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最新推荐文章于 2018-12-03 21:11:00 发布

原创最新推荐文章于 2018-12-03 21:11:00 发布 · 341 阅读

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本文介绍了一种使用矩阵快速幂方法解决特定线性关系问题的算法，包括构造矩阵、矩阵相乘和矩阵快速幂运算的过程，并通过实例展示了如何应用此方法求解给定函数的第n项。

Description
You have to find the n^th term of the following function:
f(n)      = a * f(n-1) + b * f(n-3) + c, if(n > 2)
= 0, if(n ≤ 2)
Input
Input starts with an integer T (≤ 100), denoting the number of test cases.
Each case contains four integers n (0 ≤ n ≤ 10⁸),a b c (1 ≤ a, b, c ≤ 10000).
Output
For each case, print the case number and f(n) modulo 10007.
Sample Input
2
10 1 2 3
5 1 3 9
Sample Output
Case 1: 162
Case 2: 27

#include<iostream>
#include<cstdio>
#include<string.h>
#include<string>
#include<stack>
#include<set>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>

#define LL __int64
#define lll unsigned long long
#define MAX 3000009
#define eps 1e-8
#define INF 0xfffffff
#define pi 2*acos(0.0)
#define mod 10007

using namespace std;

const int N = 4;

/*

题意：根据线性关系推到出矩阵关系

解法：矩阵快速幂

构造矩阵：

a 0 b c

1 0 0 0

0 1 0 0

0 0 0 1

初始矩阵：

f(n-1)

f(n-2)

f(n-3)

目标矩阵：

f(n)

f(n-1)

f(n-2)

*/

struct Matrix
{
    int mat[N][N];
};

Matrix mul(Matrix a, Matrix b) //矩阵相乘
{
    Matrix res;
    for(int i = 0; i < N; i++)
        for(int j = 0; j < N; j++)
        {
            res.mat[i][j] = 0;
            for(int k = 0; k < N; k++)
            {
                res.mat[i][j] += a.mat[i][k] * b.mat[k][j];
                res.mat[i][j] %= mod;
            }
        }
    return res;
}

Matrix pow_matrix(Matrix a,Matrix res,int n)  //矩阵快速幂
{
    while(n != 0)
    {
        if(n & 1)
            res = mul(res, a);
        a = mul(a, a);
        n >>= 1;
    }
    return res;
}
int main()
{
    int d = 1;
    int n, a, b, c;
    int T;
    Matrix arr;//构造矩阵
    Matrix unit_matrix;//单位矩阵
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d%d",&n,&a,&b,&c);
        if(n<=2)
        {
            printf("Case %d: ",d++);
            printf("0\n");
            continue;
        }
        memset(arr.mat, 0, sizeof(arr.mat));
        memset(unit_matrix.mat, 0, sizeof(unit_matrix.mat));

        unit_matrix.mat[0][0] = 1;
        unit_matrix.mat[1][1] = 1;
        unit_matrix.mat[2][2] = 1;
        unit_matrix.mat[3][3] = 1;

        arr.mat[0][0] = a;
        arr.mat[0][2] = b;
        arr.mat[0][3] = c;
        arr.mat[1][0] = 1;
        arr.mat[2][1] = 1;
        arr.mat[3][3] = 1;

        Matrix p = pow_matrix(arr, unit_matrix, n-2);

        printf("Case %d: ",d++);
        printf("%d\n",p.mat[0][3]);
    }
    return 0;
}