bzoj1684[Usaco2005 Oct]Close Encounter*

bzoj1684[Usaco2005 Oct]Close Encounter

题意:

找一个分数它最接近给出一个分数。你要找的分数的分子分母的范围在1..32767。

题解:

枚举所求分数的分子,用其乘上给出分数得到一个浮点数分母,比较分母向上/下取整所得分数与答案比较。

代码:

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <cmath>
 5 #define inc(i,j,k) for(int i=j;i<=k;i++)
 6 #define maxn 10010
 7 using namespace std;
 8 
 9 inline int read(){
10     char ch=getchar(); int f=1,x=0;
11     while(ch<'0'||ch>'9'){if(ch=='-')f=-1; ch=getchar();}
12     while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
13     return f*x;
14 }
15 int n,d,ansa,ansb; double ans;
16 int main(){
17     n=read(); d=read(); ans=10;
18     inc(i,1,32767){
19         int j=n*i/d,k=j+1;
20         if(fabs((double)k/(double)i-(double)n/(double)d)<ans)
21             ans=fabs((double)k/(double)i-(double)n/(double)d),ansa=k,ansb=i;
22         if(j*d!=n*i&&fabs((double)j/(double)i-(double)n/(double)d)<ans)
23             ans=fabs((double)j/(double)i-(double)n/(double)d),ansa=j,ansb=i;
24     }
25     printf("%d %d",ansa,ansb); return 0;
26 }

 

20160927

posted @ 2016-10-16 14:41  YuanZiming  阅读(139)  评论(0编辑  收藏  举报