Finding the minimal product - c++

Question

How do you find the minimal product from an array? This is the problem I have and the attempted solution isn't working. What have I done wrong?
https://www.codechef.com/problems/CHRL4
After visiting a childhood friend, Chef wants to get back to his home. Friend lives at the first street, and Chef himself lives at the N-th (and the last) street. Their city is a bit special: you can move from the X-th street to the Y-th street if and only if 1 <= Y - X <= K, where K is the integer value that is given to you. Chef wants to get to home in such a way that the product of all the visited streets' special numbers is minimal (including the first and the N-th street). Please, help him to find such a product.
Input
The first line of input consists of two integer numbers - N and K - the number of streets and the value of K respectively. The second line consist of N numbers - A1, A2, ..., AN respectively, where Ai equals to the special number of the i-th street.
Output
Please output the value of the minimal possible product, modulo 1000000007.
Constraints
1 ≤ N ≤ 10^5
1 ≤ Ai ≤ 10^5
1 ≤ K ≤ N
Example
Input:
4 2
1 2 3 4.
Output:
8
#include <iostream>
using namespace std;
int P(int A[], int N, int K) {
if (N == 1) return A[0];
int m = A[0], prod = m;
for (int i = 1; i < N; ++i) {
if (1 <= A[i]-m && A[i]-m <= K) {
prod *= A[i];
}
}
return prod;
}
int main() {
int A[] = {1, 2, 3, 4};
cout << P(A, 4, 2);
}
I get 6 instead of 8.

Answer 1

Such problems can typically be solved by Dynamic Programming:
Construct an appropriate state variable: Let the state be S = current street. Let the factor at street S be calledC_S
For each state S, collect the possible actions: a(S) = {go to any street T for which : 1 <= C_T - C_S <= K, T <=N }, a(N) = {}.
Introduce a value function V(S) = minimal product to get from S to N. Set V(N) = C_N.
Having all this together, one can now solve the Bellman equation backwards from N, where particularly the value V(0) is sought:
V(S) = min_{allowed T} { V(T)*C_S }
Example implementation:
int main()
{
int N = 4;
int K = 2;
std::vector<int> C{1,2,3,4};
std::vector<int> V(N);
V.back() = C.back();
for(int i = N - 2; i>= 0; --i)
{
int min = std::numeric_limits<int>::max(); //possible overflow here,
//better change that
for(int j=i+1; j< N; ++j)
{
double DeltaC = C[j] - C[i];
if(DeltaC <= K && DeltaC >= 1)
{
double vt = V[j] * C[i];
if(vt < min)
{
min = vt;
}
}
}
V[i] = min;
}
std::cout<<V[0]<<std::endl;
}
DEMO
The output is 8.
Please understand the code, test it and then use it with a good conscience (whatever that means).