478B. Random Teams

今天起放弃onenote开始在博客里写题解了（因为onenote打印太头疼了。。）

B. Random Teams

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)

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

水题，最大就是把人数全集中到一个队，最小就是尽量平均分配。

有个小坑就是数据溢出，把int全部换成long long即可，话说一开始不舍得全换，一个个换，结果WA了好多遍。。看来以后还是要狠心点。

#include <iostream>
using namespace std;

long long sum(long long n)
{
	return (1+n)*n/2;
}
int main(void)
{
	long long n,m;
	cin>>n>>m;
	long long a=n/m,b=n%m;
	long long  ans_min=sum(a)*b+sum(a-1)*(m-b);
	long long  ans_max=sum(n-(m-1)-1);
	cout<<ans_min<<" "<<ans_max<<"\n";
	return 0;
}