Xor-sequence

                            Solution in C :

In C++ :





#include <cstdio>
#include <cmath>
#include <iostream>
#include <set>
#include <algorithm>
#include <vector>
#include <map>
#include <cassert>
#include <string>
#include <cstring>
#include <queue>

using namespace std;

#define rep(i,a,b) for(int i = a; i < b; i++)
#define S(x) scanf("%d",&x)
#define S2(x,y) scanf("%d%d",&x,&y)
#define P(x) printf("%d\n",x)
#define all(v) v.begin(),v.end()
#define FF first
#define SS second

typedef long long int LL;
typedef pair<int, int > pii;
typedef vector<int > vi;

LL f(LL x) {
  if(x % 4 == 0) {
    return x;
  } else if (x % 4 == 1) {
    return x ^ (x - 1);
  } else if (x % 4 == 2) {
    return x ^ (x - 1) ^ (x - 2);
  } else {
    return 0;
  }
}

LL g(LL x) {
  if(x & 1) {
    return f(x) ^ g(x - 1);
  }
  return 2 * f(x / 2);
}

int main() {
  int Q;
  S(Q);
  while(Q--) {
    LL l,r;
    scanf("%lld%lld",&l,&r);
    printf("%lld\n",g(r) ^ g(l-1));
  }
  return 0;
}








In Java :





import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int Q = in.nextInt();
        
        for(int a0 = 0; a0 < Q; a0++){
            long L = in.nextLong();
            long R = in.nextLong();
            System.out.println(ant(R)^ant(L-1));
        }
        
    }
    public static long ant (long n){
        if(n%2==0){
            return 2*ans(n/2);
        }
        return ans(n)+2*ans(n/2);
    }
    public static long ans (long n){
        if(n==0) return 0;
        if(n==1) return 1;
        if(n%2==1){
            if(n%4==1){
                return 1;
            }
            return 0;
        }
        else{
            return n^(ans(n-1));
        }
    }
}








In C :





#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
long long memoize[4];

unsigned long long arrayfinder(unsigned long long);
int main(){
    int Q;
    scanf("%d",&Q);
    for(int a0 = 0; a0 < Q; a0++){
        unsigned long long L; 
        unsigned long long R;
        unsigned long long total=0;
        scanf("%llu %llu",&L,&R);
        if((R-L)<17){
            for(int i= L;i<=R;i++){
                total^=arrayfinder(i);
            }
        }
        else{
            while(L%8!=7){
                total^=arrayfinder(L);
                L++;
            }
            while(R%8!=7){
                total^=arrayfinder(R);
                R--;
            }
        }
        printf("%llu\r\n",total);
    }
    return 0;
}
unsigned long long arrayfinder(unsigned long long x){
    

       if(x%4==0)return x;
       if(x%4==1)return 1;
       if(x%4==2)return x+1;
       if(x%4==3)return 0;
    
   
    return -1;
}









In Python3 :





#!/bin/python3

import sys

def x(a):
    res = [a,1,a+1,0]
    return res[a%4]

def y(a):
    if a%2==0:
        return x(a//2)<<1
    else:
        r = 0
        if (a+1)%4==2:
            r =1
        return((x(a//2)<<1) + r)

#print(y(5))
    
Q = int(input().strip())
for a0 in range(Q):
    L,R = input().strip().split(' ')
    L,R = [int(L),int(R)]
    print(y(R)^y(L-1))