PROBLEM LINK:
Contest Division 1
Contest Division 2
Contest Division 3
Contest Division 4
Setter: Utkarsh Gupta
Tester: Satyam, Utkarsh Gupta
Editorialist: Devendra Singh
DIFFICULTY:
To be calculated
PREREQUISITES:
PROBLEM:
Chef has two arrays A and B, each of length N.
A pair (i,j) (1 \leq i \lt j \leq N) is said to be a good pair if and only if
A_i \oplus A_j = B_i \oplus B_j
Here, \oplus denotes the bitwise XOR operation.
Determine the number of good pairs.
EXPLANATION:
Let us simplify the equation given in the problem by taking XOR
with A_j\oplus B_i on both sides.
A_i \oplus A_j \oplus A_j \oplus B_i = B_i \oplus B_j \oplus A_j \oplus B_i
A_i \oplus B_i = A_j \oplus B_j
\therefore We need to find the number of pair (i,j) (1 \leq i \lt j \leq N) such that A_i \oplus B_i = A_j \oplus B_j
Iterate on the arrays from i=1 to N, keeping track of A_i\oplus B_i with the help of a map data structure. For each index i add the count of A_i\oplus B_i to the answer then update the count of A_i\oplus B_i in the map.
TIME COMPLEXITY:
O(Nlog(N)) or for each test case.
SOLUTION:
Setter's solution
//Utkarsh.25dec
#include <bits/stdc++.h>
#define ll long long int
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define vl vector <ll>
#define all(c) (c).begin(),(c).end()
using namespace std;
ll power(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll modInverse(ll a){return power(a,mod-2);}
const int N=500023;
bool vis[N];
vector <int> adj[N];
long long readInt(long long l,long long r,char endd){
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true){
char g=getchar();
if(g=='-'){
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g && g<='9'){
x*=10;
x+=g-'0';
if(cnt==0){
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);
assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd){
if(is_neg){
x= -x;
}
if(!(l <= x && x <= r))
{
cerr << l << ' ' << r << ' ' << x << '\n';
assert(1 == 0);
}
return x;
} else {
assert(false);
}
}
}
string readString(int l,int r,char endd){
string ret="";
int cnt=0;
while(true){
char g=getchar();
assert(g!=-1);
if(g==endd){
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt && cnt<=r);
return ret;
}
long long readIntSp(long long l,long long r){
return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
return readString(l,r,'\n');
}
string readStringSp(int l,int r){
return readString(l,r,' ');
}
int maxAi=((1<<30)-1);
int sumN=0;
int A[N],B[N],C[N];
void solve()
{
int n=readInt(1,300000,'\n');
sumN+=n;
assert(sumN<=300000);
for(int i=1;i<=n;i++)
{
if(i==n)
A[i]=readInt(0,maxAi,'\n');
else
A[i]=readInt(0,maxAi,' ');
}
for(int i=1;i<=n;i++)
{
if(i==n)
B[i]=readInt(0,maxAi,'\n');
else
B[i]=readInt(0,maxAi,' ');
}
set <int> s;
map <int,ll> cnt;
for(int i=1;i<=n;i++)
{
C[i]=(A[i]^B[i]);
s.insert(C[i]);
cnt[C[i]]++;
}
ll ans=0;
for(auto it:s)
ans+=((cnt[it]*(cnt[it]-1))/2);
cout<<ans<<'\n';
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
ios_base::sync_with_stdio(false);
cin.tie(NULL),cout.tie(NULL);
int T=readInt(1,100000,'\n');
while(T--)
solve();
assert(getchar()==-1);
cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}
Editorialist's Solution
#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define pb push_back
#define all(_obj) _obj.begin(), _obj.end()
#define F first
#define S second
#define pll pair<ll, ll>
#define vll vector<ll>
ll INF = 1e18;
const int N = 1e5 + 11, mod = 1e9 + 7;
ll max(ll a, ll b) { return ((a > b) ? a : b); }
ll min(ll a, ll b) { return ((a > b) ? b : a); }
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void sol(void)
{
ll n, ans = 0;
cin >> n;
map<ll, ll> cnt;
vll a(n), b(n);
for (int i = 0; i < n; i++)
{
cin >> a[i];
}
for (int i = 0; i < n; i++)
{
cin >> b[i];
ans += cnt[a[i] ^ b[i]];
cnt[b[i] ^ a[i]]++;
}
cout << ans << '\n';
return;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL), cout.tie(NULL);
int test = 1;
cin >> test;
while (test--)
sol();
}