Codeforces 2 B. The least round way

题意：

二维数组中都是非负整数，从[1,1]走到[n,n]每次可以向右或向下走，求路径上数乘积的0最少是多少，并输出路径。

 

提示：

最少0就是最少因子2或者最少因子5，分别dp一下。

注意：

如果路径里有0，那么相乘后0的个数一定是1。

 

C#10 .net6 代码

using System.Text;

int n = int.Parse(Console.ReadLine()!);
int[,,] dp = new int[n, n, 2];
char[,,] vis = new char[n, n, 2];
int zx = -1;
int zy = -1;

int GetCnt(int x, int b)
{
    if (x == 0)
    {
        return 1;
    }
    int cnt = 0;
    while (x != 0 && (x % b == 0))
    {
        cnt++;
        x /= b;
    }
    return cnt;
}

for (int i = 0; i < n; i++)
{
    var line = Console.ReadLine()!.Split(' ');
    for (int j = 0; j < n; j++)
    {
        int x = int.Parse(line[j]);
        if (x == 0)
        {
            zx = i;
            zy = j;
        }
        int cnt2 = GetCnt(x, 2);
        int cnt5 = GetCnt(x, 5);
        if (i == 0 && j == 0)
        {
            dp[i, j, 0] = cnt2;
            dp[i, j, 1] = cnt5;
        }
        else if (i == 0 && j > 0)
        {
            vis[i, j, 0] = 'R';
            vis[i, j, 1] = 'R';
            dp[i, j, 0] = dp[i, j - 1, 0] + cnt2;
            dp[i, j, 1] = dp[i, j - 1, 1] + cnt5;
        }
        else if (i > 0 && j == 0)
        {
            vis[i, j, 0] = 'D';
            vis[i, j, 1] = 'D';
            dp[i, j, 0] = dp[i - 1, j, 0] + cnt2;
            dp[i, j, 1] = dp[i - 1, j, 1] + cnt5;
        }
        else
        {
            if (dp[i, j - 1, 0] < dp[i - 1, j, 0])
            {
                vis[i, j, 0] = 'R';
                dp[i, j, 0] = dp[i, j - 1, 0] + cnt2;
            }
            else
            {
                vis[i, j, 0] = 'D';
                dp[i, j, 0] = dp[i - 1, j, 0] + cnt2;
            }
            if (dp[i, j - 1, 1] < dp[i - 1, j, 1])
            {
                vis[i, j, 1] = 'R';
                dp[i, j, 1] = dp[i, j - 1, 1] + cnt5;
            }
            else
            {
                vis[i, j, 1] = 'D';
                dp[i, j, 1] = dp[i - 1, j, 1] + cnt5;
            }
        }
    }
}

int ans = Math.Min(dp[n - 1, n - 1, 0], dp[n - 1, n - 1, 1]);

string Gao(int x, int y, int type)
{
    if (x == 0 && y == 0)
    {
        return "";
    }
    if (vis[x, y, type] == 'R')
    {
        return Gao(x, y - 1, type) + 'R';
    }
    else
    {
        return Gao(x - 1, y, type) + 'D';
    }
}

if (ans >= 1 && zx != -1)
{
    Console.WriteLine(1);
    StringBuilder sb = new();
    for (int i = 0; i < zx; i++)
    {
        sb.Append('D');
    }
    for (int j = 0; j < zy; j++)
    {
        sb.Append('R');
    }
    for (int i = zx; i < n - 1; i++)
    {
        sb.Append('D');
    }
    for (int j = zy; j < n - 1; j++)
    {
        sb.Append('R');
    }
    Console.WriteLine(sb.ToString());
}
else
{
    Console.WriteLine(ans);
    string str = string.Empty;
    if (dp[n - 1, n - 1, 0] < dp[n - 1, n - 1, 1])
    {
        str = Gao(n - 1, n - 1, 0);
    }
    else
    {
        str = Gao(n - 1, n - 1, 1);
    }
    Console.WriteLine(str);
}