Day 38 B. Random Teams

Problem
n participants of the competition were split into m teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.

Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.

Input
The only line of input contains two integers n and m, separated by a single space (1 ≤ m ≤ n ≤ 10^9) — the number of participants and the number of teams respectively.

Output
The only line of the output should contain two integers kmin and kmax — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.

Examples
input
5 1
output
10 10
input
3 2
output
1 1
input
6 3
output
3 6

Note
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.

In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.

In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.

//一道排列组合题 最大的情况是保证每支队伍有一个人 然后其他人都在一支队伍上 最小的情况是每个队伍的人竟可能接近平分

#include
#include
#include
#include
#include
#include
#include
#include
#include
#define ll long long
using namespace std;
int main()
{
    int n, m;
    cin >> n >> m;
    ll x = m - 1;
    ll nn = n - x;
    ll maxn = nn * (nn - 1) / 2;
    ll p = n / m;
    ll p1 = n % m;
    ll p2 = m - n % m;
    ll minn = p2 * ((p * (p - 1)) / 2) + p1 * ((p * (p + 1)) / 2);
    cout << minn << " " << maxn << endl;
    return 0;
}