Description
It's easy to know that arctan(1/2)+arctan(1/3)=arctan(1).The problem is,to some fixed number A,you have to 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)
Input
The first line contains a integer number T.T lines follow,each contains a single integer A, 1<=A<=60000.
Output
T lines,each contains a single integer which denotes to the minimum sum B+C.
Sample Input
1 1
Sample Output
5
Hint
Some new test data has been added on Feb.15, 2009, 36 users lose their Accepted.
Source limit: 256B
Languages : All except: C99 strict ERL JS
題意：求滿足等式的最小的B+C的值。
可以等式兩邊同時求tan值，
arctan(1/A)=arctan(1/B)+arctan(1/C) ，則左右同時取tan的值，便將等式還原爲 1/A=(1/B+1/C) / (1-1/B*1/C)
繼續化簡爲A=(B*C-1)/(B+C) 。 設 B=A+m ， C=A+n帶入上面的表達式化簡得到m*n=A*A+1， 而要求的是B+C=2*A+m+n的
最小值，知m+n>=2*sqrt(m*n)=2*sqrt(A*A+1)，要取得‘=’號需m=n=sqrt(A*A+1)，因A,B,C都爲 整數，所以m,n也必須爲整數
此時需要的是枚舉m=sqrt(A*A+1)到1，得到第一個能整除sqrt(A*A+1)的m便是需要的值。
這題代碼長度限制的很嚴。
#include <stdio.h>
#include <cmath>
int main()
{int T;long long a,b,c,n,m;scanf("%d",&T);
while(T--)
{scanf("%lld",&a);long long t=(long long)sqrt((a*a+1)*1.0);while((a*a+1)%t)t--;printf("%lld\n",2*a+t+(a*a+1)/t);
}
return 0;
}