ACM第六周竞赛题目——B CodeForces 478B

B - B

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

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

解题思路：

题意：

n个人分到m支队伍里，在同一个队伍的任意两人将成为一对朋友，问能形成最少多少对朋友，最多多少对朋友。每支队伍至少要有一个人。

分析：

假设某只队伍有k个人，那么形成的朋友对数有C(k, 2)对。

最大的对数：将所有人尽可能的集中在一只队伍里，让尽可能多的人相互认识

最小的对数：将所有人尽可能的分开，即平分

程序代码：

#include <cstdio>
using namespace std;
int main()
{
    long long  n,m;
    while(scanf("%lld%lld",&n,&m)!=EOF)
    {
    long long a=n/m;
    long long b=n%m;
    long long mi=b*a*(a+1)/2+(m-b)*a*(a-1)/2;
    long long ma=(n-m+1)*(n-m)/2;
    printf("%lld %lld
",mi,ma);
    }
    return 0;
}

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