bzoj 1684: [Usaco2005 Oct]Close Encounter【数学(?)】

枚举分母,然后离他最近的分子只有两个,分别判断一下能不能用来更新答案即可

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int a,b,aa,ab;
double mx=10;
void wk(int x,int y)
{
    if(x*b==y*a)
        return;
    if(fabs((double)x/y-(double)a/b)<mx)
    {
        mx=fabs((double)x/y-(double)a/b);
        aa=x,ab=y;
    }
}
int main()
{
    scanf("%d%d",&a,&b);
    for(int i=1;i<=32767;i++)
        wk(floor((double)a/b*i),i),wk(floor((double)a/b*i)+1,i);
    printf("%d %d\n",aa,ab);
    return 0;
}

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转载自www.cnblogs.com/lokiii/p/9159405.html