# HackerEarth ICPC Team Management problem solution

In this HackerEarth ICPC Team Management problem solution Little Chandan is an exceptional manager - apart from his role in HackerEarth - as the person who has to bug everyone, in general... and if possible, try to get some work done.

He's also offered a job as the coach of the best Russian teams participating in the ACM-ICPC World Finals. Now, Chandan is an extremely good coach, too. But he's a weird person who thrives on patterns in life, in general. So, he has decided that if there are n number of students in total, and he is supposed to divide them into camps of k students - he wants them to be arranged in such a way that the length of names of all the students in a camp is equal.

I know, totally weird, right?

## HackerEarth ICPC Team Management problem solution.

`tc = int(raw_input())assert(tc>0 and tc<51)for i in xrange(tc):    n, k = map(int, raw_input().split())    l = *1000    ok = True    assert(n>0 and n<1001)    assert(k>0 and k<1001)    assert(n>=k)    assert(n%k==0)    for j in xrange(n):        a = raw_input()        l[len(a)] +=1    for z in l:        if z%k!=0:            ok = False            break    if (ok):        print "Possible"    else:        print "Not possible"`

### Second solution

`#include<bits/stdc++.h>using namespace std;#define ll              long long int#define vi              vector<int>#define vl              vector<ll>#define pii             pair<int,int>#define pil             pair<int, ll>#define pll             pair<ll, ll>#define pli             pair<ll, int>#define pb(v, a)        v.push_back(a)#define mp(a, b)        make_pair(a, b)#define MOD             1000000007#define rep(i, a, b)    for(i=a; i<=b; ++i)#define rrep(i, a, b)   for(i=a; i>=b; --i)#define si(a)           scanf("%d", &a)#define sl(a)           scanf("%lld", &a)#define pi(a)           printf("%d", a)#define pl(a)           printf("%lld", a)#define pn              printf("\n")ll pow_mod(ll a, ll b){    ll res = 1;    while(b)    {        if(b & 1)            res = (res * a) % MOD;        a = (a * a) % MOD;        b >>= 1;    }    return res;}int cnt;int main(){    int t, i;    si(t);    rep(i, 1, t)    {        memset(cnt, 0, sizeof(cnt));        int n, k, j;        si(n);        si(k);        string str;        rep(j, 1, n)        {            cin>>str;            cnt[str.length()]++;        }        bool flag = true;        rep(j, 1, 100)        {            if(cnt[j] % k)            {                flag = false;                break;            }        }        if(flag)            cout<<"Possible\n";        else            cout<<"Not possible\n";    }    return 0;}`