HDU 3493 The Little Architect【矩阵快速幂

最新推荐文章于 2021-10-22 12:29:51 发布

Puya

最新推荐文章于 2021-10-22 12:29:51 发布

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本文探讨了一种算法挑战，即使用给定数量的立方体积木构建稳定的建筑物。通过定义稳定性并给出示例，文章进一步介绍了一种计算不同稳定建筑数量的方法，并提供了一个具体的实现代码。

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The Little Architect

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 235 Accepted Submission(s): 131

Problem Description

LY is a little architect, he likes to make all kinds of buildings with his building blocks. All the building blocks are the same cubes. Now he want to make a new building. He don’t want any floor of his building is disconnected. So sample of the building he likes and he dislikes like these:

The building he likes. The building he dislikes.

For example, the first building he likes because all the blocks in every line are not disconnected, while the second building he dislikes because in line 1, 3, 4, the blocks are disconnected.
Now LY has a question. If he has n building blocks, and how many different buildings he can build. We ensure that all the building blocks are in the same plane.

Input

There are several lines in the input. Each line is a set of input data. There is only one integer n (0 < n < 1,000,000,000) means the number of blocks buildings he has. The last line of the input is a line contains 0.

Output

For each set of input, output a single line contains one integer which is the number of different buildings he can build mod 9997.

Sample Input

Sample Output

  
  
   
   1
2
6
19
2840



   
   
    
    
     
     Hint
    
    When n = 3, there are 6 different buildings like these:

Source

2010 ACM-ICPC Multi-University Training Contest（6）——Host by BIT

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<climits>
#include<string>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;
#define rep(i,j,k)for(i=j;i<k;i++)
#define per(i,j,k)for(i=j;i>k;i++)
#define MS(x,y)memset(x,y,sizeof(x))
const int INF= 0x7FFFFFFF;
typedef  long long  LL;
const int MAX = 3;

typedef struct
{
    int m[MAX][MAX];
} Matrix;

Matrix P = {5,-7,4,
            1,0,0,
            0,1,0,
           };

Matrix I = {1,0,0,
            0,1,0,
            0,0,1,
           };

Matrix matrixmul(Matrix a,Matrix b) 
{
    int i,j,k;
    Matrix c;
    for (i = 0 ; i < MAX; i++)
        for (j = 0; j < MAX; j++)
        {
            c.m[i][j] = 0;
            for (k = 0; k < MAX; k++)
                c.m[i][j] += (a.m[i][k] * b.m[k][j])%9997;
            c.m[i][j] %= 9997;
        }
    return c;
}

Matrix quickpow(long long n)
{
    Matrix m = P, b = I;
    while (n >= 1)
    {
        if (n & 1)
            b = matrixmul(b,m);
        n = n >> 1;
        m = matrixmul(m,m);
    }
    return b;
}

int main()
{
    Matrix re;
    int f[3] = {2,6,19};
    long long n;
    while (scanf("%I64d",&n) && n != 0)
    {
        if (n == 1)
            printf("1\n");
        else if (n <= 4)
            printf("%d\n",f[n-2]);
        else
        {
            re = quickpow(n - 4);
            printf("%d\n",(((re.m[0][0]*f[2])+ (re.m[0][1]*f[1]) + (re.m[0][2]*f[0])) %9997 + 9997) % 9997);
        }
    }
    return 0;
}