Random Teams------贪心

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 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.

题意

n个人分到m个队中，每个队的队友的人都会成为好朋友，问好朋友最多最少分别有多少对

解题思路

一开始也没啥思路，就猜想了几种情况，自己找了几组数据试验了一下，发现把人集中在其中一队，其他队都放一个人这种情况好朋友的对数最多，如果每个队均分的话对数最少
容易出错的点：
注意数据范围为long long
看清楚是先输出max还是先输出min，我就给弄反了…这种蠢错误只有我会犯吧，wa了好几发

代码实现

#include<iostream>
using namespace std;
typedef long long ll;
int main()
{
	long long n, m, max = 0, min = 0;
	cin >> n >> m;
	if (m == 1)//一队的话最大值最小值相等
	{
		max = n * (n - 1) / 2;
		cout << max << " " << max << endl;
		return 0;
	}
	else
	{
		long long t = n - m + 1,x = n / m, y = n % m;
		max = t * (t - 1) / 2;
		min = (m - y) * x*(x-1)/2 + y * x*(x+1)/2;
		cout << min << " " << max << endl;
	}
	return 0;
}

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