POJ - 1980 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 200 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

Sample Output

4
7
6
2
42
1
0
9
3

题解：输入p,q,a,n;求一些分子为1的分数的和为p/q的情况，另所有分母的乘积不大于a且分数个数不大于n,

具体看代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int p,q,a,n;
int sum;
void dfs(int d,int ss,int x,int y,int num)//当前可用分母，当前所有分母乘积
{
    if(x*q==y*p&&y!=0)                    //之前所有数的和的分子，分母，当前分数的数目
    {
        sum++;//用乘数，方便分母分子不一样的情况。另初始x，y为0
        return ;
    }
    if(num>=n)//数目超出
        return ;
    if(ss>a)//乘积超出
        return ;
    if(x*q>y*p)//和超出
        return ;
    int tx,ty;
    if(y==0)//初始x，y为0，下一步变为1，1
    {
        tx=1;
        ty=d;
    }
    else
    {
        tx=d*x+y;
        ty=y*d;
    }
    if(ss*d<=a)//将1/d放进式子中
        dfs(d,ss*d,tx,ty,num+1);
    if(num+1<=n&&ss*(d+1)<=a)//不放1/d或1/d不可以，先判断之后的一个分数，剪枝效果
        dfs(d+1,ss,x,y,num);
}
int main()
{
    while(~scanf("%d%d%d%d",&p,&q,&a,&n)&&(p||q||a||n))
    {
        sum=0;
        dfs(1,1,0,0,0);
        printf("%d\n",sum);
    }
    return 0;
}