Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 2797 | Accepted: 1077 |
Description
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
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 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
#include<iostream> #include<cstdio> #include<algorithm> using namespace std; int cnt; int p,q,a,n; void dfs(int t,int pl,int tp,int tq,int fac) { if(fac>a) return ; if(p*tq==q*tp&&tq!=0) {cnt++;return ;} if(pl>=n) return ; if(p*tq<q*tp) return ; //用t 1/t+tp/tq=(tq+t*tp)/(t*tq); int x,y; if(tq){ x=tq+t*tp,y=t*tq; } else x=1,y=t; if(fac*t<=a) dfs(t,pl+1,x,y,fac*t); //不用t fac*(t+1)<=a 剪枝很重要 if(t+1<=a&&fac*(t+1)<=a) dfs(t+1,pl,tp,tq,fac); } int main() { while(scanf("%d%d%d%d",&p,&q,&a,&n)==4&&q) { cnt=0; dfs(1,0,0,0,1); printf("%d/n",cnt); } return 0; }