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

B. Random Teams

 

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

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

 题意：给你n,m，将n个人分配到m个小组，每组至少一个人，组内成员会成为朋友，问你在所有可行的分配方法中最少，最多有多少对朋友

题解：显然组成员尽量大，是最多，最分散是最少

///
#include<bits/stdc++.h>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a));
#define inf 1000000007
#define mod 1000000007
inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){
        if(ch=='-')f=-;ch=getchar();
    }
    while(ch>=''&&ch<=''){
        x=x*+ch-'';ch=getchar();
    }return x*f;
}
//************************************************
const int maxn=+;

ll n,m,ans1,ans2;
int main(){
scanf("%I64d%I64d",&n,&m);
   ans1=n-(m-);
   ans1=(ans1)*(ans1-)/;
   ans2=n/m;
   if(n%m)ans2++;
   ans2=(ans2)*(ans2-)/;
   ans2= ans2*(n%m)+(m-(n%m))*(n/m)*(n/m-)/;
   cout<<ans2<<" "<<ans1<<endl;
  return ;
}

代码