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Saturday, June 25, 2016

## Codeforces 478B | Random Teams

Question:

Solution:

#include <iostream> using namespace std; int main(){ long long n , m; // inputs cin >> n >> m;// grab em // maximum long long kmx = n- m +1; // get the max number of people in one team so forexample if 6 ppl and 3 teams the max is 4 since the other 2 teams will hv only one person kmx= kmx*(kmx-1)/2; // now match every one with his friends using the`famous formula n*(n-1)/2 which accounts for each person as n and (n-1)/2`

`are the relationships he/she might make with others (n-1) since relationships are`

undirected which means a loves b == b loves a we should divide by 2 //minimum ( real bad ) long long kmn = ((n / m) * ((n / m) - 1)) / 2; // get possible pairings PER TEAM inuniform teams example 6 ppl 3 teams 6/3 is uniform number which is 2 per team if(n % m == 0){ kmn *= m; // special case }else{ kmn *= m - (n % m); // without leftovers ppl who got no uniform team`( n%m is the number of teams who hv no uniform number of ppl) since this is the`

remainder of a unifrom division right ? // now kmn accounts for pairings in uniform teams only long long lft =(((n / m) + 1 ) * (n / m)) / 2; // now leftovers ! // lft has possible pairings kmn += lft * (n % m); // multiply possible pairings per person bythe number of lftovers and add them to kmn } cout << kmn << " " << kmx << endl; // print them ! // we done boyzz } // ez@macacm.org -- 01:03AM 23-06-2016 // Ezzeldin Adel Tahoun || McMaster University

Video Explanation:

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- Codeforces 478B [SOLVED} | Code and Video Explanation