Educational Codeforces Round 24

最新推荐文章于 2022-12-13 17:10:02 发布

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最新推荐文章于 2022-12-13 17:10:02 发布

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原文链接：http://www.cnblogs.com/BobHuang/p/7115789.html

本文介绍了一个简单的算法，用于解决竞赛中根据特定比例分配获奖学生数量的问题。输入为参赛人数及证书与文凭的比例，输出最大可能的获奖者数量及其构成。

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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

A题是个数轮，要用数学方法找出这三个数，或者根据答案找通项公式也行？

#include <stdio.h>
typedef long long LL;
int main(){
LL n,k;
scanf("%lld%lld",&n,&k);
LL a=n/2/(k+1);
LL b=k*a;
printf("%lld %lld %lld",a,b,n-a-b);
return 0;}