codeforces 478B Random Teams

codeforces   478B  Random Teams  解题报告

题目链接：cm.hust.edu.cn/vjudge/contest/view.action?cid=88890#problem/B

题目：

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

Sample Output

 

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.

题意：

n个人分成m队（每队至少一人），竞赛结束后，其中来自同一队的任意两个人可以成为朋友。求最多和最少可以组成多少队朋友？

分析：

排列组合问题。最少：n个人平均分到m个队，组成的队数最少。如果不能平分（即n%m!=0）,则将剩下的人平均分到每个组。

最多：m-1个队只有一个人，剩下的人全部在一个队中。

代码：

#include<cstdio>
#include<iostream>
using namespace std;

int main()
{
    long long n,m;
    long long kmin,kmax;
    scanf("%I64d%I64d",&n,&m);
    long long a=n/m,b=n%m;
    if(b==0)//可以平均分（最少）
        kmin=m*a*(a-1)/2;
    else //不能平均分（最少）
        kmin=(m-b)*a*(a-1)/2+b*a*(a+1)/2;
    kmax=(n-m+1)*(n-m)/2;//最多
    printf("%I64d %I64d\n",kmin,kmax);
    return 0;
}