NOIP 模拟 $83; m 树上的数$

题解 (by;zjvarphi)

直接暴力 dfs，如果一个点被扫过了，那么它的子树也一定被扫过了。

因为每个点最多只会被扫一次，复杂度 (mathcal{O m(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
",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
",ans),0;
                ans^=lft;
            } else ans^=lft;
        }
        printf("%d
",ans);
        return 0;
    }
}
int main() {return nanfeng::main();}