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

http://codeforces.com/contest/478/problem/B

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： 求n个人分到m个房间，最多能有几对朋友，最多的情况就是除了一个房间之外的其他房间都只有一个人，剩下的所有人在一个房间

最少的情况就是将所有的人尽可能平均分配到每一个房间

求组合数C(2, n)  = n * (n  - 1) / 2;

 1 #include <stdio.h>
 2 
 3 int main(){
 4     long long n, m, kmin, kmax, res;
 5     while(scanf("%I64d %I64d", &n, &m) != EOF){
 6         if(n % m == ){
 7             res = n / m;
 8             kmin = m * res * (res - ) / ;
 9         }
         else{
             res = n / m;
             kmin = (n % m) * res * (res + ) /  + (m - n % m) * res * (res - ) / ;
         }
 
         n = n - m + ;
         kmax = n * (n - ) / ;
         printf("%I64d %I64d\n", kmin, kmax);
     }
     return ;

20 }