CF2B The least round way

设 \(f_{i, j}\) 表示走到点 \((i, j)\) 获得因数 2 的最小数量。

设 \(g_{i, j}\) 表示走到点 \((i, j)\) 获得因数 5 的最小数量。

那么到点 \((i, j)\) 胃部获得的最小 0 的个数为 \(\min(f_{i,j}, g_{i, j})\)，因为如果选择数量小的那个因数，另外一个因数的个数一定多于它，就会往后加一个 0，这样就使得 0 的个数最小。

#include <bits/stdc++.h>

using namespace std;

const int N = 1010;

int n, a[N][N], y2[N][N], y5[N][N], f2[N][N], f5[N][N];
pair<int, int> lz2[N][N], lz5[N][N];
int num_0_x;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    cin >> n;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            cin >> a[i][j];
            if (!a[i][j]) {
                num_0_x = i;
            }
            else {
                while (a[i][j] % 2 == 0) {
                    y2[i][j]++;
                    a[i][j] /= 2;
                }
                while (a[i][j] % 5 == 0) {
                    y5[i][j]++;
                    a[i][j] /= 5;
                }
            }
        }
    }
    
    memset(f2, 0x3f, sizeof(f2));
    memset(f5, 0x3f, sizeof(f5));
    f2[0][1] = f5[0][1] = 0;

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if (f2[i - 1][j] < f2[i][j - 1]) {
                f2[i][j] = f2[i - 1][j] + y2[i][j];
                lz2[i][j] = {i - 1, j};
            }
            else {
                f2[i][j] = f2[i][j - 1] + y2[i][j];
                lz2[i][j] = {i, j - 1};
            }
            if (f5[i - 1][j] < f5[i][j - 1]) {
                f5[i][j] = f5[i - 1][j] + y5[i][j];
                lz5[i][j] = {i - 1, j};
            }
            else {
                f5[i][j] = f5[i][j - 1] + y5[i][j];
                lz5[i][j] = {i, j - 1};
            }
        }
    }
    int ans = min(f2[n][n], f5[n][n]);
    string s;
    if (ans == f2[n][n]) {
        int x = n, y = n;
        while (x >= 1 && y >= 1) {
            if (lz2[x][y].first == x - 1) {
                x--;
                s.push_back('D');
            }
            else {
                y--;
                s.push_back('R');
            }
        }
    }
    else {
        int x = n, y = n;
        while (x >= 1 && y >= 1) {
            if (lz5[x][y].first == x - 1) {
                x--;
                s.push_back('D');
            }
            else {
                y--;
                s.push_back('R');
            }
        }
    }
    s.pop_back();
    reverse(s.begin(), s.end());
    if (num_0_x && ans > 1) {
        cout << "1\n";
        for (int i = 2; i <= num_0_x; i++) cout << "D";
        for (int i = 2; i <= n; i++) cout << "R";
        for (int i = num_0_x + 1; i <= n; i++) cout << "D";
    }
    else {
        cout << ans << '\n';
        cout << s << '\n';
    }
    return 0;
}