Application of poj 1183 arc tangent function (push formula)

Arc tangent function application

Time limit:1000 ms		Memory limit:10000 K
Total submissions:14468		Accepted:5227

Description

The arc tangent function can be expanded into an infinite series, with the following formula

(0 <= x <= 1) formula (1)

Using the arc tangent function to calculate pi is a common method. For example, the simplest method to calculate pi is as follows:

Pi = 4 arctan (1) = 4 (1-1/3 + 1/5-1/7 + 1/9-1/11 +...) formula (2)

However, the efficiency of this method is very low, but we can use the tangent function formula based on the angle:

Tan (a + B) = [tan (A) + Tan (B)]/[1-Tan (a) * Tan (B)] formula (3)

The following is a simple transformation result:

Arctan (p) + arctan (q) = arctan [(p + q)/(1-pq)] formula (4)

Using this formula, if p = 1/2, q = 1/3, (p + q)/(1-pq) = 1

Arctan (1/2) + arctan (1/3) = arctan [(1/2 + 1/3)/(1-1/2*1/3)] = arctan (1)

Arctan (1) is calculated using the arc tangent of 1/2 and 1/3, and the speed is much faster.
The formula (4) is written as follows:

Arctan (1/a) = arctan (1/B) + arctan (1/C)

Both A, B, and C are positive integers.

Our problem is: For each given a (1 <= A <= 60000), evaluate the value of B + C. We ensure that there is an integer solution for any. If there are multiple solutions, you must provide the minimum solution of B + C.

Input

The input file only has a positive integer a, where 1 <=a <= 60000.

Output

The output file contains only one integer, which is the value of B + C.

Sample Input

 
1
 

Sample output

 
5
 

Source

Noi 01 question about a formula, pay attention to the data range

1/A = (1/B + 1/C)/(1-1/(B * C) => bc-1 = A (B + C) assume B = a + M and C = a + N (B and C is always bigger than a) (a + M) (a + n) -1 = a (a + M + A + n) => A * A + A * n + A * m + M * n-1 = 2 * a * A + M * A + N * A => M * n = A * + 1

Enumerate M (or N ).

The key is B = a + M, c = a + N. Here is a conversion.

 1 # Include <cstdio> 2 # Include <cmath>3 # Include <cstring> 4 # Include <stdlib. h> 5 # Include <algorithm> 6   # Define Ll _ int64 7   Using   Namespace  STD;  8   Int  Main ()  9   {  10      //  Freopen ("in.txt", "r", stdin );  11   Ll;  12 Scanf ( "  % I64d  " ,& A );  13       For (Ll I = A; I> = 0 ; I --) //  Enumerative m  14   { 15           If (A * A + 1 ) % I = 0 ) //  N = (A * A + 1)/I m = I  16   {  17 Printf ( "  % I64d \ n  " , A + I + A + (A * A + 1 )/ I );  18              Break  ;  19   }  20   }  21       Return   0  ;  22 }

View code

 

Application of poj 1183 arc tangent function (push formula)