codeforces Round #273(div2) B解题报告

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

题目大意：

有n个人参加某个比赛，被分成m组，赛后同组的人相互之间会成为朋友。求出在不同的分组方式中，新朋友数量的最大值和最小值。

解法：

在同组中，假设有k个人，那么可以成为朋友的数量为 k*(k-1)/2，那么让k的数量越大则朋友对数越多，k的数量越少则朋友对数越少。

采用贪心，最大值则其他组都被分1个人，最后一组将所有人都聚在一起；最小值则其他组都平均分配，多出来的也平均分配。

代码：

#include <iostream>
#define LL long long

using namespace std;

LL n, m, maxn, minn;

LL sum(LL n) {
return n*(n-1)/2;
}

int main() {
cin >> n >> m;

LL maxn = sum(n-m+1);

LL minn = (n%m)*sum(n/m+1) + (m-n%m)*sum(n/m);

cout << minn << ' ' << maxn << endl;
}