Bytelandian gold coins - c++

Question

I am solving this problem -> http://www.spoj.com/problems/COINS/ . A very simple DP problem with a very straight forward DP approach.I found enough hints in the problem statement to use DP. All the test cases are running perfectly in my compiler but I am getting a WA in SPOJ. My code is as follows :
My code
#include <cstdio>
#include <map>
#include <cstring>
#include<algorithm>
using namespace std;
map< long long,long long > data;
map < long long,long long> :: iterator p;
int max(int a,int b)
{
if(a>b)return a;
return b;
}
long long calc(int n)
{
long long c;
if(n==0 || n==1 || n==2)
return n;
p = data.find(n);
if(p==data.end())
{
c = max(n, calc(n/2) + calc(n/3) + calc(n/4));
data.insert(p, pair < long long, long long > (n, c));
return c;
}
else return (*p).second;
}
int main()
{
int t;
long long n;
scanf("%d",&t);
if(t>10)return 0;
while(t--)
{
scanf("%lld",&n);
if(n<0 || n>1000000000)
break;
data.clear();
printf("%lld",calc(n));
}
return 0;
}
I am finding it really difficult for me to figure out where I am going wrong!
A test case which contradicts my code would also do.

Answer 1

Perhaps a stack overflow in calculate. The recursion is killing your program :-)
Or simply the fact that calculate(1000000000) is too much slow.

Answer 2

Use Dynamic programming , store your result in an array . Since the value can go upto 10^9, and you cann't take array of that size , just take array of size till 10^6 and store their result and rest value calculate using simple recursion .

Answer 3

Here is a solution in python
import sys
mydict = {}
def count(n):
if n <= 5:
return n
elif n in mydict.keys():
return mydict[n]
else:
k=max(n,count(n / 2) + count(n / 3) + count(n / 4))
mydict[n]=k
return mydict[n]
for line in sys.stdin:
res = count(int(line))
print(int(res))

Answer 4

Here's my solution, uses dp :
#include<bits/stdc++.h>
using namespace std;
map<long long int,long long int> m;
long long int dp(long long int k){
long long int a;
if(k==0){
return 0;
}
a=m[k];
/* if(k<12){
return k;
} */
if(k<12){
return k;
}
else if(a==0){
a=max(k,dp(k/2)+dp(k/4)+dp(k/3));
m[k]=a;
}
return a;
}
int main(){
long long int n,t;
while(scanf("%lld",&n)>0){
t=dp(n);
cout << t << endl;
}
return 0;
}