Educational Codeforces Round 47 (Rated for Div. 2)(D. Relatively Prime Graph)

Bismillahir Rahmanir Rahim

just check every eligible pair(whose gcd is 1) until we got M

///Author: Tanvir Ahmmad
///CSE,Islamic University,Bangladesh

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<vector>
#include<iterator>
#include <functional> ///sort(arr,arr+n,greater<int>()) for decrement of array
/*every external angle sum=360 degree
  angle find using polygon hand(n) ((n-2)*180)/n*/
///Floor[Log(b)  N] + 1 = the number of digits when any number is represented in base b
using namespace std;
typedef long long ll;

int main()
{
    vector<pair<ll,ll> >vec;
    ll n,m;
    cin>>n>>m;
    ll m1=m;
    for(ll i=1; i<=n && m1>=0; i++)
        for(ll j=i+1; j<=n && m1>=0; j++)
            if(__gcd(i,j)==1)
                vec.push_back(make_pair(i,j)),m1--;
    if(m1<=0 && (m+1)>=n)
    {
        cout<<"Possible"<<endl;
        for(ll i=0; i<m; i++)
            cout<<vec[i].first<<" "<<vec[i].second<<endl;
    }
    else
        cout<<"Impossible"<<endl;
    return 0;
}
///Alhamdulillah

Problem Link:https://codeforces.com/problemset/problem/1009/D

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