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Problem Description
There’s an array that is generated by following rule.
And let us define two arrays bnandan as below.
Now, you have to print ⌊√(an)⌋ , n>1.
Your answer could be very large so print the answer modular 1000000007.
Input
The first line of input contains T (1 <= T <= 1000) , the number of test cases.
Each test case contains one integer n (1 < n <= 1015) in one line.
Output
For each test case print ⌊√(a_n )⌋ modular 1000000007.
Sample Input
3
4
7
9Sample Output
1255
324725
13185773思路
题意很简单，给了一个 的递推式，给出一个 ,求
正解好像是矩阵快速幂，不过这次用的是杜教的BM线性递推方法,全称Berlekamp Massey 算法
关于该算法的介绍可以看:算法 - 齐次线性递推求解和优化
简单的来说，就是把一个序列的前几项放进这个模板，然后他会自动递推出后面的很多项，放的越多越准确，一般来说至少放8项。。
所以这个题我们可以先用py打个表，然后放进这个板子，然后就做完了…(tql)
代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long long ll;
typedef vector<int> VI;
const int maxn = 10005;
const ll mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const double eps = 1e-9;
ll fast_mod(ll a, ll n, ll Mod)
{
ll ans = 1;
a %= Mod;
while (n)
{
if (n & 1)
ans = (ans * a) % Mod;
a = (a * a) % Mod;
n >>= 1;
}
return ans;
}
namespace linear_seq
{
ll res[maxn], base[maxn], num[maxn], md[maxn]; //数组大小约10000
vector<int> vec;
void mul(ll *a, ll *b, int k)
{
for (int i = 0; i < 2 * k; i++)
num[i] = 0;
for (int i = 0; i < k; i++)
{
if (a[i])
{
for (int j = 0; j < k; j++)
{
num[i + j] = (num[i + j] + a[i] * b[j]) % mod;
}
}
}
for (int i = 2 * k - 1; i >= k; i--)
{
if (num[i])
{
for (int j = 0; j < vec.size(); j++)
{
num[i - k + vec[j]] = (num[i - k + vec[j]] - num[i] * md[vec[j]]) % mod;
}
}
}
for (int i = 0; i < k; i++)
a[i] = num[i];
}
ll solve(ll n, VI a, VI b)
{
ll ans = 0, cnt = 0;
int k = a.size();
assert(a.size() == b.size());
for (int i = 0; i < k; i++)
md[k - 1 - i] = -a[i];
md[k] = 1;
vec.clear();
for (int i = 0; i < k; i++)
if (md[i])
vec.push_back(i);
for (int i = 0; i < k; i++)
res[i] = base[i] = 0;
res[0] = 1;
while ((1LL << cnt) <= n)
cnt++;
for (int p = cnt; p >= 0; p--)
{
mul(res, res, k);
if ((n >> p) & 1)
{
for (int i = k - 1; i >= 0; i--)
res[i + 1] = res[i];
res[0] = 0;
for (int j = 0; j < vec.size(); j++)
{
res[vec[j]] = (res[vec[j]] - res[k] * md[vec[j]]) % mod;
}
}
}
for (int i = 0; i < k; i++)
ans = (ans + res[i] * b[i]) % mod;
if (ans < 0)
ans += mod;
return ans;
}
VI BM(VI s)
{
VI B(1, 1), C(1, 1);
int L = 0, m = 1, b = 1;
for (int i = 0; i < s.size(); i++)
{
ll d = 0;
for (int j = 0; j < L + 1; j++)
d = (d + (ll)C[j] * s[i - j]) % mod;
if (d == 0)
m++;
else if (2 * L <= i)
{
VI T = C;
ll c = mod - d * fast_mod(b, mod - 2, mod) % mod;
while (C.size() < B.size() + m)
C.push_back(0);
for (int j = 0; j < B.size(); j++)
C[j + m] = (C[j + m] + c * B[j]) % mod;
L = i + 1 - L, B = T, b = d, m = 1;
}
else
{
ll c = mod - d * fast_mod(b, mod - 2, mod) % mod;
while (C.size() < B.size() + m)
C.push_back(0);
for (int j = 0; j < B.size(); j++)
C[j + m] = (C[j + m] + c * B[j]) % mod;
m++;
}
}
return C;
}
int gao(VI a, ll n)
{
VI c = BM(a);
c.erase(c.begin());
for (int i = 0; i < c.size(); i++)
c[i] = (mod - c[i]) % mod;
return solve(n, c, VI(a.begin(), a.begin() + c.size()));
}
} // namespace linear_seq
int main()
{
//填数字的时候带上模数之后的
ll t, n;
scanf("%d", &t);
while (t--)
{
scanf("%lld", &n);
printf("%lld\n", linear_seq::gao(VI{31, 197, 1255, 7997, 50959, 324725, 2069239, 13185773, 84023455, 535421093, 411853810}, n - 2));
}
return 0;
}打表代码:
import math
mod = 1000000007
h = [0 for i in range(1000)]
b = [0 for i in range(1000)]
a = [0 for i in range(1000)]
h[0] = 2
h[1] = 3
h[2] = 6
for n in range(3, 100):
h[n] = 4*h[n-1]+17*h[n-2]-12*h[n-3]-16
for n in range(1, 100):
b[n] = 3*h[n+1]*h[n]+9*h[n+1]*h[n-1]+9*h[n]*h[n] + \
27*h[n]*h[n-1]-18*h[n+1]-126*h[n]-81*h[n-1]+192
for n in range(1, 100):
a[n] = b[n]+4**n
for i in range(2, 15):
num = int(math.sqrt(a[i])) % mod
print(num, end=',')