中南大学第十一届大学生程序设计竞赛-COJ1895-Apache is late again

1895: Apache is late again

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Description
Apache is a student of CSU. There is a math class every Sunday morning, but he is a very hard man who learns late every night. Unfortunate, he was late for maths on Monday. Last week the math teacher gave a question to let him answer as a punishment, but he was easily resolved. So the math teacher prepared a problem for him to solve. Although Apache is very smart, but also was stumped. So he wants to ask you to solve the problem. Questions are as follows: You can find a m made (1 + sqrt (2)) ^ n can be decomposed into sqrt (m) + sqrt (m-1), if you can output m% 100,000,007 otherwise output No.

Input
There are multiply cases. Each case is a line of n. (|n| <= 10 ^ 18)

Output
Line, if there is no such m output No, otherwise output m% 100,000,007.

Sample Input
2
Sample Output
9
Hint
Source
中南大学第十一届大学生程序设计竞赛

题目大意：给你一个n，让你计算能否找到一个m使得
$(1+\sqrt{2})^n=\sqrt{m}+\sqrt{m-1}$
解题思路：先写出前几项
$(1+\sqrt{2})^1=\sqrt{2}+\sqrt{1}=\sqrt{1^2+1}+\sqrt{1^2}$
$(1+\sqrt{2})^2=\sqrt{9}+\sqrt{8}=\sqrt{3^2}+\sqrt{3^2-1}$
$(1+\sqrt{2})^3=\sqrt{50}+\sqrt{49}=\sqrt{7^2+1}+\sqrt{7^2}$
$(1+\sqrt{2})^4=\sqrt{289}+\sqrt{288}=\sqrt{17^2}+\sqrt{17^2-1}$
$(1+\sqrt{2})^5=\sqrt{1682}+\sqrt{1681}=\sqrt{41^2+1}+\sqrt{41^2}$
$\dots$
设数列：
$a_1=1,a_2=3,a_3=7,a_4=17,a_5=41,\cdots$
$a_n=2*a_{n-1}+a_{n-2}$
所以当n<0时，输出”No”；
当n>0时：
若n为奇数，$m=(a_n)^2+1$
若n为偶数，$m=(a_n)^2$
$\begin{bmatrix} a_n \\ a_{n-1} \\ \end{bmatrix}=\begin{bmatrix} 2&1 \\1&0 \end{bmatrix}\begin{bmatrix} a_{n-1}\\ a_{n-2}\end{bmatrix}={\begin{bmatrix} 2&1\\ 1&0\end{bmatrix} }^{n-2}\begin{bmatrix}a_2\\ a_1 \end{bmatrix}$
用矩阵快速幂加速
考察内容：数学，快速幂
复杂度： O(logn)
题目难度： ★★★★

#include<iostream>
#include<cstring>
using namespace std;
typedef long long LL;
const int MOD=1e8+7;

struct matrix
{
    LL v[2][2];
    matrix(){memset(v,0,sizeof(v));}
    matrix operator * (const matrix &m)
    {
        matrix c;
        for(int i=0;i<2;i++)
            for(int j=0;j<2;j++)
                for(int k=0;k<2;k++)
                    c.v[i][j]+=(v[i][k]*m.v[k][j])%MOD;
        return c;
    }
};

matrix M,E,ans;//E为单位矩阵,ans为M^(n-2)

void Init()//初始化
{
    for(int i=0;i<2;i++)
        E.v[i][i]=1;
    M.v[0][0]=2;M.v[0][1]=1;
    M.v[1][0]=1;M.v[1][1]=0;
}

matrix pow(matrix p,LL k)
{
    matrix tmp=E;
    while(k)
    {
        if(k&1)
        {
            tmp=tmp*p;
            k--;
        }
        k>>=1;
        p=p*p;
    }
    return tmp;
}

int main()
{
    ios::sync_with_stdio(false);
    LL n;
    Init();
    while(cin>>n)
    {
        if(n<0) cout<<"No"<<endl;
        else if(n==0) cout<<"1"<<endl;
        else if(n==1) cout<<"2"<<endl;
        else if(n==2) cout<<"9"<<endl;
        else
        {
            ans=pow(M,n-2);
            LL an=(ans.v[0][0]*3+ans.v[0][1]*1)%MOD;
            if(n&1) cout<<((an*an)%MOD+1)%MOD<<endl;
            else cout<<(an*an)%MOD<<endl;
        }
    }
    return 0;
}