[CF632E]Thief in a Shop

题目大意：有一个小偷，拿$k$个东西，有$n$种产品，每种产品都有无限多个。对于每个第$i$ 种产品，它的价值是$A_i$。可能偷走的物品价值之和。

题解：对于所有的物品构造生成函数$F(x)=\sum\limits_{i\in A}x^i$，取$k$个物品相当于取其中的$k$项相乘，输出$F^k(x)$中不为零的项就行了。（这道题模数$998244353$和$1004535809$都被$hack$了，看$Weng\_weijie\;dalao$的题解得双模数没被卡，于是就$A$了）（这道题似乎可以用$DP$，但我不怎么会）

卡点：无

C++ Code：

#include <cstdio>
#include <algorithm>
#define maxn 1 << 20 | 3
const int G = 3;
int mod, ans;
int lim, ilim, s, rev[maxn], Wn[maxn];
inline int pw(int base, long long p) {
	base %= mod, p %= mod - 1;
	int ans = 1;
	for (; p; p >>= 1, base = 1ll * base * base % mod) if (p & 1) ans = 1ll * ans * base % mod;
	return ans;
}
inline int Inv(int x) {
	return pw(x, mod - 2);
}
inline void up(int &a, int b) {if ((a += b) >= mod) a -= mod;}
inline void NTT(int *A, int op) {
	for (int i = 0; i < lim; i++) if (i < rev[i]) std::swap(A[i], A[rev[i]]);
	for (int mid = 1; mid < lim; mid <<= 1) {
		int t = lim / mid >> 1;
		for (int i = 0; i < lim; i += mid << 1) {
			for (int j = 0; j < mid; j++) {
				int W = op ? Wn[t * j] : Wn[lim - t * j];
				int X = A[i + j], Y = 1ll * A[i + j + mid] * W % mod;
				up(A[i + j], Y), up(A[i + j + mid] = X, mod - Y);
			}
		}
	}
	if (!op) for (int i = 0; i < lim; i++) A[i] = 1ll * A[i] * ilim % mod;
}
inline void init(int n, int mod) {
	::mod = mod;
	lim = 1, s = -1; while (lim < n) lim <<= 1, s++; ilim = Inv(lim);
	for (int i = 0; i < lim; i++) rev[i] = rev[i >> 1] >> 1 | (i & 1) << s;
	int W = pw(G, (mod - 1) / lim);
	Wn[0] = 1; for (int i = 1; i <= lim; i++) Wn[i] = 1ll * Wn[i - 1] * W % mod;
}
int n, k;
int a[maxn], b[maxn];
int main() {
	scanf("%d%d", &n, &k);
	for (int i = 0, tmp; i < n; i++) scanf("%d", &tmp), a[tmp] = b[tmp] = 1;
	init(1 << 20, 998244353);
	NTT(a, 1);
	for (int i = 0; i < lim; i++) a[i] = pw(a[i], k);
	NTT(a, 0);
	init(1 << 20, 1004535809);
	NTT(b, 1);
	for (int i = 0; i < lim; i++) b[i] = pw(b[i], k);
	NTT(b, 0);
	for (int i = 0; i < lim; i++) if (a[i] | b[i]) printf("%d ", i);
	return 0;
}