HDU 1005 Number Sequence(矩阵乘法+快速幂)

Problem 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.

题解

题目大意:

题目意思很好理解,给你一个类似于fabonacci数列的关系式,要求这个数列的第n项。但是注意n的范围,如果直接用数组模拟的话,可能会栈溢出。所以就需要去寻找规律,找到一个周期,这是一个思路。还有一个思路,就是将迭代式转化为矩阵乘法,取模运算,可以用快速幂算法进行优化,这种思路也很巧妙。

矩阵乘法:

迭代式为:f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7
可以将迭代式转化为矩阵乘法形式:

[f(n)f(n1)]=[1110][f(n1)f(n2)] [ f ( n ) f ( n − 1 ) ] = [ 1 1 1 0 ] [ f ( n − 1 ) f ( n − 2 ) ]

[f(n1)f(n2)]=[1110][f(n2)f(n3)] [ f ( n − 1 ) f ( n − 2 ) ] = [ 1 1 1 0 ] [ f ( n − 2 ) f ( n − 3 ) ]

[f(n)f(n1)]=[1110][1110][f(n2)f(n3)] [ f ( n ) f ( n − 1 ) ] = [ 1 1 1 0 ] [ 1 1 1 0 ] [ f ( n − 2 ) f ( n − 3 ) ]

题中的迭代式可转化为:

[f(n)f(n1)]=[A1B0][f(n1)f(n2)] [ f ( n ) f ( n − 1 ) ] = [ A B 1 0 ] [ f ( n − 1 ) f ( n − 2 ) ]

初始条件是f(1) = 1, f(2) = 1,也就转化为求中间矩阵的n-2次方,可以通过快速幂算法来处理幂运算,关于快速幂算法可以参看快速幂

[f(n)f(n1)]=[A1B0](n2)[f(2)f(1)] [ f ( n ) f ( n − 1 ) ] = [ A B 1 0 ] ( n − 2 ) [ f ( 2 ) f ( 1 ) ]

实现代码如下:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
using namespace std;
const int m = 2;
const int mod = 7;
struct Matrix
{
    int x[m][m];
};

// 矩阵乘法
Matrix multiplication(Matrix a, Matrix b)
{
    Matrix temp;
    memset(temp.x, 0, sizeof(temp.x));
    for(int i=0; i<m; i++)
    {
        for(int j=0; j<m; j++)
        {
            for(int k=0; k<m; k++)
            {
                temp.x[i][j] += a.x[i][k] * b.x[k][j];
                temp.x[i][j] %= mod;
            }
        }
    }
    return temp;
}

//矩阵的快速幂
Matrix powmatrix(Matrix matrix, int n)
{
    Matrix temp;
    memset(temp.x, 0, sizeof(temp.x));
    for(int i=0; i<m; i++)
        temp.x[i][i] = 1;     //初始化为单位阵
    while(n)
    {
        if(n%2 == 1)
        {
            temp = multiplication(temp, matrix);
        }
        matrix = multiplication(matrix, matrix);
        n = n/2;
    }
    return temp;
}

int main()
{
    int A, B;
    long long n;
    while((cin>>A>>B>>n) && A!=0 && B!=0 && n!=0)
    {
        Matrix mat;
        // 初始条件
        mat.x[0][0] = A;
        mat.x[0][1] = B;
        mat.x[1][0] = 1;
        mat.x[1][1] = 0;
        mat = powmatrix(mat, n-2);
        printf("%d\n",(mat.x[0][0] + mat.x[0][1])%mod);
    }
    return 0;
}

找规律:

题目给的n的范围比较大,所以要想到去找规律,寻找一个周期出来,然后再通过周期进行处理;题目给的迭代式是mod 7,mod 7的结果最多只有(0,1,2,3,4,5,6)7种,两个mod 7的数组合最多就是7*7=49种,所以循环节最多为49。然后再通过循环找到周期。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
using namespace std;

int main()
{
    int A, B;
    long long n;
    int f[51];
    while((cin>> A >> B >> n) && A!=0 && B!=0 && n!=0)
    {
        f[1] = 1;
        f[2] = 1;
        int i;
        for(i=3; i<51; i++)
        {
            f[i] = A*f[i-1] + B*f[i-2];
            f[i] = f[i] % 7;
            if(f[i] == 1 && f[i-1] == 1)
            {
                break;
            }
        }
        n = n%(i-2);   // i-2为一个周期
        f[0] = f[i-2];
        cout << f[n] <<endl;
    }
    return 0;
}

参考链接:

https://blog.csdn.net/qq_27508477/article/details/47663185
https://blog.csdn.net/rocklee_1227/article/details/7997659

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