Codeforces Round 273 Random Teams 解题心得

原题：

Description

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 ≤ 10⁹) — the number of participants and the number of teams respectively.

Output

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

Sample Input

Input

5 1

Output

10 10

Input

3 2

Output

1 1

Input

6 3

Output

3 6

Hint

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 2people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.

分析:

一个队所组成的朋友是Cn 2　　　　　也就是n+(n-1)+(n-2)+....+1

细心想想就会知道

保证每个队有一个人，把所有剩下的人都几种到一个队的时候朋友数最多，

尽量平均分的朋友数最少

依据这个来写代码：

#include <stdio.h>
#include <iostream>
using namespace std;
long long jie(long long x)
{
    return (x*(x - 1)) / 2;
}
int main()
{
    long long n, m, minn, maxn;

    scanf("%lld%lld", &n, &m);

    long x = n / m;
    long y = n%m;
    if (m == 1)
        minn = jie(n);
    else
        minn = y*jie(x + 1) + (m - y)*jie(x);
    maxn = jie(n - m + 1);

    printf("%lld %lld\n", minn, maxn);

    return 0;
}