打印将 N 写成两个或多个正整数之和的所有可能方式
原文:https://www . geesforgeks . org/print-所有可能的方法-将 n 写成两个或更多正整数的和/
给定一个整数 N ，任务是打印所有可能的方式，其中 N 可以写成两个或多个正整数的和。 例:
输入:N = 4 T3】输出:T5】1 1 1 1 1 1 2 1 3 2 T10】输入:N = 3 T13】输出:T15】1 1 1 1 2
方法:思路是用递归来解决这个问题。想法是考虑从 1 到 N 的每一个整数，这样在每次递归调用中，和 N 可以被这个数减少，如果在任何递归调用中，N 减少到零，那么我们将打印存储在向量中的答案。以下是递归的步骤:
- 得到数 N ，其和必须被分解成两个或多个正整数。
-
递归地从值 1 迭代到 N 作为索引 i :
-
基本情况:如果递归调用的值是 0 ，那么打印当前向量，因为这是将 N 分解成两个或更多正整数的方法之一。
if (n == 0)
printVector(arr);
- 递归调用:如果不满足基本情况，则从【I，N–I】递归迭代。将当前元素 j 推入向量(比如 arr )并递归迭代下一个索引，在这个递归结束后，弹出之前插入的元素j【T9:
for j in range[i, N]:
arr.push_back(j);
recursive_function(arr, j + 1, N - j);
arr.pop_back(j);
以下是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the values stored
// in vector arr
void printVector(vector<int>& 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<int>& 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<int> arr;
// Function Call
findWays(arr, 1, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.util.*;
class GFG{
// Function to print the values stored
// in vector arr
static void printVector(ArrayList<Integer> 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<Integer> 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<Integer> arr = new ArrayList<Integer>();
// Function call
findWays(arr, 1, n);
}
}
// This code is contributed by offbeat
Python 3
# 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<int> 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<int> 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<int> arr = new List<int>();
// Function call
findWays(arr, 1, n);
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// Javascript program for the above approach
// Function to print the values stored
// in vector arr
function printVector(arr)
{
if (arr.length != 1) {
// Traverse the vector arr
for (var i = 0; i < arr.length; i++) {
document.write( arr[i] + " ");
}
document.write("<br>");
}
}
// Recursive function to print different
// ways in which N can be written as
// a sum of at 2 or more positive integers
function findWays(arr, i, 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 (var j = i; j <= n; j++) {
// Include current element
// from representation
arr.push(j);
// Call function again
// with reduced sum
findWays(arr, j, n - j);
// Backtrack to remove current
// element from representation
arr.pop();
}
}
// Driver Code
// Given sum N
var n = 4;
// To store the representation
// of breaking N
var arr = [];
// Function Call
findWays(arr, 1, n);
</script>
Output:
1 1 1 1
1 1 2
1 3
2 2
时间复杂度:O(2N) T5】辅助空间: O(N 2 )
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