#include<bits/stdc++.h>
#define endl '\n'
#define int long long
using namespace std;
void solve()
{
int n;cin>>n;
vector<vector<int>>arr(n,vector<int>(20,0));
for(int i=0;i<n;i++)
{
int temp;cin>>temp;
int j=0;
while(temp!=0)
{
int d=(temp&1);
arr[i][j]=d;
temp=temp>>1;
j++;
}
}
vector<int>ans(20,0);
for(int i=0;i<18;i++)
{
int cnt=0;
for(int j=0;j<n;j++)
{
if(arr[j][i]==1)
cnt++;
}
if(cnt>=2)
ans[i]=1;
}
int res=0;
for(int i=0;i<20;i++)
{
res+=ans[i]*pow(2,i);
}
cout<<res<<endl;
}
int32_t main()
{
ios_base::sync_with_stdio(false);cin.tie(NULL);
int t;cin>>t;
while(t--)
{
solve();
}
}
Given that 1\le A_i\le 10^9, You’ll need at least 30 bits to represent them.
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Thank you bro you always help me!!