1 second
256 megabytes
standard input
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.
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.
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.
5 1
10 10
3 2
1 1
6 3
3 6
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.
n个人分到m组里，每一组对于答案的贡献是人数排列组合，就是n*(n-1)/2。求最大值和最小值
显然前面取m-1组的1最后一组最大的方案是答案最大
平均分的方案是答案最小
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
#define pi 3.1415926535897932384626433832795028841971
using namespace std;
inline LL read()
{
LL x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
LL a,b,c,d,e;
int main()
{
a=read();b=read();
c=(a-b+1)*(a-b)/2;
d=(a-b*(a/b));
e=(d)*(a/b+1)*(a/b)/2+(b-d)*(a/b)*(a/b-1)/2;
printf("%lld %lld\n",e,c);
}