CF2B The least round way

(large{题目链接})
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(Large extbf{Solution: } large{首先容易想到考虑因子。如果末尾有0，那么一定有一个2对应着一个5。所以只要维护2和5的个数就好。\然后需要特判矩阵中是否有0，如果有，需要和1特判一下。\这道题的输出还真是恶心，不过递归大法好，复杂问题用递归往往能轻易解决。})
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(Large extbf{Code: })

#include <bits/stdc++.h>
using namespace std;

const int N = 1e3 + 5;

int n, a[N][N], f[N][N], t[N][N], f_x, f_y;

void read(int &x) {
	x = 0;
	char c = getchar();
	while (!isdigit(c)) c = getchar();
	while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
}

int calc(int x, int y, int p) {
	int ret = 0, cur = a[x][y];
	for (; cur && !(cur % p); cur /= p) ++ret;
	return ret;
}

void print_f(int x, int y) {
	if (x == 1) { for (int i = 1; i < y; ++i) putchar('R'); return; }
	if (y == 1) { for (int i = 1; i < x; ++i) putchar('D'); return; }
	if (f[x - 1][y] + calc(x, y, 5) == f[x][y]) print_f(x - 1, y), putchar('D');
	else print_f(x, y - 1), putchar('R');
}

void print_s(int x, int y) {
	if (x == 1) { for (int i = 1; i < y; ++i) putchar('R'); return; }
	if (y == 1) { for (int i = 1; i < x; ++i) putchar('D'); return; }
	if (t[x - 1][y] + calc(x, y, 2) == t[x][y]) print_s(x - 1, y), putchar('D');
	else print_s(x, y - 1), putchar('R');
}

void print(int x, int y, int opt) {
	if (opt == 1) print_f(n, n);
	else print_s(n, n);
}

int main() {
	read(n);
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= n; ++j) {
			read(a[i][j]);
			if (!a[i][j]) f_x = i, f_y = j;
		}
	f[1][1] = calc(1, 1, 5), t[1][1] = calc(1, 1, 2);
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= n; ++j) {
			if (i == 1 && j == 1) continue;
			if (i == 1) { f[i][j] = f[i][j - 1] + calc(i, j, 5), t[i][j] = t[i][j - 1] + calc(i, j, 2); continue; }
			if (j == 1) { f[i][j] = f[i - 1][j] + calc(i, j, 5), t[i][j] = t[i - 1][j] + calc(i, j, 2); continue; }
			f[i][j] = min(f[i - 1][j], f[i][j - 1]) + calc(i, j, 5);
			t[i][j] = min(t[i][j - 1], t[i - 1][j]) + calc(i, j, 2);
		}
	}
	int ans = min(f[n][n], t[n][n]);
	if (!f_x) printf("%d
", ans), print(n, n, (f[n][n] <= t[n][n]));
	else {
		if (ans < 1) printf("%d
", ans), print(n, n, (f[n][n] <= t[n][n]));
		else {
			printf("1
");
			for (int i = 1; i < f_x; ++i) putchar('D');
			for (int i = 1; i < f_y; ++i) putchar('R');
			for (int i = f_x; i < n; ++i) putchar('D');
			for (int i = f_y; i < n; ++i) putchar('R');
		}
	}
	return 0;
}