2017 Multi-University Training Contest 7 hdu 6121

HDU - 6121

题意：有一棵n个点的有根树，标号为0到n−1，i号点的父亲是(i-1)/k号点，求所有子树大小的异或和 1<=n,k<=1e18

思路：这是一棵完全k叉树，考虑根的所有孩子，最多只有一个不是满k叉树，对这个孩子进行递归处理即可。k>1时只有log层，直接做就到底就好了，k=1时要特判。

AC代码：

#include "iostream"
#include "iomanip"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#pragma comment(linker, "/STACK:102400000,102400000")
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a,x) memset(a,x,sizeof(a))
#define step(x) fixed<< setprecision(x)<<
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define ll long long
#define endl ("
")
#define ft first
#define sd second
#define lrt (rt<<1)
#define rrt (rt<<1|1)
using namespace std;
const ll mod=1e9+7;
const ll INF = 1e18+1LL;
const int inf = 1e9+1e8;
const double PI=acos(-1.0);
const int N=1e5+100;

ll c[105],sum[105],n,k,ans;
ll dep(ll x, ll k, ll d){  //求深度
    if(x==0) return d;
    else return dep((x-1)/k, k, d+1);
}

ll fun(ll x, ll k){   //节点数为x的满k叉树的答案
    if(x==0) return 0;
    ll d=dep(x-1, k,1);
    ll ret=0, t=1;
    for(ll i=d; i>=1; --i){
        if(c[i]&1) ret^=t;
        t=t*k+1;
    }
    return ret;
}
ll fat(ll m, ll k){  //找祖先
    if(m<=k){
        return m;
    }
    return fat((m-1)/k, k);
}
void dfs(ll x, ll k){ //bug(x) char c=getchar();
    if(x==1){
        ans^=1;
        return;
    }
    ll d=dep(x-1,k,1);
    if(x==sum[d]){
        ans^=fun(sum[d], k);
        return;
    }
    ans^=x;
    ll u=fat(x-1,k);
    if((u-1)&1) ans^=fun(sum[d-1], k);
    if((k-u)&1) ans^=fun(sum[d-2], k);
    x-=(u-1)*sum[d-1]+(k-u)*sum[d-2]+1; //cout<< "x=" <<x<<endl;
    dfs(x,k);
}

int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T; cin>>T;
    while(T--){
        cin>>n>>k; mem(sum,0);
        for(ll i=1,t=1; t<=INF,i<=60; ++i,t*=k){
            c[i]=t,sum[i]+=sum[i-1]+c[i];
        }
        if(k==1){
            ll t=(n+3)/4, x=n%4;
            if(x==1) ans=1;
            if(x==3) ans=0;
            if(x==2) ans=4*t-1;
            if(x==0) ans=4*t;
            cout<<ans<<endl;
            continue;
        }
        ans=0;
        dfs(n,k);
        cout<<ans<<endl;
    }
    return 0;
}