Poj-1980 unit fraction partition **

/*
* I thought it would be difficult to prune this question. I didn't expect 1A .. 32 Ms
* The difficulty of this question is the processing of scores .. Do not use double .. Accuracy is hard to grasp ..
*
*/

# Include <cstdio>
# Include <cmath>
Using namespace STD;

Int p, q, A, N, TOT; // if the question is defined, TOT is the answer

// Compare P1/Q1 and P2/Q2
Int inline frac_cmp (INT P1, int Q1, int P2, int q2 ){
Return P1 * Q2-P2 * Q1;
}
/*
Int inline gcd (int A, int B ){
If (B = 0) return;
Return gcd (B, A % B );
}
*/
// P/Q-1/D the answer is stored in P/Q
Void inline frac_sub (Int & P, Int & Q, int d ){
P = p * D-Q;
Q = Q * D;

// Save time (600 MS-> 63 MS)
// If (P! = 0 ){
// Int G = gcd (p, q );
// P = P/g; q = Q/g;
//}
}

// The denominator of the last layer of last_d .. The numerator of the remaining left_p score .. The denominator of the remaining score of left_q .. Num layers .. Maximum Product of the Left denominator of left_a
Void DFS (INT last_d, int left_p, int left_q, int num, int left_a ){
Int next_p, next_q;
If (num = n + 1 & left_p! = 0) return;
If (Num <= n + 1 & left_p = 0 ){
TOT ++; return;
}

// If (frac_cmp (left_p, left_q, N-num + 1, last_d)> 0) return; // The remaining score is too large (which will be determined later, which can be omitted)

For (INT d = last_d; D <= left_a; D ++ ){
If (frac_cmp (left_p, left_q, 1, D) <0) continue; // 1/D is too large

Frac_sub (next_p = left_p, next_q = left_q, d); // obtain next_p, next_q
If (frac_cmp (next_p, next_q, N-num, d)> 0) break; // The score left for the next layer is too large

DFS (D, next_p, next_q, num + 1, left_a/D );
}

}


Int main (){
While (scanf ("% d", & P, & Q, & A, & N )){
If (P = 0) return 0;

TOT = 0;
DFS (1, p, q, 1, );
Printf ("% d \ n", TOT );

}

Return 0;

}