CSUOJ 1895 Apache is late again

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

题目大意：给一个n，判断是否存在m使(1 + sqrt (2)) ^ n= sqrt (m) + sqrt (m-1)先列前面几项：(1 + sqrt (2)) ^1 =sqrt (1*1+1) + sqrt (1)(1 + sqrt (2)) ^ 2=sqrt (3*3) + sqrt (8)(1 + sqrt (2)) ^ 3=sqrt (7*7+1) + sqrt (49)(1 + sqrt (2)) ^ 4=sqrt (17*17) + sqrt (288)...

an = 2 * an-1 +an-2可以推出当n为奇数时m=an*an+1n为偶数时 m=an*an

an可以用矩阵来表示

[an an-1]T =[ (2 1) (1 0)]*[an-2 an-1]T=[(2 1) (1 0)]^n-2 *[a2 a1]

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
const int MOD = 100000007;
typedef long long ll;
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 E, M, ans;
void init()
{
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 quick_pow(matrix x, ll y)
{
matrix tmp = E;
while (y)
{
if (y & 1)
{
tmp =tmp* x;
y--;
}
y >>= 1;
x = x*x;
}
return tmp;
}
int main()
{
ll n;
init();
while (~scanf("%lld", &n))
{
if (n < 0)
printf("No\n");
else if (n == 0)
printf("1\n");
else if (n == 1)
printf("2\n");
else if (n == 2)
printf("9\n");
else
{
ans = quick_pow(M, n - 2);
ll a = (ans.v[0][0] * 3 + ans.v[0][1]) % MOD;
if (n & 1)
printf("%lld\n", ((a*a)%MOD + 1) % MOD);
else
printf("%lld\n", (a*a)%MOD);
}
}
return 0;
}
/**********************************************************************
Problem: 1895
User: leo6033
Language: C++
Result: AC
Time:8 ms
Memory:2024 kb
**********************************************************************/