NOIP 模擬 $83\; \rm 樹上的數$
阿新 • • 發佈:2021-10-26
題解
題解 \(by\;zj\varphi\)
直接暴力 dfs,如果一個點被掃過了,那麼它的子樹也一定被掃過了。
因為每個點最多隻會被掃一次,複雜度 \(\mathcal{O\rm(n+m)}\),需要卡常。
Code
#include<bits/stdc++.h> #define ri signed #define pd(i) ++i #define bq(i) --i #define func(x) std::function<x> char buf[1<<21],*p1=buf,*p2=buf; #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++ #define debug1(x) std::cerr << #x"=" << x << ' ' #define debug2(x) std::cerr << #x"=" << x << std::endl #define Debug(x) assert(x) struct nanfeng_stream{ template<typename T>inline nanfeng_stream &operator>>(T &x) { bool f=false;x=0;char ch=gc(); while(!isdigit(ch)) f|=ch=='-',ch=gc(); while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=gc(); return x=f?-x:x,*this; } }cin; } using IO::cin; namespace nanfeng{ #define FI FILE *IN #define FO FILE *OUT template<typename T>inline T cmax(T x,T y) {return x>y?x:y;} template<typename T>inline T cmin(T x,T y) {return x>y?y:x;} static const int N=5e6+7; struct edge{int v,nxt;}e[N]; int first[N],t=1,fa=1,n,m,a,b,q,x,y,ans,lft; bool vis[N]; auto add=[](int u,int v) {e[t]={v,first[u]},first[u]=t++;}; func(void(int)) dfs=[](int x) { --lft; vis[x]=true; for (ri i(first[x]),v;i;i=e[i].nxt) { if (vis[v=e[i].v]) continue; dfs(v); } }; inline int main() { FI=freopen("tree.in","r",stdin); FO=freopen("tree.out","w",stdout); cin >> n >> m >> a >> b >> q >> x >> y; add(fa,2); for (ri i(3);i<=n;pd(i)) fa=((1ll*fa*a+b)^19760817ll)%(i-1)+1,add(fa,i); lft=n; dfs(q); if (!lft) return printf("%d\n",ans),0; ans^=lft; for (ri i(2);i<=m;pd(i)) { q=(((1ll*q*x+y)^19760817ll)^(i<<1))%(n-1)+2; if (!vis[q]) { dfs(q); if (!lft) return printf("%d\n",ans),0; ans^=lft; } else ans^=lft; } printf("%d\n",ans); return 0; } } int main() {return nanfeng::main();}