Happy Necklace hdu 6030 （BM算法）
Little Q wants to buy a necklace for his girlfriend. Necklaces are single strings composed of multiple red and blue beads.
Little Q desperately wants to impress his girlfriend, he knows that she will like the necklace only if for every prime length continuous subsequence in the necklace, the number of red beads is not less than the number of blue beads.
Now Little Q wants to buy a necklace with exactly n beads. He wants to know the number of different necklaces that can make his girlfriend happy. Please write a program to help Little Q. Since the answer may be very large, please print the answer modulo 109+7
.
Note: The necklace is a single string, {not a circle}.
Input
The first line of the input contains an integer T(1≤T≤10000)
, denoting the number of test cases.
For each test case, there is a single line containing an integer n(2≤n≤1018)
, denoting the number of beads on the necklace.
Output
For each test case, print a single line containing a single integer, denoting the answer modulo 109+7
.
Sample Input
2 2 3
Sample Output
3 4
#include
using namespace std;
typedef long long ll;
typedef long long ll;
typedef vector 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 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);
n--;
printf("%lld\n", linear_seq::gao(VI{3, 4, 6, 9, 13, 19, 28, 41, 60}, n-1));
}
return 0;
}