将n写为两个或多个正整数之和的方法

对于给定的n> 0，找到不同方式可以将n写入两个或多个正整数之和的数量。

例子：

Input : n = 5
Output : 6
Explanation : All possible six ways are :
4 + 1
3 + 2
3 + 1 + 1
2 + 2 + 1
2 + 1 + 1 + 1
1 + 1 + 1 + 1 + 1

Input : 4
Output : 4
Explanation : All possible four ways are :
3 + 1
2 + 2
2 + 1 + 1
1 + 1 + 1 + 1

这个问题可以用与找零问题类似的方式解决，不同之处仅在于，在这种情况下，我们应该迭代1到n-1，而不是像找零问题中那样重复特定的硬币值。

C/C++

// Program to find the number of ways, n can be
// written as sum of two or more positive integers.
#include 
using namespace std;
  
// Returns number of ways to write n as sum of
// two or more positive integers
int countWays(int n)
{
    // table[i] will be storing the number of
    // solutions for value i. We need n+1 rows
    // as the table is consturcted in bottom up
    // manner using the base case (n = 0)
    int table[n+1];
  
    // Initialize all table values as 0
    memset(table, 0, sizeof(table));
  
    // Base case (If given value is 0)
    table[0] = 1;
  
    // Pick all integer one by one and update the
    // table[] values after the index greater
    // than or equal to n
    for (int i=1; i

---

Java

// Program to find the number of ways, 
// n can be written as sum of two or 
// more positive integers.
import java.util.Arrays;
  
class GFG {
      
    // Returns number of ways to write
    // n as sum of two or more positive 
    // integers
    static int countWays(int n)
    {
          
        // table[i] will be storing the 
        // number of solutions for value
        // i. We need n+1 rows as the 
        // table is consturcted in bottom
        // up manner using the base case
        // (n = 0)
        int table[] = new int[n + 1];
      
        // Initialize all table values as 0
        Arrays.fill(table, 0);
      
        // Base case (If given value is 0)
        table[0] = 1;
      
        // Pick all integer one by one and
        // update the table[] values after 
        // the index greater than or equal 
        // to n
        for (int i = 1; i < n; i++)
            for (int j = i; j <= n; j++)
                table[j] += table[j - i];
      
        return table[n];
    }
      
    //driver code
    public static void main (String[] args)
    {
        int n = 7;
          
        System.out.print(countWays(n));
    }
}
  
// This code is contributed by Anant Agarwal.

Python

# Program to find the number of ways, n can be
# written as sum of two or more positive integers.
  
# Returns number of ways to write n as sum of
# two or more positive integers
def CountWays(n):
  
    # table[i] will be storing the number of
    # solutions for value i. We need n+1 rows
    # as the table is consturcted in bottom up
    # manner using the base case (n = 0)
    # Initialize all table values as 0
    table =[0] * (n + 1)
  
    # Base case (If given value is 0)
    # Only 1 way to get 0 (select no integer)
    table[0] = 1
  
    # Pick all integer one by one and update the
    # table[] values after the index greater
    # than or equal to n
    for i in range(1, n ):
  
        for j in range(i , n + 1):
  
            table[j] +=  table[j - i]            
  
    return table[n]
  
# driver program
def main():
  
    n = 7
  
    print CountWays(n)
  
if __name__ == '__main__':
    main()
  
#This code is contributed by Neelam Yadav

C#

// Program to find the number of ways, n can be
// written as sum of two or more positive integers.
using System;
  
class GFG {
      
    // Returns number of ways to write n as sum of
    // two or more positive integers
    static int countWays(int n)
    {
          
        // table[i] will be storing the number of
        // solutions for value i. We need n+1 rows
        // as the table is consturcted in bottom up
        // manner using the base case (n = 0)
        int []table = new int[n+1];
       
        // Initialize all table values as 0
        for(int i = 0; i < table.Length; i++)
            table[i] = 0;
       
        // Base case (If given value is 0)
        table[0] = 1;
       
        // Pick all integer one by one and update the
        // table[] values after the index greater
        // than or equal to n
        for (int i = 1; i < n; i++)
            for (int j = i; j <= n; j++)
                table[j] += table[j-i];
       
        return table[n];
    }
      
    //driver code
    public static void Main()
    {
        int n = 7;
          
        Console.Write(countWays(n));
    }
}
  
// This code is contributed by Anant Agarwal.

输出：

14

时间复杂度O(n ² )