HDU 6121 Build a tree【】

Build a tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1036    Accepted Submission(s): 416

Problem Description

HazelFan wants to build a rooted tree. The tree has n nodes labeled 0 to n−1, and the father of the node labeled i is the node labeled ⌊i−1k⌋. HazelFan wonders the size of every subtree, and you just need to tell him the XOR value of these answers.

 

Input

The first line contains a positive integer T(1≤T≤5), denoting the number of test cases.
For each test case:
A single line contains two positive integers n,k(1≤n,k≤1018).

 

Output

For each test case:
A single line contains a nonnegative integer, denoting the answer.

 

Sample Input

25 25 3

 

Sample Output

76

 

Source

2017 Multi-University Training Contest - Team 7

 

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<queue>#include<stack>#include<vector>#include<map>#include<set>#include<algorithm>using namespace std;#define ll long long#define ms(a,b)  memset(a,b,sizeof(a))const int M=1e5+10;const int MM=2e3+10;const int inf=0x3f3f3f3f;const int mod=1e9+7;;const double eps=1e-8;const double pi=acos(-1.0);ll n,m,k;ll tre[240]={1};ll ans,cur,last,x,l,r,full;void dfs(ll x){    ans^=x;    x--;    ll p=1,xx=x,ans1=0,ans2=0,pp,sum=0,g=1;    while(xx>=p*k) {xx-=p*k,p*=k;}    pp=p/k;    while(pp)    {        sum+=g;        if(pp&1) ans1^=sum;        pp/=k;        g*=k;    }    if(xx%p) dfs(sum+xx%p);    sum=0,g=1,pp=p;    while(p)    {        sum+=g;        if(p&1) ans2^=sum;        p/=k;        g*=k;    }    ll cnt2=xx/pp;    ll cnt1=k-cnt2-(xx%pp?1:0);    if(cnt2&1) ans^=ans2;    if(cnt1&1) ans^=ans1;}int main(){    int t;    scanf("%d",&t);    while(t--){        scanf("%lld%lld",&n,&k);        if(k==1){            switch(n%4){                case 1:printf("1\n");break;                case 2:printf("%lld\n",n+1);break;                case 3:printf("0\n");break;                case 4:printf("%lld\n",n);break;            }            continue;        }        ans=0;        dfs(n);        printf("%lld\n",ans);    }    return 0;}