Use of Function Arctan

 

 Use of Function Arctan

Time Limit:10000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit   Status 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 conta

ins a integer number T.T lines follow,each contains a single integer A, 1<=A<=60000.

Output T lines,each contains a single integer whic

h denotes to the minimum sum B+C.

Sample Input 1 1

 

/*由题意知:B(C-A)=1+AC;B=（1+AC）/(C-A)；B=A+（1+AA）/(C-A);C增B减；（易知B=C时B+C最小（此时B=C=a可能为小数，故只需B--枚举即可！！！））*/






#include<stdio.h>
typedef long long LL;
int main()
{
    LL B,C,A,i,j,T;
    scanf("%lld",&T);    
    for(i=1;i<=T;i++)
    {
        scanf("%lld",&A);
        LL tmp=2*A+2;
        while((tmp--)>A)
        if((A*tmp+1)%(tmp-A)==0)
        {
            B=(A*tmp+1)/(tmp-A);
            printf("%lld\n",tmp+B);
            break;
        }    
    }
}