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

512345

 

Sample Output

1/1153008/9638950623/2433796389/38252226164/77347

 

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;}