PROBLEM LINK:
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;
}