Hdu 3187 (Euler's Function + dfs)

 

/*

Given a number n, where P (k) = n is satisfied, where k has a number of prime factors <= 3; n <2 ^ 31;
Euler's function:
Phi (k) = k * (1-1/p1) (1-1/p2) (1-1/p3)
= (P1-1) * p1 ^ x * (p2-1) * p2 ^ y * (p3-1) * p3 ^ z;
Phi (1) = 1;
Because n = phi (k), enumeration pi-1, if (n % (pi-1) = 0 & is_prime (pi) is added to pi;
Then, use DFS to make phi (k) = n at most;
Note that phi (1) = 1;
When n = 1, k = 1, 2;

Note that the enumeration (pi-1) here is not the prime factor of n;

 

 

 

2013/04/22-

*/

 

 

[Cpp]
# Include "stdio. h"
# Include "string. h"
# Include "algorithm"
Using namespace std;
Int n, ans;
Int A [4010];
Int cnt;
Int ret;
Int t [3];
Int is_prime (int x)
{
Int I;
For (I = 2; I * I <= x; I ++)
If (x % I = 0) return 0;
Return 1;
}
// Starting from the number x, y has been found, and z is required in total...
Void dfs (int x, int y, int z)
{
Int I;
If (y = z)
{
Int tem = n;
Int f = 0;
For (I = 0; I <z; I ++)
{
If (tem % (t [I]-1 )! = 0)
{
F = 1; break;
}
Tem/= (t [I]-1 );
}
If (f = 0)
{
For (I = 0; I <z; I ++)
{
While (tem % t [I] = 0)
Tem/= t [I];
}
If (tem = 1) ret ++;
}
}
Else
{
For (I = x; I <cnt; I ++)
{
T [y] = A [I];
Dfs (I + 1, y + 1, z );
}
}
}
 
Int main ()
{
While (scanf ("% d", & n )! =-1)
{
If (n = 1)
{
Printf ("2 \ n ");
Continue;
}
Else
{
Int I, j;
Cnt = 0;
// Store the prime factor n into.
For (I = 1; I * I <= n; I ++)
{
If (n % I = 0 & is_prime (I + 1) = 1)
A [cnt ++] = I + 1;
If (n % I = 0 & is_prime (n/I + 1) = 1)
A [cnt ++] = n/I + 1;
}
Sort (A, A + cnt );
// There may be duplicates...
For (I = 1; I <cnt; I ++)
{
If (A [I] = A [I-1])
{
For (j = I; j <cnt-1; j ++)
A [j] = A [j + 1];
Cnt --;
}
}
Ans = 0;
For (I = 1; I <= 3; I ++)
{
Ret = 0;
Dfs (0, 0, I );
Ans + = ret;
}
Printf ("% d \ n", ans );
}
}
Return 0;
}

# Include "stdio. h"
# Include "string. h"
# Include "algorithm"
Using namespace std;
Int n, ans;
Int A [4010];
Int cnt;
Int ret;
Int t [3];
Int is_prime (int x)
{
Int I;
For (I = 2; I * I <= x; I ++)
If (x % I = 0) return 0;
Return 1;
}
// Starting from the number x, y has been found, and z is required in total...
Void dfs (int x, int y, int z)
{
Int I;
If (y = z)
{
Int tem = n;
Int f = 0;
For (I = 0; I <z; I ++)
{
If (tem % (t [I]-1 )! = 0)
{
F = 1; break;
}
Tem/= (t [I]-1 );
}
If (f = 0)
{
For (I = 0; I <z; I ++)
{
While (tem % t [I] = 0)
Tem/= t [I];
}
If (tem = 1) ret ++;
}
}
Else
{
For (I = x; I <cnt; I ++)
{
T [y] = A [I];
Dfs (I + 1, y + 1, z );
}
}
}

Int main ()
{
While (scanf ("% d", & n )! =-1)
{
If (n = 1)
{
Printf ("2 \ n ");
Continue;
}
Else
{
Int I, j;
Cnt = 0;
// Store the prime factor n into.
For (I = 1; I * I <= n; I ++)
{
If (n % I = 0 & is_prime (I + 1) = 1)
A [cnt ++] = I + 1;
If (n % I = 0 & is_prime (n/I + 1) = 1)
A [cnt ++] = n/I + 1;
}
Sort (A, A + cnt );
// There may be duplicates...
For (I = 1; I <cnt; I ++)
{
If (A [I] = A [I-1])
{
For (j = I; j <cnt-1; j ++)
A [j] = A [j + 1];
Cnt --;
}
}
Ans = 0;
For (I = 1; I <= 3; I ++)
{
Ret = 0;
Dfs (0, 0, I );
Ans + = ret;
}
Printf ("% d \ n", ans );
}
}
Return 0;
}