百练 4120: 硬币

01背包+容斥，f[j]：构成j的方案数，g[j]：除去a[i]后构成j的方案数

#define _CRT_SECURE_NO_WARNINGS
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;

const int N = 100000 + 5;

int a[N], f[N], g[N];

int main() {
	int n, x;
	scanf("%d%d", &n, &x);
	for (int i = 0; i < n; i++)
		scanf("%d", a + i);
	f[0] = 1;
	for (int i = 0; i < n; i++)
		for (int j = x; j >= a[i]; j--)
			f[j] += f[j - a[i]];
	vector<int> v;
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j <= x; j++)
		{
			if (j < a[i]) g[j] = f[j];
			else g[j] = f[j] - g[j - a[i]];
		}
		if (g[x] == 0) {
			v.push_back(a[i]);
		}
	}
	sort(v.begin(), v.end());
	printf("%d\n", v.size());
	for (int i = 0; i < v.size(); i++) {
		if (i != 0) printf(" ");
		printf("%d", v[i]);
	}
	printf("\n");
	return 0;
}