1684: [Usaco2005 Oct]Close Encounter

1684: [Usaco2005 Oct]Close Encounter

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 387  Solved: 181
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Description

Lacking even a fifth grade education, the cows are having trouble with a fraction problem from their textbook. Please help them. The problem is simple: Given a properly reduced fraction (i.e., the greatest common divisor of the numerator and denominator is 1, so the fraction cannot be further reduced) find the smallest properly reduced fraction with numerator and denominator in the range 1..32,767 that is closest (but not equal) to the given fraction. 找一个分数它最接近给出一个分数. 你要找的分数的值的范围在1..32767

Input

* Line 1: Two positive space-separated integers N and D (1 <= N < D <= 32,767), respectively the numerator and denominator of the given fraction

Output

* Line 1: Two space-separated integers, respectively the numerator and denominator of the smallest, closest fraction different from the input fraction.

Sample Input

2 3

Sample Output

21845 32767

OUTPUT DETAILS:

21845/32767 = .666676839503.... ~ 0.666666.... = 2/3.

HINT

 

Source

Silver

题解：感觉很像是NOIP2014普及组的那道题，貌似当时干掉了好多人= =（HansBug：呵呵哒我会说我第一想法是二分答案么，但是显然二分是没有办法控制分母的大小的）

于是继续脑洞，于是我想到既然分母范围那么小，那么为何不枚举分母呢？然后直接根据原来分数的大致值来估测分子，然后不断打擂台

（PS：值得注意的是要特判和原分数相同的情况，否则你会输入什么就输出什么的= =，不过敢直接这么枚举还是需要一定脑洞哒）

/**************************************************************
    Problem: 1684
    User: HansBug
    Language: Pascal
    Result: Accepted
    Time:44 ms
    Memory:224 kb
****************************************************************/
 
var
   i,j,k,l,m,n,a,b:longint;
   ans:double;
procedure check(x,y:longint);
          var t:double;
          begin
               if (x*m)=(y*n) then exit;
               t:=abs((x/y)-(n/m));
               if t<ans then
                  begin
                       ans:=t;
                       a:=x;
                       b:=y;
                  end;
          end;
begin
     readln(n,m);ans:=maxint;;
     for j:=1 to 32767 do
         begin
              i:=trunc((n/m)*j);
              check(i,j);
              check(i+1,j);
         end;
     writeln(a,' ',b);
     readln;
end.