Diplomas and Certificates (Codeforces

题目链接  http://codeforces.com/problemset/problem/818/A

A. Diplomas and Certificates

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n students who have taken part in an olympiad. Now it's time to award the students.

Some of them will receive diplomas, some wiil get certificates, and others won't receive anything. Students with diplomas and certificates are called winners. But there are some rules of counting the number of diplomas and certificates. The number of certificates must be exactly k times greater than the number of diplomas. The number of winners must not be greater than half of the number of all students (i.e. not be greater than half of n). It's possible that there are no winners.

You have to identify the maximum possible number of winners, according to these rules. Also for this case you have to calculate the number of students with diplomas, the number of students with certificates and the number of students who are not winners.

Input

The first (and the only) line of input contains two integers n and k (1 ≤ n, k ≤ 1012), where n is the number of students and k is the ratio between the number of certificates and the number of diplomas.

Output

Output three numbers: the number of students with diplomas, the number of students with certificates and the number of students who are not winners in case when the number of winners is maximum possible.

It's possible that there are no winners.

Examples

input

18 2

output

3 6 9

input

9 10

output

0 0 9

input

1000000000000 5

output

83333333333 416666666665 500000000002

input

1000000000000 499999999999

output

1 499999999999 500000000000

题目大意：

有一场比赛，一共有n个人参加，得奖人数有两条限制：1 得第一种奖的人数 是 得第二中奖人数 的k倍；2 总共得奖人数<=总人数的一半。题目问：在得奖人数最大的情况下，得第一种奖项有几人，得第二种奖项有几人，未得奖有几人。

解题思路：

这道题我主张直接看代码，然后自己进行总结。需要注意的是ans1、ans2、ans3，要分开算，要充分利用结果算，要不然结果不一样；还有，得奖人数是0，也是有可能的。

代码：

#include<iostream>using namespace std;int main(){    long long int n,k,half,ans1,ans2,ans3;    while(cin>>n>>k)    {        half=n/2;        ans1=half/(k+1);        ans2=k*ans1;        ans3=n-ans1-ans2;        cout<<ans1<<' '<<ans2<<' '<<ans3<<endl;    }    return 0;}