Educational Codeforces Round 24
1 second
256 megabytes
standard input
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.
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 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.
18 2
3 6 9
9 10
0 0 9
1000000000000 5
83333333333 416666666665 500000000002
1000000000000 499999999999
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//(k+);
LL b=k*a;
printf("%lld %lld %lld",a,b,n-a-b);
return ;}
Educational Codeforces Round 24的更多相关文章
- Educational Codeforces Round 24 A 水 B stl C 暴力 D stl模拟 E 二分
A. Diplomas and Certificates time limit per test 1 second memory limit per test 256 megabytes input ...
- Educational Codeforces Round 24 CF 818 A-G 补题
6月快要结束了 期末也过去大半了 马上就是大三狗了 取消了小学期后20周的学期真心长, 看着各种北方的学校都放假嗨皮了,我们这个在北回归线的学校,还在忍受酷暑. 过年的时候下定决心要拿块ACM的牌子, ...
- Educational Codeforces Round 24 E
Vova again tries to play some computer card game. The rules of deck creation in this game are simple ...
- codeforces Educational Codeforces Round 24 (A~F)
题目链接:http://codeforces.com/contest/818 A. Diplomas and Certificates 题解:水题 #include <iostream> ...
- Educational Codeforces Round 24 D
Alice and Bob got very bored during a long car trip so they decided to play a game. From the window ...
- Educational Codeforces Round 24 B
n children are standing in a circle and playing a game. Children's numbers in clockwise order form a ...
- Educational Codeforces Round 24 A
There are n students who have taken part in an olympiad. Now it's time to award the students. Some o ...
- Educational Codeforces Round 24 题解
A: 考你会不会除法 //By SiriusRen #include <bits/stdc++.h> using namespace std; #define int long long ...
- Educational Codeforces Round 32
http://codeforces.com/contest/888 A Local Extrema[水] [题意]:计算极值点个数 [分析]:除了第一个最后一个外,遇到极值点ans++,包括极大和极小 ...
随机推荐
- 開玩樹莓派(一):安裝Raspbian系統
目錄: 開玩樹莓派(一):安裝Raspbian系統 開玩樹莓派(二):配置IP,實現無顯示器局域網內Putty連接和RDP遠程 開玩樹莓派(三):Python編程 開玩樹莓派(四):GPIO控制和遠程 ...
- Webservice相关的知识
一.利用jdk web服务api实现,这里使用基于 SOAP message 的 Web 服务 1.首先建立一个Web services EndPoint: package Hello; import ...
- Android 两个ArrayList找出相同元素及单个ArrayList删除元素
//从一个ArrayList中删除重复元素 List<String> arrayList1 = new ArrayList<String>(); arrayList1.add( ...
- C 函数库 (libc,glibc,uClibc,newlib)
glibc glibc和libc都是Linux下的C函数库,libc是Linux下的ANSI C的函数库:glibc是Linux下的GUN C的函数库:GNU C是一种ANSI C的扩展实现.ANSI ...
- iOS 如何解决并发请求时,只接受最后一个请求返回的结果
大致意思是 虽然NSOperation 的cancel 并不能取消请求,但是可以对这个NSOperation进行标记. 当cancel 属性是YES时,表明 NSOperation虽然已经执行,并 ...
- c语言中的->代表什么意思
c语言中 ->符号是什么意思? 比如c=a->b a为结构体或联合体的指针,->表示调用其成员
- Jenkins怎么启动和停止服务
笔者没有把Jenkins配置到tomcat中,每次都是用命令行来启动Jenkins.但是遇到一个问题:Jenkins一直是开着的,想关闭也关闭不了.百度了一些资料,均不靠谱(必须吐槽一下百度).于是进 ...
- UVA 10572 Black & White (状压DP)
题意:有一个n*m的矩阵,其中部分格子已经涂黑,部分涂白,要求为其他格子也上黑/白色,问有多少种涂法可以满足一下要求: (1)任意2*2的子矩阵不可以同色. (2)所有格子必须上色. (3)只能有两个 ...
- MVC视图特性
在主界面的视图中可以使用viewdata,引用主界面的分布视图界面也可以调用主界面的分部视图,但是分部视图不可以定义viewdata并使用 例子如下: // // GET: /Home/ public ...
- Tensorflow_入门学习_1
1.0 TensorFlow graphs Tensorflow是基于graph based computation: 如: a=(b+c)∗(c+2) 可分解为 d=b+c e=c+2 a=d∗e ...