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一个节点的 \(G\) 值为子树内子集数减去子树内 lca 不是它的子集数
有点好奇原题是啥？搬题人忘记删的数据范围貌似暗示了是个有多次询问的题

点击查看代码

#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;
int head[N], ecnt;
const ll mod=998244353;
struct edge{int to, next;}e[N<<1];
inline void add(int s, int t) {e[++ecnt]={t, head[s]}; head[s]=ecnt;}

namespace task1{
	int siz[N];
	ll pw[N], g[N], sum[N];
	void dfs(int u, int fa) {
		siz[u]=1;
		for (int i=head[u],v; ~i; i=e[i].next) {
			v = e[i].to;
			if (v==fa) continue;
			dfs(v, u);
			siz[u]+=siz[v];
			sum[u]=(sum[u]+sum[v])%mod;
		}
		g[u]=(pw[siz[u]]-1-sum[u])%mod;
		sum[u]=(sum[u]+g[u])%mod;
	}
	void solve() {
		pw[0]=1;
		for (int i=1; i<=n; ++i) pw[i]=(pw[i-1]<<1)%mod;
		dfs(1, 0);
		for (int i=1; i<=n; ++i) printf("%lld\n", (g[i]%mod+mod)%mod);
	}
}

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

	n=read();
	memset(head, -1, sizeof(head));
	for (int i=1,u,v; i<n; ++i) {
		u=read(); v=read();
		add(u, v); add(v, u);
	}
	task1::solve();

	return 0;
}