hdu 6209 The Intersection

Problem Description

A given coefficient K leads
an intersection of two curves f(x) and gK(x).
In the first quadrant, the curve f is
a monotone increasing function that f(x)=x−−√.
The curve g is
decreasing and g(x)=K/x.

To calculate the x-coordinate
of the only intersection in the first quadrant is the following question. For accuracy, we need the nearest rational number to x and
its denominator should not be larger than 100000.

 

Input

The first line is an integer T (1≤T≤100000) which
is the number of test cases.

For each test case, there is a line containing the integer K (1≤K≤100000),
which is the only coefficient.

 

Output

For each test case, output the nearest rational number to x.
Express the answer in the simplest fraction.

 

Sample Input

5
1
2
3
4
5

 

Sample Output

1/1
153008/96389
50623/24337
96389/38252
226164/77347

 

Source

2017 ACM/ICPC Asia Regional Qingdao Online

比赛时想用连分数直接算出答案的,结果精度死活卡不过去

赛后换了一种做法

假设答案是x，整数部分是a

一开始取区间[a/1,(a+1)/1]

假设当前区间是[a/b,c/d]，那么取(a+b)/(c+d)作为中间值

和x比较后把边界修改，一直到c+d大于10w就break掉

一开始c++没过，改成g++过的

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<cassert>
#include<iostream>
#include<algorithm>
using namespace std;
inline int gcd(int x,int y)
{
int m=x%y;
while(m!=0)
{
x=y;
y=m;
m=x%y;
}
return y;
}

long double ffabs(long double x) {
if (x < 0) x = -x;
return x;
}

int main()
{
// freopen("1004.in","r",stdin);
// freopen("1004.out","w",stdout);
int T;
scanf("%d",&T);
while(T>0)
{
T--;
int k;
scanf("%d",&k);
// long double x=pow((long double)k*k,1.0/3.0);
long double x=pow((long double)(k),(long double)2.0/(long double)3.0);
//long double l1=0,l2=1,r1=1,r2=0;
int xx=round(x);
if(1LL*xx*xx*xx==1LL*k*k)
{
printf("%d/1\n",xx);
continue;
}
int l1=int(x),l2=1,r1=int(x)+1,r2=1,g;
long double ans1=0,ans2=0;
while(1)
{
int xt1=l1+r1,xt2=l2+r2;
g=gcd(xt1,xt2);
xt1/=g;
xt2/=g;
if(xt2>100000)
break;
long double mid1=xt1,mid2=xt2;
if(ans2==0||ffabs(mid1/mid2-x)<ffabs(ans1/ans2-x))
{
ans1=mid1;
ans2=mid2;
}
if(x<mid1/mid2)
{
// if(l2+mid2>100000)
// break;
r1=mid1;
r2=mid2;
}
else if(x>mid1/mid2)
{
//if(r2+mid2>100000)
// break;
l1=mid1;
l2=mid2;
}
else
break;
}
int mid1=ans1,mid2=ans2;
g=gcd(mid1,mid2);
printf("%d/%d\n",mid1/g,mid2/g);
}
return 0;
}