A partition of an integer n is a way of writing n as a sum of positive integers. For
example, for n=7, a partition is 1+1+5. I need a program that finds all the
partitions of an integer 'n' using 'r' integers. For example, all the partitions of n=7
using r=3
integers are 1+1+5
, 1+2+4
, 1+3+3
, 2+2+3
.
This is what I have so far:
#include <iostream>
#include <vector>
using namespace std;
void print (vector<int>& v, int level){
for(int i=0;i<=level;i++)
cout << v[i] << " ";
cout << endl;
}
void part(int n, vector<int>& v, int level){
int first; /* first is before last */
if(n<1) return ;
v[level]=n;
print(v, level);
first=(level==0) ? 1 : v[level-1];
for(int i=first;i<=n/2;i++){
v[level]=i; /* replace last */
part(n-i, v, level+1);
}
}
int main(){
int num;
cout << "Enter a number:";
cin >> num;
vector<int> v(num);
part(num, v, 0);
}
The output of this program is:
Enter a number:5
5
1 4
1 1 3
1 1 1 2
1 1 1 1 1
1 2 2
2 3
Process returned 0 (0x0) execution time : 1.837 s
Press any key to continue.
How can I change my code so I can have that 'r' variable?
EDIT:
In case it was not clear, the 'r' value represents the number of integers per partition. So in the case above, if r=2, then the partitions can only have two integers in them. The partitions would be 4+1, and 3+2. The 'r' value should be entered by the user.
Essentially what Codor said, plus you don't need to recurse further into part()
once you found a partition of the target length since they would be longer:
#include <iostream>
#include <vector>
using namespace std;
void print (vector<int>& v, int level){
for(int i=0;i<=level;i++)
cout << v[i] << " ";
cout << endl;
}
void part(int n, vector<int>& v, int level, int r){
int first; /* first is before last */
if(n<1) return ;
v[level]=n;
if( level+1 == r ) {
print(v, level);
return;
}
first=(level==0) ? 1 : v[level-1];
for(int i=first;i<=n/2;i++){
v[level]=i; /* replace last */
part(n-i, v, level+1, r);
}
}
int main(){
int num,r;
cout << "Enter a number:";
cin >> num;
cout << "Enter size (r):";
cin >> r;
vector<int> v(num);
part(num, v, 0, r);
}
Output:
Enter a number:5
Enter size (r):2
1 4
2 3
A sort of "hack" would be to make r
an argument of part
, pass it along recursively an just print the output if level
equals r
.
How about this? Have an additional argument passed as reference for r, and increment r each time within the recursion block?
#include <iostream>
#include <vector>
using namespace std;
void print (vector<int>& v, int level){
for(int i=0;i<=level;i++)
cout << v[i] << " ";
cout << endl;
}
void part(int n, vector<int>& v, int level, int &r){
int first; /* first is before last */
if(n<1) return ;
v[level]=n;
print(v, level);
first=(level==0) ? 1 : v[level-1];
for(int i=first;i<=n/2;i++){
v[level]=i; /* replace last */
r++;
part(n-i, v, level+1, r);
}
}
int main(){
int num;
cout << "Enter a number:";
cin >> num;
int r = 0;
vector<int> v(num);
part(num, v, 0, r);
cout << "r = " << r << endl;
}
Output comes as:
Enter a number:5
1 4
1 1 3
1 1 1 2
1 1 1 1 1
1 2 2
2 3
r = 6
Is this what you are looking for?