Question A

Very simple and naive question. One thing worth noticing is that $b > 0$.

int main(){
    int sum = 0;
    int tp;
    for(int i = 0; i < 5; ++ i){        
        cin >> tp;
        sum += tp;
    }
    if (sum%5 == 0 && sum)
        cout << sum/5 << endl;
    else
        cout << -1 << endl;
    return 0;
}

Question B

It is not hard to guess the two extreme cases with minimum and maximum number of relationship. For minimum number of relationships, we strive to distribute all n people to m groups. For maximum number of relationships, we should try our best to centralize all people in one group. Then the computation is trivial. One important thing is that n can be $1e9$, which results in a very large number of relationships. Thus we should use long long (even double fails because of precision..).

int main(){
    long long n, m;
    cin >> n >> m;
    long long mx = (n-m+1)*(n-m)/2;
    long long mod = n%m;
    long long quo = n/m;
    long long mi = (m-mod) * (quo * (quo-1))/2 + mod * (quo * (quo+1))/2;
    printf("%I64d %I64d\n", mi, mx);    

    return 0;
}

Question C

The solution to this problem is a greedy solution, which is very short. The restriction is that we cannot have three balloons with the same color at the same table.

The key ideas are

• As every table has three balloons, the number of the table must be lower than $(r+g+b)/3$
• We cannot have a number of tables that exceeds any of $(r+g)$, $(g+b)$, or $(r+b)$. This is because, for example, if there is an additional table besides the $r+g$ tables, this table must be of three blue balloons

int main()
{
    long long r,g,b;
    cin>>r>>g>>b;
    cout << min(min(min((r+g+b)/3,r+g),g+b),r+b);
}

Question D

This is a DP problem. I have not figured it out yet. To be continued…