how can i separate number by plus other unique number?
up vote
-3
down vote
favorite
i have a problem that i have to split that number to others positive number(all are unique)
for example 6=6=5+1=1+2+3=4+2 so we have 4 ways, 5=5=4+1=3+2 so we have 3 ways, i think about
recursive
for example is 6 my idea is from a starter point is 0, i have 0+(6,0) then recursive 0+(5,1) then 0+(4,2) then 0+ (3,3) 3,3 is wrong because numbers are unique, then move to 1 like
1+ (5,1) but is same as above,like 1+4+1 is wrong how can i do this problem?, please give an idea or for example for me, thanks a lots. i have been making this problem for 4 days but nothing happen,I'm using c++
i didn't write code yet, because i though it wrong, but here is my fast code:
#include <iostream>
using namespace std;
int count=1;
int recursive (int n,int k)
{
if(k>=n)
{
return 0;
}
count++;
for(int i=k;i<n;i++) {
recursive(n-i,i+1);
}
return count;
}
int main()
{
int n;
cin>>n;
cout<< recursive (n,0);
}
c++ recursion
asked Nov 7 at 9:25
Quốc Khánh Lê
184
|
show 8 more comments
up vote
-3
down vote
favorite
i have a problem that i have to split that number to others positive number(all are unique)
for example 6=6=5+1=1+2+3=4+2 so we have 4 ways, 5=5=4+1=3+2 so we have 3 ways, i think about
recursive
for example is 6 my idea is from a starter point is 0, i have 0+(6,0) then recursive 0+(5,1) then 0+(4,2) then 0+ (3,3) 3,3 is wrong because numbers are unique, then move to 1 like
1+ (5,1) but is same as above,like 1+4+1 is wrong how can i do this problem?, please give an idea or for example for me, thanks a lots. i have been making this problem for 4 days but nothing happen,I'm using c++
i didn't write code yet, because i though it wrong, but here is my fast code:
#include <iostream>
using namespace std;
int count=1;
int recursive (int n,int k)
{
if(k>=n)
{
return 0;
}
count++;
for(int i=k;i<n;i++) {
recursive(n-i,i+1);
}
return count;
}
int main()
{
int n;
cin>>n;
cout<< recursive (n,0);
}
c++ recursion
asked Nov 7 at 9:25
Quốc Khánh Lê
184
2
seems like you already have an idea for an algorithm, why dont you write some code to see how it does?
– user463035818
Nov 7 at 9:27
Your starter point should be 1 and the next number must be bigger than the current number to ensure uniqueness
– Thomas Sablik
Nov 7 at 9:28
i have but problem is my idea is probably not effective, because it must have unique number
– Quốc Khánh Lê
Nov 7 at 9:29
efficiency should be your last concern when implmenting version zero. If you dont want solutions where a number appears twice, just skip them
– user463035818
Nov 7 at 9:30
As I if your numbers are strictly increasing than your solutions will be unique. 1+5 will be a valid solution but 5+1 won't.
– Thomas Sablik
Nov 7 at 9:31
|
show 8 more comments
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
i have a problem that i have to split that number to others positive number(all are unique)
for example 6=6=5+1=1+2+3=4+2 so we have 4 ways, 5=5=4+1=3+2 so we have 3 ways, i think about
recursive
for example is 6 my idea is from a starter point is 0, i have 0+(6,0) then recursive 0+(5,1) then 0+(4,2) then 0+ (3,3) 3,3 is wrong because numbers are unique, then move to 1 like
1+ (5,1) but is same as above,like 1+4+1 is wrong how can i do this problem?, please give an idea or for example for me, thanks a lots. i have been making this problem for 4 days but nothing happen,I'm using c++
i didn't write code yet, because i though it wrong, but here is my fast code:
#include <iostream>
using namespace std;
int count=1;
int recursive (int n,int k)
{
if(k>=n)
{
return 0;
}
count++;
for(int i=k;i<n;i++) {
recursive(n-i,i+1);
}
return count;
}
int main()
{
int n;
cin>>n;
cout<< recursive (n,0);
}
c++ recursion
asked Nov 7 at 9:25
Quốc Khánh Lê
184
i have a problem that i have to split that number to others positive number(all are unique)
for example 6=6=5+1=1+2+3=4+2 so we have 4 ways, 5=5=4+1=3+2 so we have 3 ways, i think about
recursive
for example is 6 my idea is from a starter point is 0, i have 0+(6,0) then recursive 0+(5,1) then 0+(4,2) then 0+ (3,3) 3,3 is wrong because numbers are unique, then move to 1 like
1+ (5,1) but is same as above,like 1+4+1 is wrong how can i do this problem?, please give an idea or for example for me, thanks a lots. i have been making this problem for 4 days but nothing happen,I'm using c++
i didn't write code yet, because i though it wrong, but here is my fast code:
#include <iostream>
using namespace std;
int count=1;
int recursive (int n,int k)
{
if(k>=n)
{
return 0;
}
count++;
for(int i=k;i<n;i++) {
recursive(n-i,i+1);
}
return count;
}
int main()
{
int n;
cin>>n;
cout<< recursive (n,0);
}
c++ recursion
c++ recursion
asked Nov 7 at 9:25
Quốc Khánh Lê
184
asked Nov 7 at 9:25
Quốc Khánh Lê
184
edited Nov 7 at 9:38
asked Nov 7 at 9:25
Quốc Khánh Lê
184
asked Nov 7 at 9:25
Quốc Khánh Lê
184
asked Nov 7 at 9:25
Quốc Khánh Lê
184
184
2
seems like you already have an idea for an algorithm, why dont you write some code to see how it does?
