INTSEQ - Editorial

PROBLEM LINK:

Contest Division 1
Contest Division 2
Contest Division 3

Setter: Utkarsh Gupta
Tester: Aryan Choudhary
Editorialist: Lavish Gupta

DIFFICULTY:

Simple

PREREQUISITES:

Observations

PROBLEM:

You are given an integer K.

Consider an integer sequence A = [A_1, A_2, \ldots, A_N].

Define another sequence S of length N, such that S_i = A_1 + A_2 + \ldots + A_i for each 1 \leq i \leq N.

A is said to be interesting if A_i + S_i = K for every 1 \leq i \leq N.

Find the maximum length of an interesting sequence. If there are no interesting sequences, print 0.

QUICK EXPLANATION:

  • Calculate A_1, A_2 and so on in terms of K by substituting i = 1, 2 and so on in the equation A_i + S_i = K
  • A_i will come out to be K/2^i.
  • So the largest value of N such that A_N is an integer will be equal to the largest power of 2 that divided K.

EXPLANATION:

In the problem statement, it is given that:

  1. S_i = A_1 + A_2 + \cdots + A_i for each 1 \leq i \leq N.
  2. S_i + A_i = K for each 1 \leq i \leq N.

Let us see what happens when i = 1:
S_1 = A_1
\implies S_1 + A_1 = 2 \cdot A_1 = K
\implies A_1 = K/2

Now, Let us substitute i = 2, and use the value of A_1:
S_2 = A_1 + A_2
\implies S_2 + A_2 = A_1 + 2 \cdot A_2 = K
\implies 2\cdot A_2 = K/2
\implies A_2 = K/4

We can continue the process, and see that A_i = K/2^i. This can be rigorously proved using Strong Induction.

We have the constraint that A_i is an integer for all 1 \leq i \leq N. This means that we can create the sequence as long as K/2^i is an integer. In other words, the maximum value of N is equal to the highest power of 2 that divides K.

Let the highest power of 2 that divides K be M.
2^M \leq K \implies M \leq \log_2K

To calculate this highest power, we can continue dividing K by 2 as long as it is divisible.

TIME COMPLEXITY:

To calculate the highest power of 2 that divides K, we will require O(M) time.
Therefore, Time Complexity for each test case will be O(\log_2K).

SOLUTION:

Tester's Solution
/* in the name of Anton */

/*
  Compete against Yourself.
  Author - Aryan (@aryanc403)
  Atcoder library - https://atcoder.github.io/ac-library/production/document_en/
*/

#ifdef ARYANC403
    #include <header.h>
#else
    #pragma GCC optimize ("Ofast")
    #pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
    //#pragma GCC optimize ("-ffloat-store")
    #include<bits/stdc++.h>
    #define dbg(args...) 42;
#endif

// y_combinator from @neal template https://codeforces.com/contest/1553/submission/123849801
// http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2016/p0200r0.html
template<class Fun> class y_combinator_result {
    Fun fun_;
public:
    template<class T> explicit y_combinator_result(T &&fun): fun_(std::forward<T>(fun)) {}
    template<class ...Args> decltype(auto) operator()(Args &&...args) { return fun_(std::ref(*this), std::forward<Args>(args)...); }
};
template<class Fun> decltype(auto) y_combinator(Fun &&fun) { return y_combinator_result<std::decay_t<Fun>>(std::forward<Fun>(fun)); }

using namespace std;
#define fo(i,n)   for(i=0;i<(n);++i)
#define repA(i,j,n)   for(i=(j);i<=(n);++i)
#define repD(i,j,n)   for(i=(j);i>=(n);--i)
#define all(x) begin(x), end(x)
#define sz(x) ((lli)(x).size())
#define pb push_back
#define mp make_pair
#define X first
#define Y second
#define endl "\n"

typedef long long int lli;
typedef long double mytype;
typedef pair<lli,lli> ii;
typedef vector<ii> vii;
typedef vector<lli> vi;

const auto start_time = std::chrono::high_resolution_clock::now();
void aryanc403()
{
#ifdef ARYANC403
auto end_time = std::chrono::high_resolution_clock::now();
std::chrono::duration<double> diff = end_time-start_time;
    cerr<<"Time Taken : "<<diff.count()<<"\n";
#endif
}

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;
            }
            assert(l<=x&&x<=r);
            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,' ');
}

void readEOF(){
    assert(getchar()==EOF);
}

vi readVectorInt(int n,lli l,lli r){
    vi a(n);
    for(int i=0;i<n-1;++i)
        a[i]=readIntSp(l,r);
    a[n-1]=readIntLn(l,r);
    return a;
}

// #include<atcoder/dsu>
// vector<vi> readTree(const int n){
//     vector<vi> e(n);
//     atcoder::dsu d(n);
//     for(lli i=1;i<n;++i){
//         const lli u=readIntSp(1,n)-1;
//         const lli v=readIntLn(1,n)-1;
//         e[u].pb(v);
//         e[v].pb(u);
//         d.merge(u,v);
//     }
//     assert(d.size(0)==n);
//     return e;
// }

const lli INF = 0xFFFFFFFFFFFFFFFL;

lli seed;
mt19937 rng(seed=chrono::steady_clock::now().time_since_epoch().count());
inline lli rnd(lli l=0,lli r=INF)
{return uniform_int_distribution<lli>(l,r)(rng);}

class CMP
{public:
bool operator()(ii a , ii b) //For min priority_queue .
{    return ! ( a.X < b.X || ( a.X==b.X && a.Y <= b.Y ));   }};

void add( map<lli,lli> &m, lli x,lli cnt=1)
{
    auto jt=m.find(x);
    if(jt==m.end())         m.insert({x,cnt});
    else                    jt->Y+=cnt;
}

void del( map<lli,lli> &m, lli x,lli cnt=1)
{
    auto jt=m.find(x);
    if(jt->Y<=cnt)            m.erase(jt);
    else                      jt->Y-=cnt;
}

bool cmp(const ii &a,const ii &b)
{
    return a.X<b.X||(a.X==b.X&&a.Y<b.Y);
}

const lli mod = 1000000007L;
// const lli maxN = 1000000007L;

    lli T,n,i,j,k,in,cnt,l,r,u,v,x,y;
    lli m;
    string s;
    vi a;
    //priority_queue < ii , vector < ii > , CMP > pq;// min priority_queue .

int main(void) {
    ios_base::sync_with_stdio(false);cin.tie(NULL);
    // freopen("txt.in", "r", stdin);
    // freopen("txt.out", "w", stdout);
// cout<<std::fixed<<std::setprecision(35);
T=readIntLn(1,5e3);
while(T--)
{

    n=readIntLn(1,1e9);
    lli c=0;
    while(n%2==0){
        n/=2;
        c++;
    }
    cout<<c<<endl;
}   aryanc403();
    readEOF();
    return 0;
}

Editorialist's Solution
#include<bits/stdc++.h>
using namespace std ;

int main()
{
    int t ;
    cin >> t ;
    while(t--)
    {
        long long k ;
        cin >> k ;

        int ans = 0 ;

        while(k%2 == 0)
        {
            ans++ ;
            k /= 2 ;
        }

        cout << ans << '\n' ;
    }
 
    return 0;
}
4 Likes

Sir, what if the k= 9 or 15 etc,
k=9 the answer obtain is 0 bu there is an sequence containing {1,2,3}
where last element ai+si = 3+3+2+1 =9
could you explain this

2 Likes

Please, help me to find the sequence A and S for any odd k. I am not able to understand properly

1 Like

But it should be true for all i, not just i=3 case right?

Bro Check for each term