【BZOJ2614】梦的困境【质因数分解】【背包】【GCD】

【题目链接】

%段少，【lowsfish的题解】

学习一发神奇的背包。

不知道为什么要开LL。

/* Pigonometry */
#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;

typedef long long LL;

const int maxn = 205, maxm = 200005;

int cnt;
LL fact[maxn], hash[maxm], n, m, k, p, val[maxn][maxn], f[maxn][maxn], dp[maxn][maxn];

inline LL gcd(LL a, LL b) {
	for(; b; b ^= a ^= b ^= a %= b);
	return a;
}

inline void findfact(LL x) {
	LL _sqrtx = sqrt(x + 0.5);
	for(int i = 1; i <= _sqrtx; i++) if(x % i == 0) {
		fact[++cnt] = i; hash[i] = cnt;
		if(i * i != x) fact[++cnt] = x / i, hash[x / i] = cnt;
	}
}

int main() {
	scanf("%lld%lld%lld%lld", &n, &m, &k, &p);

	cnt = 0;
	findfact(k);

	for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) {
		LL x; scanf("%lld", &x);
		x = hash[gcd(x, k)];
		val[i][x]++; f[i][x]++;
	}

	for(int i = 2; i < n; i++) for(int j = 1; j <= cnt; j++) for(int t = 1; t <= cnt; t++) {
		int x = hash[gcd(fact[j] * fact[t], k)];
		if(j == t) (f[i][x] += val[i][j] * (val[i][j] - 1) % p) %= p;
		else (f[i][x] += val[i][j] * val[i][t] % p) %= p;
	}

	for(int j = 1; j <= cnt; j++) dp[1][j] = val[1][j];
	for(int i = 2; i <= n; i++) for(int j = 1; j <= cnt; j++) for(int t = 1; t <= cnt; t++) {
		int x = hash[gcd(fact[j] * fact[t], k)];
		(dp[i][x] += dp[i - 1][j] * f[i][t] % p) %= p;
	}

	printf("%lld\n", dp[n][hash[k]]);
	return 0;
}