首先打了个暴力递归代码，获得了30pts

#include <bits/stdc++.h>
using namespace std;
const int N = 805;
int n, m, k, a[N][N], dp[N][N][2], ans;
int dfs(int x, int y, int w, int v1, int v2) {
	if (x > n || y > m) return 0;
	if (!w) v1 += a[x][y], v1 %= (k + 1);
	else v2 += a[x][y], v2 %= (k + 1);
	if (!w || (w && v1 != v2)) return dfs(x + 1, y, !w, v1, v2) + dfs(x, y + 1, !w, v1, v2);
	return 1 + dfs(x + 1, y, !w, v1, v2) + dfs(x, y + 1, !w, v1, v2);
}
int main() {
	scanf("%d%d%d", &n, &m, &k);
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j)
			scanf("%d", &a[i][j]);
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j)
			ans += dfs(i, j, 0, 0, 0);
	printf("%d", ans);
	return 0;
}

然后直接套上个记忆化，结果变成了20pts

#include <bits/stdc++.h>
using namespace std;
const int N = 805, MOD = 1e9 + 7;
int n, m, k, a[N][N], dp[N][N][2], ans;
int dfs(int x, int y, int w, int v1, int v2) {
	if (x > n || y > m) return 0;
	if (dp[x][y][w] != -1) return dp[x][y][w];
	if (!w) v1 += a[x][y], v1 %= (k + 1);
	else v2 += a[x][y], v2 %= (k + 1);
	if (!w || (w && v1 != v2)) return dp[x][y][w] = (dfs(x + 1, y, !w, v1, v2) + dfs(x, y + 1, !w, v1, v2)) % MOD;
	return dp[x][y][w] = (1 + dfs(x + 1, y, !w, v1, v2) + dfs(x, y + 1, !w, v1, v2)) % MOD;
}
int main() {
	scanf("%d%d%d", &n, &m, &k);
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j)
			scanf("%d", &a[i][j]);
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j) {
			memset(dp, -1, sizeof(dp));
			ans += dfs(i, j, 0, 0, 0) % MOD;
		}
	printf("%d", ans);
	return 0;
}

暴力递归评测记录

记忆化搜索评测记录