Programmer interview question selection (44): integer segmentation (that is, all combinations of a number N from the sum of numbers less than or equal to N)

Question Description: for example, given an integer 4, it has the following situation: 4 = 4;

4 = 3 + 1;

4 = 2 + 2;

4 = 2 + 1 + 1;

4 = 1 + 1 + 1 + 1;

The following is the split implementation code of the two versions.

# Include "stdio. H"

Int compute (INT number, int maximum)
{
If (number = 1 | maximum = 1)
Return 1;
Else if (number <maximum)
Return compute (number, number );
Else if (number = maximum)
Return 1 + compute (number, maximum-1 );
Else
Return compute (number, maximum-1) + compute (number-maximum, maximum );
}
 
Int intpartionno (int n) // calculates the total number of versions of the combination;
{
Return compute (N, N );
}

Int integerpartition (int n) // calculates the total number of combinations and prints out all versions;
{
Int * partition = new int [N] ();
Int * repeat = new int [N] ();

Partition [1] = N;
Repeat [1] = 1;
Int COUNT = 1;
Int part = 1;

Int last, smaller, remainder;

Printf ("% 3d", partition [1]);
Do
{
Last = (partition [Part] = 1 )? (Repeat [part --] + 1): 1;
Smaller = partition [Part]-1;
If (Repeat [Part]! = 1)
-- Repeat [Part ++];
Partition [Part] = smaller;
Repeat [Part] = 1 + last/smaller;

If (remainder = last % smaller )! = 0)
{
Partition [++ part] = remainder;
Repeat [Part] = 1;
}

++ Count;

Printf ("/N ");
For (INT I = 1; I <= part; ++ I)
For (Int J = 1; j <= repeat [I]; ++ J)
Printf ("% 3d", partition [I]);

} While (Repeat [Part]! = N );

If (partition)
{
Delete [] partition;
Partition = 0;
}
If (Repeat)
{
Delete [] Repeat;
Repeat = 0;
}

Return count;
}

Int main ()
{
Printf ("% d/N", intpartionno (4 ));
Integerpartition (4 );
Getchar ();
}