#include <iostream>
using namespace std;
long ways [10000];
void count ()
{
int coin [] = {1,5,10,25,50},i,j;
ways [0] = 1;
for (i = 0; i < 5; i++ ) {
for (j = coin [i]; j < 10000; j++ )
ways [j] += ways [j - coin [i]];
}
}
int main ()
{
count ();
int n;
while ( cin>>n )
cout<<ways[n]<<endl;
return 0;
}
good work (y)!
ReplyDeletePlease write dp approach also.
ReplyDeleteThanks.