How will I solve this using DP?

Question

Question link: http://codeforces.com/contest/2/problem/B

There is a square matrix n × n, consisting of non-negative integer numbers. You should find such a way on it that

starts in the upper left cell of the matrix; each following cell is to the right or down from the current cell; the way ends in the bottom right cell. Moreover, if we multiply together all the numbers along the way, the result should be the least "round". In other words, it should end in the least possible number of zeros.

Input The first line contains an integer number n (2 ≤ n ≤ 1000), n is the size of the matrix. Then follow n lines containing the matrix elements (non-negative integer numbers not exceeding 10^9).

Output In the first line print the least number of trailing zeros. In the second line print the correspondent way itself.

I thought of the following: In the end, whatever the answer will be, it should contain minimum powers of 2's and 5's. Therefore, what I did was, for each entry in the input matrix, I calculated the powers of 2's and 5's and stored them in separate matrices.

    for (i = 0; i < n; i++)
    {
        for ( j = 0; j < n; j++)
        {
            cin>>foo;
            matrix[i][j] = foo;
            int n1 = calctwo(foo);   // calculates the number of 2's in factorisation of that number
            int n2 = calcfive(foo); // calculates number of 5's
            two[i][j] = n1;
            five[i][j] = n2;
        }
    }

After that, I did this:

for (i = 0; i < n; i++)
{
    for ( j = 0; j < n; j++ )
    {
        dp[i][j] = min(two[i][j],five[i][j]);  // Here, dp[i][j] will store minimum number of 2's and 5's. 
    }
}

But the above doesn't really a valid answer, I don't know why? Have I implemented the correct approach? Or, is this the correct way of solving this question?

Edit: Here are my functions of calculating the number of two's and number of five's in a number.

int calctwo (int foo)
{
    int counter = 0;
    while (foo%2 == 0)
    {
        if (foo%2 == 0)
        {
            counter++;
            foo = foo/2;
        }
        else
            break;
    }
    return counter;
}

int calcfive (int foo)
{
    int counter = 0;
    while (foo%5 == 0)
    {
        if (foo%5 == 0)
        {
            counter++;
            foo = foo/5;
        }
        else
            break;
    }
    return counter;
}

Edit2: I/O Example as given in the link:

Input:

3
1 2 3
4 5 6
7 8 9

Output:

0
DDRR

Answer 1

Since you are interested only in the number of trailing zeroes you need only to consider the powers of 2, 5 which you could keep in two separate nxn arrays. So for the array

1 2 3
4 5 6
7 8 9

you just keep the arrays

the powers of 2    the powers of 5       
0 1 0              0 0 0
2 0 1              0 1 0
0 3 0              0 0 0

The insight for the problem is the following. Notice that if you find a path which minimizes the sum of the powers of 2 and a path which minimizes the number sum of the powers of 5 then the answer is the one with lower value of those two paths. So you reduce your problem to the two times application of the following classical dp problem: find a path, starting from the top-left corner and ending at the bottom-right, such that the sum of its elements is minimum. Again, following the example, we have:

 minimal path for the 
 powers of 2                 value
 * * -                         2
 - * * 
 - - *

 minimal path for the 
 powers of 5                 value
 * - -                         0
 * - - 
 * * *

so your answer is

 * - -                      
 * - - 
 * * *

with value 0

Note 1

It might seem that taking the minimum of the both optimal paths gives only an upper bound so a question that may rise is: is this bound actually achieved? The answer is yes. For convenience, let the number of 2's along the 2's optimal path is a and the number of 5's along the 5's optimal path is b. Without loss of generality assume that the minimum of the both optimal paths is the one for the power of 2's (that is a < b). Let the number of 5's along the minimal path is c. Now the question is: are there as much as 5's as there are 2's along this path (i.e. is c >= a?). Assume that the answer is no. That means that there are less 5's than 2's along the minimal path (that is c < a). Since the optimal value of 5's paths is b we have that every 5's path has at least b 5's in it. This should also be true for the minimal path. That means that c > b. We have that c < a so a > b but the initial assumption was that a < b. Contradiction.

Note 2

You might also want consider the case in which there is an element 0 in your matrix. I'd assume that number of trailing zeroes when the product is 1. In this case, if the algorithm has produced a result with a value more than 1 you should output 1 and print a path that goes through the element 0.