上题目- -

Ratio

Description

If you ever see a televised report on stock market activity, you’ll hear the anchorperson say something like ``Gainers outnumbered losers 14 to 9,’’ which means that for every 14 stocks that increased in value that day, approximately 9 other stocks declined in value. Often, as you hear that, you’ll see on the screen something like this:

Gainers 1498

Losers   902

As a person with a head for numbers, you’ll notice that the anchorperson could have said Gainers outnumbered losers 5 to 3'', which is a more accurate approximation to what really happened.  After all, the exact ratio of winners to losers is (to the nearest millionth) 1.660754, and he reported a ratio of 14 to 9, which is 1.555555, for an error of 0.105199; he could have said5 to 3’’, and introduced an error of only 1.666667-1.660754=0.005913.  The estimate 5 to 3'' is not as accurate as1498 to 902’’ of course; evidently, another goal is to use small integers to express the ratio. So, why did the anchorperson say ``14 to 9?’‘  Because his algorithm is to lop off the last two digits of each number and use those as the approximate ratio.

What the anchorman needs is a list of rational approximations of increasing accuracy, so that he can pick one to read on the air. Specifically, he needs a sequence {a_1, a_2, …, a_n} where a_1 is a rational number with denominator 1 that most exactly matches the true ratio of winners to losers (rounding up in case of ties), a_{i+1} is the rational number with least denominator that provides a more accurate approximation than a_i, and a_n is the exact ratio, expressed with the least possible denominator.  Given this sequence, the anchorperson can decide which ratio gives the best tradeoff between accuracy and simplicity.

For example, if 5 stocks rose in price and 4 fell, the best approximation with denominator 1 is 1/1; that is, for every stock that fell, about one rose.  This answer differs from the exact answer by 0.25 (1.0 vs 1.25).  The best approximations with two in the denominator are 2/2 and 3/2, but neither is an improvement on the ratio 1/1, so neither would be considered.  The best approximation with three in the denominator 4/3, is more accurate than any seen so far, so it is one that should be reported.  Finally, of course, 5/4 is exactly the ratio, and so it is the last number reported in the sequence.

Can you automate this process and help the anchorpeople?

Input

input contains several pairs of positive integers. Each pair is on a line by itself, beginning in the first column and with a space between the two numbers.  The first number of a pair is the number of gaining stocks for the day, and the second number is the number of losing stocks for the day.   The total number of stocks never exceeds 5000.

Output

For each input pair, the standard output should contain a series of approximations to the ratio of gainers to losers.  The first approximation has ‘1’ as denominator, and the last is exactly the ratio of gainers to losers, expressed as a fraction with least possible denominator.  The approximations in between are increasingly accurate and have increasing denominators, as described above.

The approximations for a pair are printed one to a line, beginning in column one, with the numerator and denominator of an approximation separated by a slash (``/’’).  A blank line separates one sequence of approximations from another.

Sample Input

5 4 
1498 902

Sample Output

1/1 
4/3 
5/4 

2/1 
3/2 
5/3 
48/29 
53/32 
58/35 
63/38 
68/41 
73/44 
78/47 
83/50 
88/53 
93/56 
377/227 
470/283 
563/339 
656/395 
749/451

题目有点长- -一看来源是USA的题目- -怪不得英语读起来感觉就不一样- -凑合还是能够看懂题目 题目的意思是给你两个数字n,m，然后要求得到一串数字a1到ak，其中a1为分母为1时a/1最接近n/m的整数，然后a2是分母比1大并且比a1/1更加接近n/m的那个整数，以此类推，ak表示刚好等于n/m的情况，对于同一个分母有两个分子同时符合要求时取大的 一开始想到二分啥的- -其实不需要二分直接算就可以，不过还是用二分写了，写出来之后发现即使是4/10跟6/15相比都会有浮点误差，于是稍微修改了一下判断条件,加了个eps，本地调试通过，提交，WA了…… 接下来修改了一些无关紧要的东西，继续WA中………… 看看时间有点迟了，急着睡觉，干脆看Discuss，发现Discuss里面跟我不一样的主要是有些人用整数比较来避免浮点误差，另外还有就是直接计算而不是二分- -话说这个时段脑子果然不行- - 于是我在纸上推导了一下，得到一个式子，改好，提交，继续WA………… 再看程序，发现比较的地方可能会溢出- -（其实是不会溢出的），于是干脆改成int64，自己写了一个abs，提交，WA…… 悲情……………… 突然发现读入的地方感觉可能会出错，改了一下判断文件结束的代码，居然就AC了……瞎了………… 我之前的代码， scanf(“%d%d”, &n, &m); if(feof(stdin)) return false; 改之后的代码 if(scanf(“%d%d”, &n, &m) == EOF) return false; 前面的代码用了很久了，主要原因是遇到除了m，n之外还有其他数据的时候这样子判断比较方便，但是偏偏今天只有n，m，于是就悲剧了…… 回过头，拿了原来WA的几个程序，发现全部都能AC 看来一开始对精度的估计还是比较准的- -可惜对精度估计不够自信反而对错误的输入方式太自信了……悲剧…… 以后一定得小心- -现场赛的时候这样子那可就太悲剧了…… 话说刷AC率- -1A的题目好少……2A的题目更少……大部分都是要么1A，要么N个提交A……+U