[POJ1980]Unit Fraction Partition(搜索)
Unit Fraction Partition
Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 4571 Accepted: 1809 Description
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 0Sample Output
4
7
6
2
42
1
0
9
3Source
沦落到要做普及组题目的地步了吗。。发现自己连搜索都不会写了。
几个可行性剪枝就可以了:乘积不超限,个数不超限,分数和不超过目标。
起先一直TLE,把循环中的除法提到外面就卡过了。
这种题目竟然也要做1h。。
#include<cstdio>
#include<algorithm>
#define rep(i,l,r) for (int i=l; i<=r; i++)
using namespace std; int p,q,a,n,ans; void dfs(int mn,int num,int deno,int mul,int dq){
if (mul>a) return;
if (num*q==deno*p) ans++;
if (num*q>deno*p || dq==n) return;
int t=a/mul;
rep(i,mn,t) dfs(i,num*i+deno,deno*i,mul*i,dq+);
} int main(){
freopen("poj1980.in","r",stdin);
freopen("poj1980.out","w",stdout);
while (~scanf("%d%d%d%d",&p,&q,&a,&n)){
if (q==) return ;
ans=; dfs(,,,,); printf("%d\n",ans);
}
return ;
}
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