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__Princess Farida__

Link to the question : FARIDA

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__HINT :__

The question asks us to take the maximum number of gold coins from the monsters such that if m_1, m_2, m_3......m_n are the monsters then you can take gold coin from monster m_i only if you had not taken from m_(i-1).

This can be solved very easily by dp. Let us take an array b[] such that b[i] stores the maximum number of gold coins you can get from m_1 to m_i monsters. So,

b[i] = max( gold coins taken from ith monster + b[j]) where 0<j<i-1 .

This can be solved very easily by dp. Let us take an array b[] such that b[i] stores the maximum number of gold coins you can get from m_1 to m_i monsters. So,

b[i] = max( gold coins taken from ith monster + b[j]) where 0<j<i-1 .

####
__RECOMMENDED QUESTION :__

After solving this dp question, you would love solving this question as well.

###
__SOURCE CODE :__

```
/* Princess Farida */
/* Sushant Gupta */
#include<iostream>
using namespace std;
int main()
{
int t,k=1;
cin>>t;
while(t--)
{
int n,i,j;
cin>>n;
long long int a[n],b[n],max=0;
for(i=0;i<n;i++)
{
cin>>a[i];
b[i]= 0;
}
b[0]= a[0];
b[1]= a[1];
for(i=2;i<n;i++)
{
for(j=0;j<i-1;j++)
{
if(b[i]< b[j] + a[i])
b[i]= b[j] + a[i];
}
}
for(i=0;i<n;i++)
{
if(max< b[i])
max = b[i];
}
cout<<"Case "<<k++<<": "<<max<<endl;
}
return 0;
}
```

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