MAKEZERO-Editorial

PROBLEM LINK:

Contest Division 1
Contest Division 2
Contest Division 3
Contest Division 4

Setter: Yash Gandhi
Tester: Manan Grover, Lavish Gupta
Editorialist: Devendra Singh

DIFFICULTY:

1320

PREREQUISITES:

None

PROBLEM:

You are given an array A of length N.

You can perform the following operation on the array any number of times:

  • Choose any subsequence S of the array A and a positive integer X such that X is a power of 2 and subtract X from all the elements of the subsequence S.

Find the minimum number of operations required to make all the elements of the array equal to 0.

EXPLANATION:

Any two operations have different values of X in the optimal strategy

Let us suppose the subsequences chosen in two operations having same value of X be S_1 and S_2. Select all the elements which are present in both the subsequences and put them in new\:S_2 and and apply the operation on them taking 2\cdot X and the elements which are present only in either S_1 or only in S_2 into new\: S_1 and apply the operation by taking X as the integer. Keep doing this until there are no two operations with same value of X.

This means for each bit i which is set to 1 in the binary representation of any element K of the array A , 2^i will be subtracted, from all the elements having i^{th} bit set to 1, exactly once.
Therefore the answer is the number of different bits which are set to 1 in any element of the array.
Let OrValue be the bitwise\:OR of all the elements of the array.
Then the answer = countofsetbits(OrValue).

TIME COMPLEXITY:

O(N) 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 sumN=0;
int MaxAi=1000000000;
void solve()
{
    int n=readInt(1,100000,'\n');
    int A[n+1]={0};
    for(int i=1;i<=n;i++)
    {
    	if(i==n)
    		A[i]=readInt(0,MaxAi,'\n');
    	else
    		A[i]=readInt(0,MaxAi,' ');
    }
    ll orval=0;
    for(int i=1;i<=n;i++)
    	orval|=A[i];
    cout<<(__builtin_popcount(orval))<<'\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(1,1000,'\n');
    //cin>>T;
    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>
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)
{
    int n, ans = 0;
    cin >> n;
    vll v(n);
    int bit[30] = {0};
    for (int i = 0; i < n; i++)
    {
        cin >> v[i];
        for (int j = 0; j < 30; j++)
            if ((1 << j) & v[i])
                bit[j]++;
    }
    for (int j = 0; j < 30; j++)
        ans += (bit[j] > 0);
        cout<<ans<<'\n';
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL), cout.tie(NULL);
    int test = 1;
    cin >> test;
    while (test--)
        sol();
}

for (int j = 0; j < 30; j++)
if ((1 << j) & v[i])
bit[j]++;

in this part of the code.,why do we only shift bits till the 29th position is actually an integer in a 32-bit number?