题解 子集和

传送门

发现我们可以很容易地确定集合中最小的元素
然后从大到小用类似背包的逆过程处理掉这个元素的贡献
于是转化为子问题，递归做即可

Code:

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100010
#define ll long long
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n, m;
int a[N], b[N], tem[N];

namespace task1{
	void solve(int k) {
		if (k>n) return ;
		int now;
		for (int i=1; i<=m; ++i) if (b[i]) {
			a[k]=i; --b[i]; now=i;
			break;
		}
		for (int i=1; i<=m; ++i) tem[i]=0;
		for (int i=m; i; --i) if (b[i]) {
			// assert(b[i-now]);
			b[i-now]-=b[i]; tem[i-now]+=b[i];
			b[i]=0;
		}
		m-=now;
		for (int i=1; i<=m; ++i) b[i]=tem[i];
		solve(k+1);
	}
}

signed main()
{
	freopen("subset.in", "r", stdin);
	freopen("subset.out", "w", stdout);

	n=read(); m=read();
	for (int i=0; i<=m; ++i) b[i]=read();
	task1::solve(1);
	sort(a+1, a+n+1);
	for (int i=1; i<=n; ++i) printf("%d%c", a[i], " \n"[i==n]);

	return 0;
}