C++ 解法, 执行用时: 1267ms, 内存消耗: 3256K, 提交时间: 2021-10-15 21:34:37

#pragma GCC optimize("3", "inline", "Ofast")
#include<bits/stdc++.h>
using namespace std;
   
#define broken fprintf(stderr, "running on %s %d\n", __FUNCTION__, __LINE__)
#define sz(a) int((a).size())
#define x first
#define y second
#define mp make_pair
typedef long long i64;
typedef unsigned long long u64;
const int N = 1e2 + 10, M = 1e3 + 10, T = 1e5 + 10, V = T - 10;
vector<int> a[N];

int inv[T], fac[T], ifac[T], n, m, p;
map<vector<int>, vector<int> > g;

void Mod(int &x) { x += x >> 31 & p; }

int power(int a, int b, int c = 1) {
	for(; b; b >>= 1, a = 1ll * a * a % p)
		if(b & 1)
			c = 1ll * c * a % p;
	return c;
}

int C(int n, int m) {return n < m ? 0 : 1ll * fac[n] * ifac[m] % p * ifac[n - m] % p; }

int binom(int n, int m) {
	if(n < m) return 0;
	if(m == 0) return 1;
	int res = binom(n / p, m / p);
	n %= p, m %= p;
	if(n < m) return 0;
	if(n <= V) return 1ll * C(n, m) * res % p;
	for(int i = n; i >= n - m + 1; i--) res = 1ll * res * i % p;
	return 1ll * res * ifac[m] % p;
}

void initmath(int n) {
	fac[0] = ifac[0] = inv[1] = 1;
	for(int i = 2; i <= n; i++) inv[i] = 1ll * inv[p % i] * (p - p / i) % p;
	for(int i = 1; i <= n; i++) {
		fac[i] = 1ll * fac[i - 1] * i % p;
		ifac[i] = 1ll * ifac[i - 1] * inv[i] % p;
	}
	return;
}

int main() {
	scanf("%d%d%d", &n, &m, &p);
	initmath(V);
	for(int i = 0; i < n; i++) {
		a[i].resize(m);
		for(int j = 0; j < m; j++) {
			scanf("%d", &a[i][j]);
		}
		vector<int> b(m);
		int pos = 0;
		while(pos < m && !a[i][pos]) pos++;
		if(pos < m - 1) {
			int x = 0, base = 1;
			b = a[i];
			for(int j = 1; pos + j < m; j++) {
				int t = j;
				for(; t % p == 0; t /= p);
				if(t == 1) {
					x += 1ll * b[pos + j] * power(a[i][pos], p - 2) % p * base;
					base *= p;
				}
				int coef = binom(x, j);
				if(j & 1) coef = (p - coef) % p;
				for(int k = pos; k + j < m; k++) {
					Mod(b[k + j] += 1ll * coef * a[i][k] % p - p);
				}
			}
		} else {
			b = a[i];
		}
		g[b].push_back(i);
	}
	printf("%d\n", sz(g));
	for(auto v : g) {
		printf("%d\n", sz(v.y));
		for(int a : v.y) printf("%d ", a);
		printf("\n");
	}
	return 0;
}