ARCTAN - Use of Function Arctan

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ARCTAN

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Description

相信大家都看得懂英文吧，我就不翻译了吧…………

It is easy to know that

arctan(1/2)+arctan(1/3)=arctan(1)

Given number A, write a program to calculate the minimum sum B+C.
A，B and C are all positive integers and satisfy the equation below:

arctan(1/A)=arctan(1/B)+arctan(1/C)

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Input

The first line contains a integer number T(about 1000).
T lines follow,each contains a single integer A，1≤A≤60000.

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Output

T lines，each contains a single integer which denotes to the minimum sum B+C.

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Sample Input

1

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Sample Output

5

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Solution

解法一：附上扫雷神犇Cai的坑爹解法。
解法二：
由

tan(α+β)=1A,tanβ=1B,tanα=1C

得

A=BC−1B+C

设 B+C=k，则

BC=Ak+1

，
则 B、C 为一元二次方程

x2−kx+Ak+1=0

的两个根。
那么

k2−4Ak−4

为完全平方数。
设 k2−4Ak−4=s2，

k=4A+16A2+16+4s2−−−−−−−−−−−−−√2=2A+4A2+s2+4−−−−−−−−−−−√

则 4A2+s2+4−−−−−−−−−−−√ 为完全平方数。
设 4A2+s2+4=r2，则 4A2+4=(r+s)(r−s)
枚举 4A2+4 的约数即可。

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Code

因为有代码长度限制，所以读者在复制代码的时候请注意删除缩进以及空格…………

#include<iostream>#include<cmath>typedef unsigned long long L;using namespace std;L t,i,T,a,b,k;main(){    cin>>t;    while(t--){        cin>>a;        T=a*a*4+4;        b=sqrt(T);        i=b+1;        while(--i)            if(!(T%i)&&!((i+T/i)&1)){                k=(T/i-i)/2;                break;            }        cout<<2*a+L(sqrt(T+k*k)+0.5)<<endl;    }    return 0;}