给定一个整数N ,任务是打印所有可能的方式,其中N可以被写为两个或多个正整数之和。
例子:
Input: N = 4
Output:
1 1 1 1
1 1 2
1 3
2 2
Input: N = 3
Output:
1 1 1
1 2
方法:想法是使用递归来解决此问题。这个想法是考虑从1到N的每个整数,这样在每次递归调用时,总和N都可以减少此数字,如果在任何递归调用中N都减少为零,那么我们将打印存储在向量中的答案。以下是递归的步骤:
- 获取必须将其总和分解为两个或更多个正整数的数字N。
- 从值1递归地迭代到N作为索引i :
- 基本情况:如果递归调用的值为0 ,则打印当前向量,因为这是将N分解为两个或多个正整数的方法之一。
if (n == 0)
printVector(arr);
- 递归调用:如果不满足基本条件,则从[i,N – i]递归迭代。将当前元素j推入vector(say arr )中,并递归地迭代下一个索引,在此递归结束后,弹出先前插入的元素j :
for j in range[i, N]:
arr.push_back(j);
recursive_function(arr, j + 1, N - j);
arr.pop_back(j);
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 values stored
// in vector arr
void printVector(vector& arr)
{
if (arr.size() != 1) {
// Traverse the vector arr
for (int i = 0; i < arr.size(); i++) {
cout << arr[i] << " ";
}
cout << endl;
}
}
// Recursive function to print different
// ways in which N can be written as
// a sum of at 2 or more positive integers
void findWays(vector& arr, int i, int n)
{
// If n is zero then print this
// ways of breaking numbers
if (n == 0)
printVector(arr);
// Start from previous element
// in the representation till n
for (int j = i; j <= n; j++) {
// Include current element
// from representation
arr.push_back(j);
// Call function again
// with reduced sum
findWays(arr, j, n - j);
// Backtrack to remove current
// element from representation
arr.pop_back();
}
}
// Driver Code
int main()
{
// Given sum N
int n = 4;
// To store the representation
// of breaking N
vector arr;
// Function Call
findWays(arr, 1, n);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to print the values stored
// in vector arr
static void printVector(ArrayList arr)
{
if (arr.size() != 1)
{
// Traverse the vector arr
for(int i = 0; i < arr.size(); i++)
{
System.out.print(arr.get(i) + " ");
}
System.out.println();
}
}
// Recursive function to print different
// ways in which N can be written as
// a sum of at 2 or more positive integers
static void findWays(ArrayList arr,
int i, int n)
{
// If n is zero then print this
// ways of breaking numbers
if (n == 0)
printVector(arr);
// Start from previous element
// in the representation till n
for(int j = i; j <= n; j++)
{
// Include current element
// from representation
arr.add(j);
// Call function again
// with reduced sum
findWays(arr, j, n - j);
// Backtrack to remove current
// element from representation
arr.remove(arr.size() - 1);
}
}
// Driver code
public static void main(String[] args)
{
// Given sum N
int n = 4;
// To store the representation
// of breaking N
ArrayList arr = new ArrayList();
// Function call
findWays(arr, 1, n);
}
}
// This code is contributed by offbeat
Python3
# Python3 program for the above approach
# Function to print the values stored
# in vector arr
def printVector(arr):
if (len(arr) != 1):
# Traverse the vector arr
for i in range(len(arr)):
print(arr[i], end = " ")
print()
# Recursive function to prdifferent
# ways in which N can be written as
# a sum of at 2 or more positive integers
def findWays(arr, i, n):
# If n is zero then prthis
# ways of breaking numbers
if (n == 0):
printVector(arr)
# Start from previous element
# in the representation till n
for j in range(i, n + 1):
# Include current element
# from representation
arr.append(j)
# Call function again
# with reduced sum
findWays(arr, j, n - j)
# Backtrack to remove current
# element from representation
del arr[-1]
# Driver Code
if __name__ == '__main__':
# Given sum N
n = 4
# To store the representation
# of breaking N
arr = []
# Function Call
findWays(arr, 1, n)
# This code is contributed by mohit kumar 29
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to print the values stored
// in vector arr
static void printList(List arr)
{
if (arr.Count != 1)
{
// Traverse the vector arr
for(int i = 0; i < arr.Count; i++)
{
Console.Write(arr[i] + " ");
}
Console.WriteLine();
}
}
// Recursive function to print different
// ways in which N can be written as
// a sum of at 2 or more positive integers
static void findWays(List arr,
int i, int n)
{
// If n is zero then print this
// ways of breaking numbers
if (n == 0)
printList(arr);
// Start from previous element
// in the representation till n
for(int j = i; j <= n; j++)
{
// Include current element
// from representation
arr.Add(j);
// Call function again
// with reduced sum
findWays(arr, j, n - j);
// Backtrack to remove current
// element from representation
arr.RemoveAt(arr.Count - 1);
}
}
// Driver code
public static void Main(String[] args)
{
// Given sum N
int n = 4;
// To store the representation
// of breaking N
List arr = new List();
// Function call
findWays(arr, 1, n);
}
}
// This code is contributed by 29AjayKumar
Javascript
输出:
1 1 1 1
1 1 2
1 3
2 2
时间复杂度: O(2 N )
辅助空间: O(N 2 )