All possible combinations of coins - c++

Question

I need to write a program which displays all possible change combinations given an array of denominations [1 , 2, 5, 10, 20, 50, 100, 200] // 1 = 1 cent
Value to make the change from = 300
I'm basing my code on the solution from this site http://www.geeksforgeeks.org/dynamic-programming-set-7-coin-change/
#include<stdio.h>
int count( int S[], int m, int n )
{
int i, j, x, y;
// We need n+1 rows as the table is consturcted in bottom up manner using
// the base case 0 value case (n = 0)
int table[n+1][m];
// Fill the enteries for 0 value case (n = 0)
for (i=0; i<m; i++)
table[0][i] = 1;
// Fill rest of the table enteries in bottom up manner
for (i = 1; i < n+1; i++)
{
for (j = 0; j < m; j++)
{
// Count of solutions including S[j]
x = (i-S[j] >= 0)? table[i - S[j]][j]: 0;
// Count of solutions excluding S[j]
y = (j >= 1)? table[i][j-1]: 0;
// total count
table[i][j] = x + y;
}
}
return table[n][m-1];
}
// Driver program to test above function
int main()
{
int arr[] = {1, 2, 5, 10, 20, 50, 100, 200}; //coins array
int m = sizeof(arr)/sizeof(arr[0]);
int n = 300; //value to make change from
printf(" %d ", count(arr, m, n));
return 0;
}
The program runs fine. It displays the number of all possible combinations, but I need it to be more advanced. The way I need it to work is to display the result in following fashion:
1 cent: n number of possible combinations.
2 cents:
5 cents:
and so on...
How can I modify the code to achieve that ?

Answer 1

Greedy Algorithm Approach
Have this denominations in an int array say, int den[] = [1 , 2, 5, 10, 20, 50, 100, 200]
Iterate over this array
For each iteration do the following
Take the element in the denominations array
Divide the change to be allotted number by the element in denominations array number
If the change allotted number is perfectly divisible by the number in denomination array then you are done with the change for that number.
If the number is not perfectly divisible then check for the remainder and do the same iteration with other number
Exit the inner iteration once you get the value equal to the change number
Do the same for the next denomination available in our denomination array.
Explained with example
den = [1 , 2, 5, 10, 20, 50, 100, 200]
Change to be alloted : 270, let take this as x
and y be the temporary variable
Change map z[coin denomination, count of coins]
int y, z[];
First iteration :
den = 1
x = 270
y = 270/1;
if x is equal to y*den
then z[den, y] // z[1, 270]
Iteration completed
Second Iteration:
den = 2
x = 270
y = 270/2;
if x is equal to y*den
then z[den , y] // [2, 135]
Iteration completed
Lets take a odd number
x = 217 and den = 20
y= 217/20;
now x is not equal to y*den
then update z[den, y] // [20, 10]
find new x = x - den*y = 17
x=17 and identify the next change value by greedy it would be 10
den = 10
y = 17/10
now x is not equal to y*den
then update z[den, y] // [10, 1]
find new x = x - den*y = 7
then do the same and your map would be having following entries
[20, 10]
[10, 1]
[5, 1]
[2, 1]