速幂

本文链接：https://blog.youkuaiyun.com/West___wind/article/details/52422692

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本文介绍了一种高效计算特定递推数列f(n)的方法，该数列定义为f(1)=1,f(2)=1,f(n)=(A*f(n-1)+B*f(n-2))mod7。通过矩阵快速幂算法，文章展示了如何处理大规模输入n时的计算问题，并给出了具体的C++代码实现。

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Description

A number sequence is defined as follows:

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.

Given A, B, and n, you are to calculate the value of f(n).

Input

The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.

Output

For each test case, print the value of f(n) on a single line.

Sample Input

1 1 3
1 2 10
0 0 0

Sample Output

2
5

Fn+3 Fn+2 Fn+2 Fn+1 a 1

= *

Fn+2 Fn+1 Fn+1 Fn b 0

#include<cstdio>
using namespace std;
int n,a,b;
struct Matrix
{
    int m[2][2];
    void init()
    {
        m[0][0]=a;
        m[0][1]=1;
        m[1][0]=b;
        m[1][1]=0;
    }
    void init0()
    {
        m[0][0]=a+b;
        m[0][1]=1;
        m[1][0]=1;
        m[1][1]=1;
    }
    void init2()
    {
        m[0][0]=1;
        m[0][1]=0;
        m[1][0]=0;
        m[1][1]=1;
    }
};
Matrix multiply(Matrix x,Matrix y)
{
    Matrix res;
    for (int i = 0 ;i < 2; i++)
    {
        for (int j = 0 ;j < 2 ;j++)
        {
            res.m[i][j] = 0;
            for (int k = 0 ;k < 2 ;k++)
                res.m[i][j] += (x.m[i][k] * y.m[k][j]);
            res.m[i][j] %= 7;
        }
    }
    return res;
}
Matrix quickpow(Matrix s,int k)
{
    Matrix res;
    res.init2();
    while(k)
    {
        if(k&1)
            res = multiply(res ,s);
        k >>= 1;
        s = multiply(s ,s);
    }
    return res;
}
int main()
{
    while (scanf("%d%d%d",&a,&b,&n)&&(a||b||n))
    {
        if(n==1||n==2)
        {
            printf("1\n");
            continue;
        }
        Matrix ans;
        ans.init0();
        Matrix t;
        t.init();
        ans = multiply(ans ,quickpow(t ,n-2));
        printf("%d\n",ans.m[0][1] % 7);
    }
    return 0;
}