给定数字N。任务是将所有数字从1到N进行分组，因此每组中所有数字的GCD为1，因为必须将组的数量减到最少。
例子：
Input: N = 3
Output:
1 2 3
Explanation:
gcd(1, 2, 3) = 1
Input: N = 6
Output:
1 2
3 4
5 6
Explanation:
gcd(1, 2) = 1
gcd(3, 4) = 1
gcd(5, 6) = 1
方法：我们知道每个连续数字的GCD为1的事实，因此我们将使用此事实来解决此问题。步骤如下：
- 如果数字为3，则GCD(1、2、3)只能为1的一组{1,2,3} 。
- 在[1，N / 2]上进行其他迭代(说i ) ，并使用(2 * i – 1，2 * i)形成组，并将其存储在向量数组中(说arr [] [] )。
- 存储在arr [] []中的组是所需的组。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to print the groups
void printGroups(vector A[], int N)
{
// Interate over [1, N/2];
for (int i = 1; i <= N / 2 + 1; i++) {
// Print all pairs
for (int j = 0; j < A[i].size(); j++) {
cout << A[i][j] << ' ';
}
cout << endl;
}
}
// Function to form a groups with GCD 1
void formGroups(int N)
{
int i;
// Corner case
if (N == 3) {
cout << "1 2 3";
return;
}
// Array of vectors to store the
// possible pairs
vector arr[N / 2 + 2];
for (i = 1; i <= N / 2; i++) {
// Push the pair (2*i - 1, 2*i)
arr[i].push_back(2 * i - 1);
arr[i].push_back(2 * i);
}
// If N is odd then push the
// last element N
if (N & 1) {
arr[i].push_back(N);
}
// Function Call
printGroups(arr, N);
return;
}
// Driver Code
int main()
{
int N = 10;
// Function Call
formGroups(N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to print the groups
static void printGroups(Vector A[],
int N)
{
// Interate over [1, N/2];
for(int i = 1; i <= N / 2 + 1; i++)
{
// Print all pairs
for(int j = 0; j < A[i].size(); j++)
{
System.out.print(A[i].get(j) + " ");
}
System.out.println();
}
}
// Function to form a groups with GCD 1
static void formGroups(int N)
{
int i;
// Corner case
if (N == 3)
{
System.out.print("1 2 3");
return;
}
// Array of vectors to store the
// possible pairs
@SuppressWarnings("unchecked")
Vector []arr = new Vector[N / 2 + 2];
for(int k = 0; k < arr.length; k++)
arr[k] = new Vector();
for(i = 1; i <= N / 2; i++)
{
// Push the pair (2*i - 1, 2*i)
arr[i].add(2 * i - 1);
arr[i].add(2 * i);
}
// If N is odd then push the
// last element N
if ((N & 1) != 0)
{
arr[i].add(N);
}
// Function call
printGroups(arr, N);
return;
}
// Driver Code
public static void main(String[] args)
{
int N = 10;
// Function call
formGroups(N);
}
}
// This code is contributed by Rohit_ranjan Python3
# Python3 program for
# the above approach
# Function to print the groups
def printGroups(A, N):
# Interate over [1, N / 2];
for i in range (0, N // 2 ):
# Print all pairs
for j in range (len(A[i])):
print (A[i][j], end = ' ')
print ()
# Function to form a
# groups with GCD 1
def formGroups(N):
# Corner case
if (N == 3):
print ("1 2 3")
return
# Array of vectors
# to store the
# possible pairs
arr = []
for i in range (1, N // 2 + 1):
P = []
# Push the pair
# (2 * i - 1, 2 * i)
P.append(2 * i - 1)
P.append(2 * i)
arr.append(P)
# If N is odd then push the
# last element N
if (N & 1):
arr.append(N)
# Function Call
printGroups(arr, N)
return
# Driver Code
if __name__ == "__main__":
N = 10
# Function Call
formGroups(N)
# This code is contributed by ChitranayalC#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to print the groups
static void printGroups(List []A,
int N)
{
// Interate over [1, N/2];
for(int i = 1; i <= N / 2 + 1; i++)
{
// Print all pairs
for(int j = 0; j < A[i].Count; j++)
{
Console.Write(A[i][j] + " ");
}
Console.WriteLine();
}
}
// Function to form a groups with GCD 1
static void formGroups(int N)
{
int i;
// Corner case
if (N == 3)
{
Console.Write("1 2 3");
return;
}
// Array of vectors to store the
// possible pairs
List []arr = new List[N / 2 + 2];
for(int k = 0; k < arr.Length; k++)
arr[k] = new List();
for(i = 1; i <= N / 2; i++)
{
// Push the pair (2*i - 1, 2*i)
arr[i].Add(2 * i - 1);
arr[i].Add(2 * i);
}
// If N is odd then push the
// last element N
if ((N & 1) != 0)
{
arr[i].Add(N);
}
// Function call
printGroups(arr, N);
return;
}
// Driver Code
public static void Main(String[] args)
{
int N = 10;
// Function call
formGroups(N);
}
}
// This code is contributed by amal kumar choubey 输出：
1 2
3 4
5 6
7 8
9 10
时间复杂度： O(N)，其中N是给定的数字。
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