假设有N个人坐在一个从1到N的循环队列中,那么任务就是计算选择其中一个子集的方式的数目,这样就不会有两个连续的人坐在一起。答案可能很大,因此请以10 9 + 7为模计算答案。请注意,空子集也是有效的子集。
例子:
Input: N = 2
Output: 3
All possible subsets are {}, {1} and {2}.
Input: N = 3
Output: 4
方法:让我们找到N的较小值的答案。
N = 1- >所有可能的子集为{},{1} 。
N = 2- >所有可能的子集是{},{1},{2} 。
N = 3- >所有可能的子集为{},{1},{2},{3} 。
N = 4- >所有可能的子集为{},{1},{2},{3},{4},{1、3},{2、4} 。
因此顺序将是2、3、4、7…
当N = 5时,计数将为11 ;如果N = 6,则计数将为18 。
现在可以观察到,该序列类似于从第二项开始的斐波那契数列,前两项为3和4。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
#define ll long long
const ll N = 10000;
const ll MOD = 1000000007;
ll F[N];
// Function to pre-compute the sequence
void precompute()
{
// For N = 1 the answer will be 2
F[1] = 2;
// Starting two terms of the sequence
F[2] = 3;
F[3] = 4;
// Comute the rest of the sequence
// with the relation
// F[i] = F[i - 1] + F[i - 2]
for (int i = 4; i < N; i++)
F[i] = (F[i - 1] + F[i - 2]) % MOD;
}
// Driver code
int main()
{
int n = 8;
// Pre-compute the sequence
precompute();
cout << F[n];
return 0;
}
Java
// Java implementation of the approach
class GFG
{
static int N = 10000;
static int MOD = 1000000007;
static int []F = new int[N];
// Function to pre-compute the sequence
static void precompute()
{
// For N = 1 the answer will be 2
F[1] = 2;
// Starting two terms of the sequence
F[2] = 3;
F[3] = 4;
// Comute the rest of the sequence
// with the relation
// F[i] = F[i - 1] + F[i - 2]
for (int i = 4; i < N; i++)
F[i] = (F[i - 1] + F[i - 2]) % MOD;
}
// Driver code
public static void main(String []args)
{
int n = 8;
// Pre-compute the sequence
precompute();
System.out.println(F[n]);
}
}
// This code is contributed by PrinciRaj1992
Python3
# Python implementation of the approach
N = 10000;
MOD = 1000000007;
F = [0] * N;
# Function to pre-compute the sequence
def precompute():
# For N = 1 the answer will be 2
F[1] = 2;
# Starting two terms of the sequence
F[2] = 3;
F[3] = 4;
# Comute the rest of the sequence
# with the relation
# F[i] = F[i - 1] + F[i - 2]
for i in range(4,N):
F[i] = (F[i - 1] + F[i - 2]) % MOD;
# Driver code
n = 8;
# Pre-compute the sequence
precompute();
print(F[n]);
# This code is contributed by 29AjayKumar
C#
// C# implementation of the approach
using System;
class GFG
{
static int N = 10000;
static int MOD = 1000000007;
static int []F = new int[N];
// Function to pre-compute the sequence
static void precompute()
{
// For N = 1 the answer will be 2
F[1] = 2;
// Starting two terms of the sequence
F[2] = 3;
F[3] = 4;
// Comute the rest of the sequence
// with the relation
// F[i] = F[i - 1] + F[i - 2]
for (int i = 4; i < N; i++)
F[i] = (F[i - 1] + F[i - 2]) % MOD;
}
// Driver code
public static void Main(String []args)
{
int n = 8;
// Pre-compute the sequence
precompute();
Console.WriteLine(F[n]);
}
}
// This code is contributed by 29AjayKumar
输出:
47