Codeforces 185A Plant

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本文介绍了一道 Codeforces 的竞赛题目，通过矩阵快速幂的方法来高效计算一种特殊三角形植物随时间增长的数量。具体地，这种植物每年会分裂成四个新的植物，其中三个朝向与母体相同，一个相反。文章提供了 C++ 代码实现，并解释了如何求解任意年份指向特定方向的植物数量。

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Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.

$\text{[math]}$

Help the dwarfs find out how many triangle plants that point "upwards" will be in n years.

Input

The first line contains a single integer n (0 ≤ n ≤ 10¹⁸) — the number of full years when the plant grew.

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.

Output

Print a single integer — the remainder of dividing the number of plants that will point "upwards" in n years by 1000000007 (10⁹ + 7).

Sample test(s)

Input

Output

Input

Output

Note

The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.

这题很简单，矩阵乘法。

很显然

正n+1=3*正n+倒n

倒n+1=3*倒n+正n

代码

#include <stdio.h>

#define MOD 1000000007

typedef struct
{
    __int64 a[2][2];
}Matrix;

Matrix a,b;

Matrix Mul(Matrix x,Matrix y)
{
    int i,j,k;
    Matrix z;
    for (i=0;i<2;i++)
    {
        for (j=0;j<2;j++)
        {
            z.a[i][j]=0;
            for (k=0;k<2;k++)
            {
                z.a[i][j]=(z.a[i][j]+x.a[i][k]*y.a[k][j])%MOD;
            }
        }
    }
    return z;
}

int main()
{
    int i,j;
    __int64 n;
    scanf("%I64d",&n);
    a.a[0][0]=a.a[1][1]=3;
    a.a[1][0]=a.a[0][1]=1;
    b.a[0][0]=1;
    b.a[1][0]=0;
    while(n)
    {
        if (n&1) b=Mul(a,b);
        a=Mul(a,a);
        n>>=1;
    }
    printf("%I64d\n",b.a[0][0]);
    return 0;
}