MUSICHAIR - Editorial

PROBLEM LINK:

Practice

Div-3 Contest

Div-2 Contest

Div-1 Contest

Author: Utkarsh Gupta

Tester: Aryan Choudhary

DIFFICULTY:

SIMPLE

PREREQUISITES:

Math

PROBLEM:

There are N friends who decided to play the game of musical chairs one day.

  • Initially they decide a number K and with each passing round they remove exactly K chairs. Hence, in each round exactly K players are eliminated. The game stops when less than or exactly K players remain in the game.
  • Now, Chef wants to figure out for how many possible values of K there will be exactly one winner in the game.

EXPLANATION:

To solve this problem, let’s look at the final state. There should be exactly 1 chair remaining. In the turn before that, we would have had K + 1 chairs (since every turn, we remove exactly K chairs). And so on. We can see that N must be of the form 1 + xK, where x is any natural number. We want to find all the possible values of K.

Let’s solve for K.

N = 1 + xK
\implies K = (N - 1) / x

Since x can be any natural number, we just want to find all the possible divisors of (N - 1).
This can be done in \sqrt{N} by iterating over 1 to \sqrt{N}, since divisors form pairs. x \cdot y = N implies x <= \sqrt{N}.

For every x, we have 2 values (x and n / x) as the possible divisors. Check for the edge case where x = N / x, i.e., x = \sqrt{N}.

SOLUTIONS:

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

int main() {
    int tt;
    cin >> tt;
    while (tt--) {
        int n; cin >> n;
        n--;
        int ans = 0;
        for (int i = 1; i * i <= n; i++) {
            if (n % i == 0) {
                ans++;
                if (i * i != n)
                    ans++;
            }
        }
        cout << ans << '\n';
    }
}
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(1,1e2);
while(T--)
{

    n=readIntLn(2,1e9)-1;
    lli ans=0;
    for(lli i=1;i*i<=n;++i){
        if(n%i)
            continue;
        ans++;
        if(i*i!=n)
            ans++;
    }
    cout<<ans<<endl;
}   aryanc403();
    readEOF();
    return 0;
}

in tester solution I cant able to understand the line
if (i * i != n)
ans++;
can anybody help.?

Let’s take an example for N=17 but in the end only one will be the winner. So we have to find a total number from 1 to 16 which can divide 16. Manually we can say that 1,2,4,8,16 are the number which can divide 16.

Our loop is iterating from 1 to 4, but 8 and 16 are also the no’s which can divide 16. So by using that if condition these outer no are also get added to our ans. Because i*y=N, so if i can divide 16, N/i also can divide 16. For example,

For iteration i=1, 16%1==0, so other no which can divide 16 is N/i i.e 16/1=16(ans=2)

For iteration i=2, 16%2==0, so other no which can divide 16 is N/i i.e 16/2=8(ans=4)

For iteration i=4, 16%4==0, so other no which can divide 16 is N/i i.e 16/4=4 (But we can’t take 4 twice so we don’t increment it twice and that is the use of that if statement)(ans=5)

3 Likes