human
Solve in Python:
This is a rather simple problem to describe. You will be given three numbers $S, P$ and $k$. Your task is to find if there are integers $n_1, n_2,...,n_k$ such that $n_1 + n_2 +...+ n_k = S$, $n_1 \cdot n_2 \cdot ... \cdot n_k = P$. If such integers exist, print them out. If no such sequence of integers exist, then print "NO".
For example if $S=11, P=48$ and $k=3$ then $3, 4$ and $4$ is a solution. On the other hand, if $S=11, P=100$ and $k=3$, there is no solution and you should print "NO".

-----Input:-----
A single line with three integers $S, P$ and $k$.

-----Output:-----
A single word "NO" or a seqence of $k$ integers $n_1, n_2,..., n_k$ on a single line. (The $n_i$'s must add up to $S$ and their product must be $P$).

-----Constraints:-----
- $1 \leq k \leq 4$.
- $1 \leq S \leq 1000$.
- $1 \leq P \leq 1000$.

-----Sample input 1:-----
11 48 3

-----Sample output 1:-----
3 4 4 

-----Sample input 2:-----
11 100 3

-----Sample output 2:-----
NO
 ... What is a Python program to find if there are integers $n_1, n_2,?
assistant
#Here is Python program to find if there are integers $n_1, n_2,.

import math

def part(n, k):
    def _part(n, k, pre):
        if n <= 0:
            return []
        if k == 1:
            if n <= pre:
                return [[n]]
            return []
        ret = []
        for i in range(min(pre, n), 0, -1):
            ret += [[i] + sub for sub in _part(n-i, k-1, i)]
        return ret
    return _part(n, k, n)

S, P, k=map(int, input().split())
factors=[]

for i in range(1, int(math.sqrt(P))+1):
    if(P%i==0 and P//i!=i):
        factors.append(i)
        factors.append(P//i)
    elif(P%i==0 and P//i==i):
        factors.append(i)
        
arr=part(S, k)
flag=-1
#print(part(S, k))

for i in arr:
    temp=i
    cnt=0;  prod=1
    
    for j in temp:
        if(j in factors):
            cnt+=1
        prod*=j
           
    if(cnt==k and prod==P):
        print(*temp)
        flag=0
        break
    else:
        flag=1
        
if(flag!=0):
    print("NO")