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：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define inc(i,j,k) for(int i=j;i<=k;i++)
#define maxn 10010
using namespace std;

inline int read(){
    char ch=getchar(); int f=1,x=0;
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1; ch=getchar();}
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return f*x;
}
int n,d,ansa,ansb; double ans;
int main(){
    n=read(); d=read(); ans=10;
    inc(i,1,32767){
        int j=n*i/d,k=j+1;
        if(fabs((double)k/(double)i-(double)n/(double)d)<ans)
            ans=fabs((double)k/(double)i-(double)n/(double)d),ansa=k,ansb=i;
        if(j*d!=n*i&&fabs((double)j/(double)i-(double)n/(double)d)<ans)
            ans=fabs((double)j/(double)i-(double)n/(double)d),ansa=j,ansb=i;
    }
    printf("%d %d",ansa,ansb); return 0;
}

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