hdoj 4291 A Short problem 【找循环节 + 矩阵快速幂】

最新推荐文章于 2019-02-21 14:45:09 发布

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最新推荐文章于 2019-02-21 14:45:09 发布

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分类专栏：矩阵

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本文介绍了一种通过矩阵快速幂解决特定递归问题的方法，并给出了详细的实现步骤与代码示例。

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A Short problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2333 Accepted Submission(s): 817

Problem Description

　　According to a research, VIM users tend to have shorter fingers, compared with Emacs users.
　　Hence they prefer problems short, too. Here is a short one:
　　Given n (1 <= n <= 10¹⁸), You should solve for
g(g(g(n))) mod 10⁹ + 7
　　where
g(n) = 3g(n - 1) + g(n - 2)
g(1) = 1
g(0) = 0

Input

　　There are several test cases. For each test case there is an integer n in a single line.
　　Please process until EOF (End Of File).

Output

　　For each test case, please print a single line with a integer, the corresponding answer to this case.

Sample Input

0
1
2

Sample Output

0
1
42837

题意不说了。。。

思路：打表找出每一层循环节，然后矩阵快速幂。

打表找循环节代码：

#include <cstdio>
#define LL long long
#define MOD1 1000000007
#define MOD2 222222224
#define MOD3 183120
int main()
{
    LL x = 0, y = 1;
    for(int i = 1; ; i++)
    {
        LL x1 = 3*y%MOD2 + x;
        x1 %= MOD2;
        x = y, y = x1;
        if(x == 0 && y == 1)
        {
            printf("%d\n", i);
            break;
        }
    }
    return 0;
}

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
#define MOD1 1000000007
#define MOD2 222222224
#define MOD3 183120
using namespace std;
struct Matrix
{
    LL a[3][3];
};
Matrix ori, res;
void init()
{
    memset(ori.a, 0, sizeof(ori.a));
    memset(res.a, 0, sizeof(res.a));
    ori.a[0][1] = ori.a[1][0] = 1;
    ori.a[1][1] = 3;
    for(int i = 0; i < 2; i++)
        res.a[i][i] = 1;
}
Matrix muitl(Matrix x, Matrix y, LL M)
{
    Matrix z;
    memset(z.a, 0, sizeof(z.a));
    for(int i = 0; i < 2; i++)
    {
        for(int k = 0; k < 2; k++)
        {
            if(x.a[i][k] == 0) continue;
            for(int j = 0; j < 2; j++)
                z.a[i][j] = (z.a[i][j] + (x.a[i][k] * y.a[k][j]) % M) % M;
        }
    }
    return z;
}
LL ans, F[3];
LL solve(LL n, LL M)
{
    if(n <= 1)
        return n;
    init();
    n -= 1;//自减一
    while(n)
    {
        if(n & 1)
            res = muitl(ori, res, M);
        ori = muitl(ori, ori, M);
        n >>= 1;
    }
    return F[1] * res.a[1][1] % M;
}
int main()
{
    F[0] = 0, F[1] = 1;
    LL N;
    while(scanf("%lld", &N) != EOF)
    {
        ans = solve(N, MOD3);//最里层
        ans = solve(ans, MOD2);//第二层
        ans = solve(ans, MOD1);//最外层
        printf("%lld\n", ans);
    }
    return 0;
}