CFRAC2 - Continuous Fractions Again
A simple continuous fraction has the form:
where the ai’s are integer numbers.
The previous continuous fraction could be noted as [a1, a2, ..., an]. It is not difficult to show that any rational number p / q, with integers p > q > 0, can be represented in a unique way by a simple continuous fraction with n terms, such that p / q = [a1, a2, ..., an−1, 1], where n and the ai’s are positive natural numbers.
Now given a simple continuous fraction, your task is to calculate a rational number which the continuous fraction most corresponds to it.
Input for each case will consist of several lines. The first line is two integer m and n, which describe a char matrix, then followed m lines, each line cantain n chars. The char matrix describe a continuous fraction The continuous fraction is described by the following rules:
- Horizontal bars are formed by sequences of dashes '-'.
- The width of each horizontal bar is exactly equal to the width of the denominator under it.
- Blank characters should be printed using periods '.'
- The number on a fraction numerator must be printed center justified. That is, the number of spaces at either side must be same, if possible; in other case, one more space must be added at the right side.
The end of the input is indicated by a line containing 0 0.
Output will consist of a series of cases, each one in a line corresponding to the input case. A line describing a case contains p and q, two integer numbers separated by a space, and you can assume that 10^20 > p > q > 0.
Input: 9 17 ..........1...... 2.+.------------- ............1.... ....4.+.--------- ..............1.. ........1.+.----- ................1 ............5.+.- ................1 5 10 ......1... 1.+.------ .........1 ....11.+.- .........1 0 0 Output: 75 34 13 12
Like CFRAC, fun while breaking down, very easy afterwards. Thumbs up!
to make it clear just get the numbers separately and think for a formula/pattern the result follows
can anyone explain the problem please? I am not getting it.Last edit: 2017-06-23 10:32:09
very easy accept in 1go use long long int
for numerator / denominator use long in java
For those getting WA, here's a nice test case:
Very easy problem..