题解 树上的数

传送门

两种做法
第一种可以按题意模拟，dfs染色
如果一个点已经被染过了就return
这样每个点只会被染色一次，均摊 \(O(n)\)

第二种考虑每个点被染色的时间
在时间轴上差分计算贡献，也是 \(O(n)\)

Code:

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 5000010
#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, q;
ll a, b, x, y;
int head[N], size, val[N];
struct edge{int to, next;}e[N];
inline void add(int s, int t) {e[++size].to=t; e[size].next=head[s]; head[s]=size;}

namespace force{
	ll ans, sum;
	void dfs(int u) {
		if (!val[u]) return ;
		val[u]=0; --sum;
		for (int i=head[u],v; ~i; i=e[i].next) {
			v = e[i].to;
			dfs(v);
		}
	}
	void solve() {
		sum=n;
		dfs(q);
		ans=sum;
		for (int i=2; i<=m; ++i) {
			q=(((1ll*q*x+y)^19760817)^(i<<1))%(n-1)+2;
			dfs(q);
			ans^=sum;
		}
		printf("%lld\n", ans);
		exit(0);
	}
}

namespace task1{
	ll ans, sum;
	int sta[N], cur[N], top;
	void dfs(int u) {
		if (!val[u]) return ;
		top=0;
		sta[++top]=u;
		while (top) {
			u=sta[top];
			if (val[u]) val[u]=0, --sum;
			for (int i=cur[u],v; ~i; i=e[i].next) {
				cur[u]=e[i].next;
				v = e[i].to;
				if (val[v]) {sta[++top]=v; goto jump;}
			}
			--top;
			jump: ;
		}
	}
	void solve() {
		sum=n;
		for (int i=1; i<=n; ++i) cur[i]=head[i];
		dfs(q);
		ans=sum;
		for (int i=2; i<=m; ++i) {
			q=(((1ll*q*x+y)^19760817)^(i<<1))%(n-1)+2;
			dfs(q);
			ans^=sum;
		}
		printf("%lld\n", ans);
		exit(0);
	}
}

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

	// cout<<double(sizeof(val)*6)/1024/1024<<endl;
	n=read(); m=read();
	memset(head, -1, sizeof(head));
	for (int i=1; i<=n; ++i) val[i]=1;
	a=read(); b=read();
	add(1, 2);
	int lst=1;
	for (int i=3; i<=n; ++i) {
		lst=((lst*a+b)^19760817)%(i-1)+1;
		add(lst, i);
	}
	q=read(); x=read(); y=read();
	// force::solve();
	task1::solve();
	
	return 0;
}