1684: [Usaco2005 Oct]Close Encounter

1684: [Usaco2005 Oct]Close Encounter

Time Limit: 5 Sec  Memory Limit:
64 MB
Submit: 423  Solved: 199

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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

 

题意是找一个最接近但不相等的分数。。然后注意精度

判断(a/b)==(x/y)使用ay==bx...

#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;

typedef double DB;

DB v,now,M;
int a,b,i,j,a1,a2;

void Judge(int x,int y)
{
DB V = (DB)(x) / (DB)(y);
if (abs(V - v) < M)
{
if (x * b == y * a) return;
M = abs(V-v);
a1 = x; a2 = y;
}
}

int main()
{
cin >> a >> b;
v = (DB)(a) / (DB)(b);
M = 1E5;
for (i = 1; i <= 32767; i++)
{
int k = (DB)(a) / (DB)(b) * (DB)(i);
Judge (k-1,i); Judge(k,i); Judge(k+1,i);
}
printf("%d %d",a1,a2);
return 0;
}