掃地機器人

題目背景：

5.19 模擬 JSOI2018D2T2

分析：結論分析 + DP

首先我們來明確兩個結論，對於長寬n, m，令d = gcd(n, m)那麼，

1、我們只需要固定前d步，那麼後面的每d步都應該和前d步是相同的;

2、這d步中有i步向下，d - i步向右，那麼gcd(i, n) == 1,gcd(d - i, m) == 1;

滿足以上兩個條件，那麼一定是一個合法方案，至於證明什麼的很顯然啦啦啦，本寶寶不會啊，神特喵的顯然。我們姑且當做是對的吧······

那麼對於n * m個格子，一定是會走n * m / d輪，每一輪走到的格子是一個(i + 1) * (d - i +1)的矩形，i為向下走的步數，每一輪的矩形的左上角是上一輪的右下角，如果超過右邊界就回到左邊，超過下邊界就回到上邊，那麼考慮如何計算貢獻，對於一個格子，顯然如果要有貢獻，那麼在經過他之前不能經過任何一個有障礙的格子，那麼對於第

Source:

/*
	created by scarlyw
*/
#include <cstdio>
#include <string>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <cctype>
#include <vector>
#include <set>
#include <queue>
#include <ctime>
#include <deque>
#include <iterator>
#include <map>

inline char read() {
	static const int IN_LEN = 1024 * 1024;
	static char buf[IN_LEN], *s, *t;
	if (s == t) {
		t = (s = buf) + fread(buf, 1, IN_LEN, stdin);
		if (s == t) return -1;
	}
	return *s++;
}

// /*
template<class T>
inline void R(T &x) {
	static char c;
	static bool iosig;
	for (c = read(), iosig = false; !isdigit(c); c = read()) {
		if (c == -1) return ;
		if (c == '-') iosig = true;	
	}
	for (x = 0; isdigit(c); c = read()) 
		x = ((x << 2) + x << 1) + (c ^ '0');
	if (iosig) x = -x;
}
//*/

const int OUT_LEN = 1024 * 1024;
char obuf[OUT_LEN], *oh = obuf;
inline void write_char(char c) {
	if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf;
	*oh++ = c;
}

template<class T>
inline void W(T x) {
	static int buf[30], cnt;
	if (x == 0) write_char('0');
	else {
		if (x < 0) write_char('-'), x = -x;
		for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;
		while (cnt) write_char(buf[cnt--]);
	}
}

inline void flush() {
	fwrite(obuf, 1, oh - obuf, stdout);
}

/*
template<class T>
inline void R(T &x) {
	static char c;
	static bool iosig;
	for (c = getchar(), iosig = false; !isdigit(c); c = getchar()) {
		if (c == -1) return ;
		if (c == '-') iosig = true;	
	}
	for (x = 0; isdigit(c); c = getchar()) 
		x = ((x << 2) + x << 1) + (c ^ '0');
	if (iosig) x = -x;
}
// */

const int MAXN = 50 + 3;
const int mod = 998244353;

int t, n, m, d;
long long ans;
char s[MAXN];
int f[MAXN][MAXN], g[MAXN][MAXN];
bool a[MAXN << 1 | 1][MAXN << 1 | 1], ban[MAXN][MAXN];

inline void add(int &x, int t) {
    ((x += t) >= mod) && (x -= mod);
}

inline int gcd(int a, int b) {
    return b ? gcd(b, a % b) : a;
}

inline void solve(int p, int q) {
    int sx = 1, sy = 1;
    for (int i = 1; i <= p; ++i)
        for (int j = 1; j <= q; ++j)
            ban[i][j] = false;
    // memset(ban, 0, sizeof(ban));
    for (int c = 1; c <= n * m / d; ++c) {
        for (int i = 1; i <= p; ++i)
            for (int j = 1; j <= q; ++j)
                f[i][j] = g[i][j] = 0;
        f[p][q] = 1 - ban[p][q];
        for (int i = p; i >= 1; --i)
            for (int j = q; j >= 1; --j)
                if (ban[i][j] == false) 
                    add(f[i - 1][j], f[i][j]), add(f[i][j - 1], f[i][j]);
        for (int i = 1; i <= p; ++i)
            for (int j = 1; j <= q; ++j)
                ban[i][j] |= a[sx + i - 1][sy + j - 1];
        sx += p - 1, ((sx > n) ? (sx -= n) : sx);
        sy += q - 1, ((sy > m) ? (sy -= m) : sy);
        g[1][1] = 1 - ban[1][1]; 
        for (int i = 1; i <= p; ++i)
            for (int j = 1; j <= q; ++j)
                if (ban[i][j] == false)
                    add(g[i + 1][j], g[i][j]), add(g[i][j + 1], g[i][j]);
        for (int i = 1; i <= p; ++i)
            for (int j = 1; j <= q; ++j)
                if (ban[i][j] == false && (i != p || j != q))
                    ans = (ans + (long long)f[i][j] * g[i][j]) % mod;
    }
}

inline void solve() {
    R(n), R(m), d = gcd(n, m), ans = 0;
    for (int i = 1; i <= n; ++i, read())
        for (int j = 1; j <= m; ++j)
            a[i][j] = a[i + n][j] = a[i][j + m]
                 = a[i + n][j + m] = read() - '0';
    for (int i = 1; i <= d; ++i)
        if (gcd(i, n) == 1 && gcd(d - i, m) == 1)
            solve(i + 1, d - i + 1);
    std::cout << ans << '\n';
}

int main() {
    // freopen("in.in", "r", stdin);
    R(t);
    while (t--) solve();
    return 0;
}