Problem:
Problem C: Unit Fraction Partition

A fraction whose numerator is 1
and whose denominator is a positive integer
is called a unit fraction.
A representation of a positive rational number
p/q
as the sum of finitely many unit fractions is called a partition of
p/q
into unit fractions.
For example,
1/2 + 1/6
is a partition of
2/3
into unit fractions.
The difference in the order of addition is disregarded.
For example, we do not distinguish
1/6 + 1/2
from
1/2 + 1/6.

For given four positive integers
p,
q,
a, and
n,
count the number
of partitions of
p/q
into unit fractions satisfying the following two conditions.

The partition is the sum of at most
n
many unit fractions.

The product of the denominators of the unit fractions in the
partition is less than or equal to
a.

For example, if
(p,q,a,n) = (2,3,120,3),
you should report 4 since

enumerates all of the valid partitions.

Input

The input is a sequence of at most 1000 data sets followed by a terminator.

A data set is a line containing four positive integers
p,
q,
a, and
n
satisfying
p,q <= 800,
a <= 12000
and
n <= 7.
The integers are separated by a space.

The terminator is composed of just one line
which contains four zeros separated by a space.
It is not a part of the input data but a mark for the end of the input.

Output

The output should be composed of lines
each of which contains a single integer.
No other characters should appear in the output.

The output integer corresponding to a data set
p,
q,
a,
n
should be the number of all partitions of
p/q
into at most
n
many unit fractions such that the product of the denominators of the
unit fractions is less than or equal to
a.

Sample Input

2 3 120 3
2 3 300 3
2 3 299 3
2 3 12 3
2 3 12000 7
54 795 12000 7
2 3 300 1
2 1 200 5
2 4 54 2
0 0 0 0

Output for the Sample Input

4
7
6
2
42
1
0
9
3

Solution in C++14 (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define rep(i, a, b) for (int i = (a); i < (b); ++i)
using pii = pair<int, int>;
using vi=vector<int>;
using vvi=vector<vector<int>>;

int p,q,a,n;
map<vector<int>,int>dp;

int gcd(int a, int b) { return a ? gcd(b % a, a) : b; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }

int dfs(int x,int y,int z,int w,int last){
    vi arg={x,y,z,w,last};
    //if(dp.count(arg))return dp[arg];
    if(z==n){
        if(x==0)return 1;
        else return 0;
    }
    if(x<0)return 0;
    if(x==0)return 1;
    int ret=0;
    for(int i=last;i*w<=a;i++){
        // x/y - 1/i
        int bunbo=lcm(y,i);
        int bunshi=bunbo/y*x - bunbo/i*1;
        int g=gcd(bunbo,bunshi);
        bunbo/=g;
        bunshi/=g;
        ret+=dfs(bunshi,bunbo,z+1,i*w,i);
    }

    //return dp[arg]=ret;
    return ret;
}

signed main() {
    while(cin>>p>>q>>a>>n, p or q or a or n){
        dp.clear();
        int g=gcd(p,q);
        p/=g;
        q/=g;
        cout<<dfs(p,q,0,1,1)<<endl;
    }
}