Money Matters Solution: A recursive approach

Problem Statement:
The Government of Byteland has decide to issue new currency notes with special protection features to commemorate a great mathematician. 

It has decided to issue notes summing up to N such that all the sums from 1 to N should be possible to be made by selecting some of the notes. Also this has to be done in way such that if any note from 1 to N can be made in more than one way by combining other selected notes then that set of notes is not valid.

For example, with N = 7,

The valid sets are:

Invalid sets are:
{1,1,1,2,2} beacuse the sum adds up to 7 but 2 can be made by {1,1} and {2}, 3 can be made by {1,1,1} and {1,2}, 4 can be made by {1,1,2} and {2,2} and similarly 5, 6 and 7 can also be made in multiple ways using the same set.
{1,1,3,6} beacuse all from 1 to 7 can be uniquely made but the sum is not 7 (its 11).

Before implementing this idea of issuing these new set of notes, the Government wants to know how many possible valid sets are there for a given N. Your program should help the Government to find this out.


First line will contain the number of test cases T (1<=T<=6666).
Each test case will have one line containing an integer N (<= 2^31-1).


For each value of N output the total number of valid sets on a separate line.





//Solved by Sanjay Verma
  1. #include<stdio.h>
  2. #define memSize 3333
  3. int noofsol(int);
  4. int count[memSize];
  5. main()
  6. {
  7. int t,x;
  8. count[0]=0;
  9. count[1]=1;
  10. count[2]=1;
  11. count[3]=2;
  12. count[4]=1;
  13. scanf("%d",&t);
  14. while(t--)
  15. {
  16. scanf("%d",&x);
  17. printf("%d\n",noofsol(x));
  18. }
  19. return 0;
  20. }
  21. int noofsol(int x)
  22. {
  23. int sum=1,i,j;
  24. if((x<memSize-1) && (count[x]!=0 || x==0))
  25. return count[x];
  26. for(i=1;i<=x;i++)
  27. {
  28. for(j=1;j<=x/i;j++)
  29. {
  30. if(j*i + i-1 == x)
  31. sum+=noofsol(i-1);
  32. }
  33. }
  34. if(x<memSize-1)
  35. count[x]=sum;
  36. return sum;
  37. }
The original problem on codechef can be found here.
Note: This is a recursive solution. For achieving faster speed, you may convert it into iterative code.
If you want any clarification, please comment.
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