zhber 有好多做过的题没写下来，如果我还能记得就补吧

## 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.

2 3

## Sample Output

21845 32767

OUTPUT DETAILS:

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

#include<cstdio>
#include<cmath>
int a,b,i,aa,bb;
double save=10;
inline double abs(double x)
{if (x<0)x=-x;return x;}
void jud(int x,int y)
{
if (x*b==y*a) return;
double work=abs((double)x/y-(double)a/b);
if (work<save)
{
save=work;
aa=x;
bb=y;
}
}
int main()
{
scanf("%d%d",&a,&b);
for (int j=1;j<=32767;j++)
{
i=(int)floor((double)a/b*j);
jud(i,j);
jud(i+1,j);
}
printf("%d %d\n",aa,bb);
}


posted on 2014-07-25 22:45  zhber  阅读(258)  评论(0编辑  收藏  举报