Need help in a dp problem

Atleast paste the question link. and try int instead of long long

Here is My code ...... Use Long long int to be on safe side

void Here_We_Go() { int n,x; cin >> n >> x; vector v(n); for (int i = 0; i < n;i++) cin >> v[i]; vector dp(x + 1, 0); dp[0] = 1;

for(int e : v )
{
    for (int i = e; i <= x;i++)
    {
        (dp[i] += dp[i - e]) %= mod;
    }
}
cout << dp[x] << endl;

In the Coin Combinations I problem, let's assume that dp[k] is the number of ways that we can create a sum of k from given coins. It is clear that dp[0] = 1, which means we always can create sum = 0 with 0 coin.

Thus, to do this task, iterate sum value from 1 to X, then for each sum value, we find all combinations possible with given n coins, each combination of that coin is caculated in to the dp[] once. The code will look like this:

In Coin Combinations II problem, let's recall the define of dp[k]: number of ways we can create sum of k from given coins where each coin has infinite count. For example, if there is a coin with value '2' and we want to create sum of 14, there are possible sum could be created: 2,4,6,8,10,12 and 14

As can be seen, there is a little different from the previous problem: with each coin, we have to find all possible value could be created from that coin, with each combination of that coin could be caculated in to dp[] more than once. Then the code should look like this:

I hope this would help you.