– user463035818
Nov 7 at 9:27
Your starter point should be 1 and the next number must be bigger than the current number to ensure uniqueness
– Thomas Sablik
Nov 7 at 9:28
i have but problem is my idea is probably not effective, because it must have unique number
– Quốc Khánh Lê
Nov 7 at 9:29
efficiency should be your last concern when implmenting version zero. If you dont want solutions where a number appears twice, just skip them
– user463035818
Nov 7 at 9:30
As I if your numbers are strictly increasing than your solutions will be unique. 1+5 will be a valid solution but 5+1 won't.
– Thomas Sablik
Nov 7 at 9:31
|
show 8 more comments
2
seems like you already have an idea for an algorithm, why dont you write some code to see how it does?
– user463035818
Nov 7 at 9:27
Your starter point should be 1 and the next number must be bigger than the current number to ensure uniqueness
– Thomas Sablik
Nov 7 at 9:28
i have but problem is my idea is probably not effective, because it must have unique number
– Quốc Khánh Lê
Nov 7 at 9:29
efficiency should be your last concern when implmenting version zero. If you dont want solutions where a number appears twice, just skip them
– user463035818
Nov 7 at 9:30
As I if your numbers are strictly increasing than your solutions will be unique. 1+5 will be a valid solution but 5+1 won't.
– Thomas Sablik
Nov 7 at 9:31
2
2
seems like you already have an idea for an algorithm, why dont you write some code to see how it does?
– user463035818
Nov 7 at 9:27
seems like you already have an idea for an algorithm, why dont you write some code to see how it does?
– user463035818
Nov 7 at 9:27
Your starter point should be 1 and the next number must be bigger than the current number to ensure uniqueness
– Thomas Sablik
Nov 7 at 9:28
Your starter point should be 1 and the next number must be bigger than the current number to ensure uniqueness
– Thomas Sablik
Nov 7 at 9:28
i have but problem is my idea is probably not effective, because it must have unique number
– Quốc Khánh Lê
Nov 7 at 9:29
i have but problem is my idea is probably not effective, because it must have unique number
– Quốc Khánh Lê
Nov 7 at 9:29
efficiency should be your last concern when implmenting version zero. If you dont want solutions where a number appears twice, just skip them
– user463035818
Nov 7 at 9:30
efficiency should be your last concern when implmenting version zero. If you dont want solutions where a number appears twice, just skip them
– user463035818
Nov 7 at 9:30
As I if your numbers are strictly increasing than your solutions will be unique. 1+5 will be a valid solution but 5+1 won't.
– Thomas Sablik
Nov 7 at 9:31
As I if your numbers are strictly increasing than your solutions will be unique. 1+5 will be a valid solution but 5+1 won't.
– Thomas Sablik
Nov 7 at 9:31
|
show 8 more comments
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2
seems like you already have an idea for an algorithm, why dont you write some code to see how it does?
– user463035818
Nov 7 at 9:27
Your starter point should be 1 and the next number must be bigger than the current number to ensure uniqueness
– Thomas Sablik
Nov 7 at 9:28
i have but problem is my idea is probably not effective, because it must have unique number
– Quốc Khánh Lê
Nov 7 at 9:29
efficiency should be your last concern when implmenting version zero. If you dont want solutions where a number appears twice, just skip them
– user463035818
Nov 7 at 9:30
As I if your numbers are strictly increasing than your solutions will be unique. 1+5 will be a valid solution but 5+1 won't.
– Thomas Sablik
Nov 7 at 9:31