2021-10-24: cross the river with least steps

 The outputs:

Inputs
5           // the number of elements in array    
2 0 1 1 1   // the values in array
outputs:
4            // that means we need at least 4 steps to cross river

 The codes:

# include <iostream>
# include <iomanip>
# include <climits>
using namespace std;
void create_array(int* &a, int n);
int cross_river(int a[], int n);
int main ()
{
    int n; // the length of the array or the interval length of river
    cin >> n;
    int* a;
    create_array (a, n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    cout << cross_river(a, n);
    return 0;
}
void create_array (int* &a, int n)
{
    a = new int[n];
    for(int i = 0; i < n; i ++){
        *(a + i) = INT_MAX;
    }
}
int cross_river(int a[], int n)
{
    int* dp;
    create_array (dp, n+1); // create the dynamic programming array
    dp[0] = 0; // that means from bank to bank no needs step
    for (int i = 1; i <= n; i ++){
        for (int j = 0; j < i; ++j){
            if (a[j] + j >= i){
                dp[i] = min (dp[j]+1, dp[i]);
            }
        }
    }
    return dp[n] < INT_MAX ? dp[n] : -1; // it is obvious that dp[n] store the least steps from bank to the other.  
}