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
题意:给了一个分数p/q,求有多少种单位分数之和等于p/q,且满足所有单位分数的分母之积小于等于a,个数小于n。
思路:分母从1开始枚举,每次记录一下枚举到哪个数了,如果满足分母之ji积小于等于a,且个数不超过n,从此数继续让下搜,因为单位分数之间不分先后顺序,所以这样就涵盖了所有情况。
代码如下:
#include<cstdio>
#include<cstring>
int a,n,ans;
void dfs(int p,int q,int mod,int x,int t) //p 此时分子,q分母,mod值,x位置(那个数),t还有多少个
{
if(p==0&&mod<=a) //拼成了 且分母不超过a
{
ans++;
return;
}
if(t==0||p*x>q*t||mod*x>a) //剪枝
return;
for(int i=x;i<=a;i++) //从x开始往后尝试1
{
if(mod*i<=a) //分母满足条件
{
int xt=p*i-q; //减过后分子
if(xt>=0) //符合条件
dfs(xt,q*i,mod*i,i,t-1);
}
else //不满足<=a
break;
}
}
int main()
{
int p,q;
while(~scanf("%d%d%d%d",&p,&q,&a,&n)&&(p||q||a||n))
{
ans=0;
dfs(p,q,1,1,n);
printf("%d\n",ans);
}
}