【BZOJ】1684: [Usaco2005 Oct]Close Encounter(暴力+c++)

http://www.lydsy.com/JudgeOnline/problem.php?id=1684

这货完全在考精度啊。。

比如奇葩

(llf)a/b*i

(llf)(a/b*i)和(llf)(a/b)*i和(llf)(a/b)*(llf)i

这两货竟然不通????上边的能对,下边的就错了??

噗。

全部都要。。(llf)a/(llf)b*(llf)i。。。。。

这样才不会错。。

T_T

教训吸取了。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }
typedef long double llf;
llf eps=1e30;
int a, b, ans1, ans2;
void cal(int x, int y) {
	if(a*y==b*x) return;
	llf c=abs((llf)x/y-(llf)a/b);
	if(c<eps) {
		eps=c;
		ans1=x; ans2=y;
	}
}
int main() {
	a=getint(), b=getint();
	for1(i, 1, 32767) {
		int j=(llf)a/(llf)b*(llf)i;
		cal(j, i); cal(j+1, i);
	}
	printf("%d %d", ans1, ans2);
	return 0;
}

 

 


 

 

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

 

posted @ 2014-09-09 14:04  iwtwiioi  阅读(242)  评论(0编辑  收藏  举报