joj 1839: Arctan

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	5s	8192K	262	42	Standard

Arctan function can expand to an infinite progression:

　

People often caculate pi using arctan functions.For example,the easiest way to calculate π is :

　

however,it's too inefficient,but we can use this formula:

　

that is

　

now，let ， then ，so

　

It'll be more faster to calculate arctan(1) using and     :

　

a , b and c are positive integers。

The problem is：For a given a()，caculate the value of (b+c)。If there are several solutions,just give the minimum one.

Input

The input contains several positive integers,one per line.

Output

For each a, print the minimum (b+c).

Sample Input

1

Sample Output

5

/*

败在数据类型了，long long会TLE，int 会WA,仔细算了下可能会有50000到60000的a 会使a^2+1是素数，则要一个unsigned型变量

*/

#include <cstdio>
unsigned m,ans,i,a;
int main ()
{
    while(scanf("%u",&a)!=EOF)
    {
        m=a*a+1;
        for (i=a ; i>0 ; i--)
        {
             if(!(m%i))
              {
                  ans=a*2+i+m/i;break;//倒序搜索第一个就是b+c最小的答案，数学的公式推导很重要
              }
        }
        printf("%u/n",ans );
    }
    return 0;
}