HDU4447 Yuanfang, What Do You Think?

2012 Asia JinHua Regional Contest
Problem F. Yuanfang, What Do You Think?
Official solution report:
Manual or written brute-force programs calculate less n, and we can see that the decomposition formula of (x ^ N-1) contains the decomposition formula of (x ^ ni-1.
Among them, ni is the approximate number of n (except n itself ).
In addition to the decomposition formula (x ^ ni-1), (x ^ N-1) also contains a unique decomposition formula, recorded as P (n ).
So we get the decomposition of (x ^ N-1): (x ^ N-1) = P (n1) P (n2) P (n3)... P (n)
Among them, n1, n2, n3,... is the approximate number of n (except for n itself ).
From n = 1, P (1) = X-1.
All n, P (n) can be calculated using the preceding result: P (n) = (x ^ N-1)/P (n1)/P (n2) /P (n3 )/...
Finally, sort P (n1) P (n2) P (n3)... P (n.
Time Complexity: O (n ^ 2 * logn ).

[Cpp]
# Include <cstdio>
# Include <cstring>
# Include <algorithm>
Using namespace std;
Const int MAXN = 1111;
 
Inline void printX (int x, int v, bool show)
{
If (0 = v)
{
Return;
}
If (show)
{
If (v> 0)
{
Printf ("+ ");
}
Else
{
Printf ("-");
}
}
If (0 = x)
{
Printf ("1 ");
}
Else
{
If (v! = 1 & v! =-1)
{
Printf ("% d", abs (v ));
}
Printf ("x ");
If (x> 1)
{
Printf ("^ % d", x );
}
}
}
 
Struct Polynomial
{
Int n;
Int value [MAXN];
 
Void output () const
{
PrintX (n, value [n], false );
For (int I = n-1; I> = 0; -- I)
{
PrintX (I, value [I], true );
}
}
} Intrinsic [MAXN], sets [MAXN];
Int setNumber;
Bool visit [MAXN];
 
Bool operator <(const Polynomial & a, const Polynomial & B)
{
If (a. n = B. n)
{
For (int I = a. n; I> = 0; -- I)
{
If (abs (a. value [I]) = abs (B. value [I])
{
If (a. value [I]! = B. value [I])
{
Return a. value [I] <B. value [I];
}
}
Else
{
Return abs (a. value [I]) <abs (B. value [I]);
}
}
}
Return a. n <B. n;
}
 
Polynomial operator/(Polynomial a, const Polynomial & B)
{
Polynomial c;
C. n = a. n-B. n;
For (int I = c. n; I> = 0; -- I)
{
If (a. n-B. n <I)
{
C. value [I] = 0;
Continue;
}
C. value [I] = a. value [a. n]/B. value [B. n];
Int d = a. n-B. n;
A. n-= B. n;
For (int j = B. n; j> = 0; -- j)
{
A. value [j + d]-= B. value [j] * c. value [I];
If (a. value [j + d])
{
A. n = max (a. n, j + d );
}
}
}
Return c;
}
 
Void makeSet (int n)
{
SetNumber = 0;
For (int I = 1; I <= n; ++ I)
{
If (0 = n % I)
{
If (! Visit [I])
{
Visit [I] = true;
Intrinsic [I]. n = I;
Intrinsic [I]. value [I] = 1;
Intrinsic [I]. value [0] =-1;
For (int j = 1; j <I; ++ j)
{
If (0 = I % j)
{
Intrinsic [I] = intrinsic [I]/intrinsic [j];
}
}
}
Sets [setNumber ++] = intrinsic [I];
}
}
Sort (sets, sets + setNumber );
}
 
Int main ()
{
Int n;
While (scanf ("% d", & n), n)
{
If (1 = n)
{
Printf ("X-1 \ n ");
}
Else
{
MakeSet (n );
For (int I = 0; I <setNumber; ++ I)
{
Printf ("(");
Sets [I]. output ();
Printf (")");
}
Printf ("\ n ");
}
}
Return 0;
}