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描述
如果一个二进制数包含连续的两个1，我们就称这个二进制数是非法的。
小Hi想知道在所有 n 位二进制数（一共有2n个）中，非法二进制数有多少个。
例如对于 n = 3,有 011, 110, 111 三个非法二进制数。
由于结果可能很大，你只需要输出模109+7的余数。
输入
一个整数 n (1 ≤ n ≤ 100)。
输出
n 位非法二进制数的数目模109+7的余数。
样例输入
3
样例输出
3
思路
首先用python打表，把符合条件的数前20个找到。然后BM板子过去或者。找规律，一个数的值等于他的前一项的值*2+一个斐波那契数
代码
py打表:
def get0(n):
ans = 0
for i in range(2**n):
s = bin(i)[2:].zfill(n)
for i in range(1, len(s)):
if s[i] == '1' and s[i-1] == '1':
ans += 1
break
return ans
for i in range(1, 20):
print(i, get0(i))
bm算法暴力过:
#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()
{
ll t, n;
scanf("%lld", &n);
printf("%lld\n", linear_seq::gao(VI{0, 1, 3, 8, 19, 43, 94, 201, 423, 880}, n - 1));
return 0;
}
找规律:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9 + 7;
ll a[110], b[110];
int main()
{
ll n;
scanf("%lld", &n);
a[1] = 0, a[2] = 1, a[3] = 3;
b[2] = 1, b[3] = 1;
for (ll i = 4; i <= n; i++)
{
b[i] = (b[i - 1] + b[i - 2]) % mod;
a[i] = (a[i - 1] * 2 + b[i]) % mod;
}
printf("%lld\n", a[n]);
return 0;
}