假设硬币被抛了N次。任务是找到抛掷序列的总数，以使从左起第一个头部之后，右侧的所有交替位置仅由头部占据。除交替位置外，其他任何位置均可被头部或尾部占据。
例如，如果您抛硬币10次(N = 10)，并且第一个头部出现在第3个位置，那么右边的所有其他交替位置是5、7、9…
例子：
Input: N = 2
Output: 4
All possible combinations will be TT, TH, HT, HH.
Input: N = 3
Output: 6
All possible combination will be TTT, TTH, THT, THH, HTH, HHH.
In this case, HHT & HTT is not possible because in this combination
condition of alternate heads does not satisfy. Hence answer will be 6.
方法：
如果序列以尾部开始，则长度为N-1的所有可能序列均有效。现在，如果序列从头开始，则序列中的每个奇数索引(假设序列基于1)将为头，其余N / 2位可以为头或尾。因此，上述问题的递归公式为：
f(N) = f(N-1) + 2floor(N/2)数学计算：
Let fo(N) and fe(N) be the functions
that are defined for the odd and even values of N respectively.
fo(N) = fe(N-1) + 2(N-1)/2
fe(N) = fo(N-1) + 2N/2
From above equation compute
fo(N) = fo(N-2) + 2(N-1)/2 + 2(N-1)/2
fe(N) = fe(N-2) + 2N/2 + 2(N-2)/2
Base Case:
fo(1) = 2
fe(0) = 1
By using the above equation, compute the following results :
fo(N) - fo(N-2) = 2(N-1)/2 + 2(N-1)/2
fo(N) - fo(N-2) = 2(N+1)/2
By taking the sum of above equation for all odd values of N,
below thing is computed.
fo(N) - fo(N-2) + fo(N-1) - fo(N-3) + ...... + fo(3) - fo(1) =
22 + 23 + 24 + ..... + 2(N+1)/2
Hence on summation,
fo(N) - fo(1) = SUM[ n = 0 to (N+1)/2 ] 2n - 21 - 20
By using sum of geometric progression
fo(N) = 2( N + 1 ) / 2 + 2( N + 1 ) / 2 - 2
Similarly, find fe(N) :
fe(N) = fe(N-2) + 2N/2 + 2(N-2)/2
fe(N) - fe(0) = SUM[ n = 0 to N/2 ] 2n - 20 - SUM[ n = 0 to (N-1)/2 ] 2n
By using sum of geometric progression
fe(N) = 2 (N / 2) + 1 + 2 N / 2 - 2最终公式如下：
f(N) = 2(N+1)/2 + 2(N+1)/2 – 2, if N is odd
f(N) = 2(N/2) + 1 + 2N/2 – 2, if N is even
下面是上述方法的实现：
C++
// C++ program to find number of sequences
#include
using namespace std;
// function to calculate total sequences possible
int findAllSequence(int N)
{
// Value of N is even
if (N % 2 == 0) {
return pow(2, N / 2 + 1) + pow(2, N / 2) - 2;
}
// Value of N is odd
else {
return pow(2, (N + 1) / 2) + pow(2, (N + 1) / 2) - 2;
}
}
// Driver code
int main()
{
int N = 2;
cout << findAllSequence(N) << endl;
return 0;
} Java
// Java program to find
// number of sequences
import java.io.*;
class GFG
{
// function to calculate
// total sequences possible
static int findAllSequence(int N)
{
// Value of N is even
if (N % 2 == 0)
{
return (int)(Math.pow(2, N / 2 + 1) +
Math.pow(2, N / 2) - 2);
}
// Value of N is odd
else
{
return (int)(Math.pow(2, (N + 1) / 2) +
Math.pow(2, (N + 1) / 2) - 2);
}
}
// Driver code
public static void main (String[] args)
{
int N = 2;
System.out.print( findAllSequence(N));
}
}
// This code is contributed
// by anuj_67.Python3
# Python3 program to find number
# of sequences
# function to calculate total
# sequences possible
def findAllSequence(N):
# Value of N is even
if (N % 2 == 0):
return (pow(2, N / 2 + 1) +
pow(2, N / 2) - 2);
# Value of N is odd
else:
return (pow(2, (N + 1) / 2) +
pow(2, (N + 1) / 2) - 2);
# Driver code
N = 2;
print(int(findAllSequence(N)));
# This code is contributed by mitsC#
// C# program to find
// number of sequences
using System;
public class GFG{
// function to calculate
// total sequences possible
static int findAllSequence(int N)
{
// Value of N is even
if (N % 2 == 0)
{
return (int)(Math.Pow(2, N / 2 + 1) +
Math.Pow(2, N / 2) - 2);
}
// Value of N is odd
else
{
return (int)(Math.Pow(2, (N + 1) / 2) +
Math.Pow(2, (N + 1) / 2) - 2);
}
}
// Driver code
public static void Main ()
{
int N = 2;
Console.WriteLine( findAllSequence(N));
}
}
// This code is contributed by 29AjayKumarPHP
输出：
4
时间复杂度：O(LogN)