给定一个正整数n ，任务是生成所有可能的唯一方式，将n表示为正整数之和。
例子：
Input: 4
Output:
4
3 1
2 2
2 1 1
1 1 1 1
Input: 3
Output:
3
2 1
1 1 1
方法：我们已经在本文中讨论了生成唯一分区的实现。这篇文章使用递归包含了另一个解决上述问题的直观方法。
这个想法很简单，并且与此处使用的方法相同。我们必须从n递归移动到1，并继续在数组中附加用于形成sum的数字。当总和等于n时，我们将打印该数组并返回。在实现中考虑的基本情况是： remSum == 0：形成了必需的n，因此打印数组。
然后，我们开始使用上一个分区中使用的最大值形成所需的总和。如果该数字大于n，我们将其忽略，否则我们将该数字附加到数组中，然后递归移至下一个迭代，形成sum =(remSum – i) ，其中最大值
可以使用的数字是i或小于i。这样，我们可以生成所需的唯一分区。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Array to store the numbers used
// to form the required sum
int dp[200];
int count = 0;
// Function to print the array which contains
// the unique partitions which are used
// to form the required sum
void print(int idx)
{
for (int i = 1; i < idx; i++) {
cout << dp[i] << " ";
}
cout << endl;
}
// Function to find all the unique partitions
// remSum = remaining sum to form
// maxVal is the maximum number that
// can be used to make the partition
void solve(int remSum, int maxVal, int idx, int& count)
{
// If remSum == 0 that means the sum
// is achieved so print the array
if (remSum == 0) {
print(idx);
count++;
return;
}
// i will begin from maxVal which is the
// maximum value which can be used to form the sum
for (int i = maxVal; i >= 1; i--) {
if (i > remSum) {
continue;
}
else if (i <= remSum) {
// Store the number used in forming
// sum gradually in the array
dp[idx] = i;
// Since i used the rest of partition
// cant have any number greater than i
// hence second parameter is i
solve(remSum - i, i, idx + 1, count);
}
}
}
// Driver code
int main()
{
int n = 4, count = 0;
solve(n, n, 1, count);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Array to store the numbers used
// to form the required sum
static int[] dp = new int[200];
static int count = 0;
// Function to print the array which contains
// the unique partitions which are used
// to form the required sum
static void print(int idx)
{
for (int i = 1; i < idx; i++)
{
System.out.print(dp[i] + " ");
}
System.out.println("");
}
// Function to find all the unique partitions
// remSum = remaining sum to form
// maxVal is the maximum number that
// can be used to make the partition
static void solve(int remSum, int maxVal,
int idx, int count)
{
// If remSum == 0 that means the sum
// is achieved so print the array
if (remSum == 0)
{
print(idx);
count++;
return;
}
// i will begin from maxVal which is the
// maximum value which can be used to form the sum
for (int i = maxVal; i >= 1; i--)
{
if (i > remSum)
{
continue;
}
else if (i <= remSum)
{
// Store the number used in forming
// sum gradually in the array
dp[idx] = i;
// Since i used the rest of partition
// cant have any number greater than i
// hence second parameter is i
solve(remSum - i, i, idx + 1, count);
}
}
}
// Driver code
public static void main(String[] args)
{
int n = 4, count = 0;
solve(n, n, 1, count);
}
}
// This code has been contributed by 29AjayKumarPython3
# Python 3 implementation of the approach
# Array to store the numbers used
# to form the required sum
dp = [0 for i in range(200)]
count = 0
# Function to print the array which contains
# the unique partitions which are used
# to form the required sum
def print1(idx):
for i in range(1,idx,1):
print(dp[i],end = " ")
print("\n",end = "")
# Function to find all the unique partitions
# remSum = remaining sum to form
# maxVal is the maximum number that
# can be used to make the partition
def solve(remSum,maxVal,idx,count):
# If remSum == 0 that means the sum
# is achieved so print the array
if (remSum == 0):
print1(idx)
count += 1
return
# i will begin from maxVal which is the
# maximum value which can be used to form the sum
i = maxVal
while(i >= 1):
if (i > remSum):
i -= 1
continue
elif (i <= remSum):
# Store the number used in forming
# sum gradually in the array
dp[idx] = i
# Since i used the rest of partition
# cant have any number greater than i
# hence second parameter is i
solve(remSum - i, i, idx + 1, count)
i -= 1
# Driver code
if __name__ == '__main__':
n = 4
count = 0
solve(n, n, 1, count)
# This code is contributed by
# Surendra_GangwarC#
// C# implementation of the approach
using System;
class GFG
{
// Array to store the numbers used
// to form the required sum
static int[] dp = new int[200];
// Function to print the array which contains
// the unique partitions which are used
// to form the required sum
static void print(int idx)
{
for (int i = 1; i < idx; i++)
{
Console.Write(dp[i] + " ");
}
Console.WriteLine("");
}
// Function to find all the unique partitions
// remSum = remaining sum to form
// maxVal is the maximum number that
// can be used to make the partition
static void solve(int remSum, int maxVal,
int idx, int count)
{
// If remSum == 0 that means the sum
// is achieved so print the array
if (remSum == 0)
{
print(idx);
count++;
return;
}
// i will begin from maxVal which is the
// maximum value which can be used to form the sum
for (int i = maxVal; i >= 1; i--)
{
if (i > remSum)
{
continue;
}
else if (i <= remSum)
{
// Store the number used in forming
// sum gradually in the array
dp[idx] = i;
// Since i used the rest of partition
// cant have any number greater than i
// hence second parameter is i
solve(remSum - i, i, idx + 1, count);
}
}
}
// Driver code
public static void Main()
{
int n = 4, count = 0;
solve(n, n, 1, count);
}
}
// This code is contributed by AnkitRai01输出：
4
3 1
2 2
2 1 1
1 1 1 1