Codeforces Round #273 (Div. 2) --B Random Teams

题目链接：Random Teams

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.

解题思路：n个人，m个队，仅仅有在一个队的人能够两两成为朋友。问能比完赛后，朋友的对数的最大值和最小值为多少。

非常显然。最大值的分法是有n-1个队中仅仅放一个人。 把剩下的人全放到剩下的那个队。

最少的分法是尽可能的均分n个人到m个队。详细的朋友对数是简单的组合，k个人两两之间的朋友对数为  C（k,2）  =  k*(k-1)/2 .

AC代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define INF 0x7fffffff

long long n, m, minn, maxn;

long long c(long long x){
    return (x*(x-1))/2;
}

int main()
{
    #ifdef sxk
        freopen("in.txt","r",stdin);
    #endif
    while(scanf("%lld%lld",&n, &m)!=EOF)
    {
        long x = n/m;
        long y = n%m;
        if(m == 1) minn = c(n);
        else   minn = y*c(x+1) + (m-y)*c(x);
        maxn = c(n-m+1);
        printf("%lld %lld
", minn, maxn);
    }
    return 0;
}