[C++]Yellow Cards - GYM - 102348A(Practice *) - CodeForces
1 Problem Description
Problem
The final match of the Berland Football Cup has been held recently. The referee has shown n yellow cards throughout the match. At the beginning of the match there were a1 players in the first team and a2 players in the second team. The rules of sending players off the game are a bit different in Berland football. If a player from the first team receives k1 yellow cards throughout the match, he can no longer participate in the match — he’s sent off. And if a player from the second team receives k2 yellow cards, he’s sent off. After a player leaves the match, he can no longer receive any yellow cards. Each of n yellow cards was shown to exactly one player. Even if all players from one team (or even from both teams) leave the match, the game still continues. The referee has lost his records on who has received each yellow card. Help him to determine the minimum and the maximum number of players that could have been thrown out of the game.Input
The first line contains one integer a1 (1 ≤ a1 ≤ 1 000) — the number of players in the first team.The second line contains one integer a2 (1 ≤ a2 ≤ 1 000) — the number of players in the second team.The third line contains one integer k1 (1 ≤ k1 ≤ 1 000) — the maximum number of yellow cards a player from the first team can receive (after receiving that many yellow cards, he leaves the game).The fourth line contains one integer k2 (1 ≤ k2 ≤ 1 000) — the maximum number of yellow cards a player from the second team can receive (after receiving that many yellow cards, he leaves the game).The fifth line contains one integer n (1 ≤ n ≤ a1 · k1 + a2 · k2) — the number of yellow cards that have been shown during the match.Output
Print two integers — the minimum and the maximum number of players that could have been thrown out of the game.Examples

- Note
In the first example it could be possible that no player left the game, so the first number in the output is 0. The maximum possible number of players that could have been forced to leave the game is 4 — one player from the first team, and three players from the second.
In the second example the maximum possible number of yellow cards has been shown (3 · 6 + 1 · 7 = 25),so in any case all players were sent of
2 Problem Analysis

3 Code
/*
Yellow Cards - GYM - 102348 - Problem:A - 2019-09-29
[Problem Description] https://codeforces.com/gym/102348/problem/A
[Contest Material] https://codeforces.com/gym/102348/attachments/download/9362/statements.pdf
[Contest Information]
[Name] Southern and Volga Russia Qualifier 2019-2020 (GYM)
[Start] Sep/17/2019 00:00UTC+8
[Length] 04:00 (Hour)
[Other] Final standings / Prepared by BledDest Official ICPC Contest Northeastern Europe Region Russia, Saratov, 2019-2020 Statements: in English, in Russian
*/
#include<stdio.h>
#include<iostream>
using namespace std;
int min(int a1, int a2, int k1, int k2, int n){
int total = a1*(k1-1) + a2*(k2-1);
if(n <= total){
return 0;
} else {
return (n - total);
}
}
int max(int a1, int a2, int k1, int k2, int n){
int minTeamOfYellowCardsNum[2];
minTeamOfYellowCardsNum[0] = k1<k2?a1:a2;
minTeamOfYellowCardsNum[1] = k1<k2?k1:k2;
int temp1 = minTeamOfYellowCardsNum[0]*minTeamOfYellowCardsNum[1];
int temp2 = a1*k1 + a2*k2;
//printf("Amin*Kmin: %d\n", temp1);
//printf("A1*K1+A2*K2: %d\n", temp2);
if(n <= temp1){
return n/minTeamOfYellowCardsNum[1];// n < temp1
} else {
//printf("n/minTeamOfYellowCardsNum[1]: %d\n", n/minTeamOfYellowCardsNum[1]);
//printf("n: %d | Amin*Kmin(temp2): %d\n",n, temp2);
int maxTeamOfYellowCardsNum[2];
maxTeamOfYellowCardsNum[0] = k1>k2?a1:a2;
maxTeamOfYellowCardsNum[1] = k1>k2?k1:k2;
return minTeamOfYellowCardsNum[0] + ( ( n - temp1 )/maxTeamOfYellowCardsNum[1]);
}
}
int main(){
int a1,a2;
int k1,k2;
int n;
scanf("%d\n%d\n%d\n%d\n%d", &a1, &a2, &k1, &k2, &n);
//printf("%d\n%d\n%d\n%d\n%d\n", a1, a2, k1, k2, n);//test output
printf("%d\n", min(a1, a2, k1, k2, n));
printf("%d", max(a1, a2, k1, k2, n));
return 0;
}

Congratulations~

4 Problem Related Information
- [Problem Description]
- [Contest Material]
- [Contest Information]
- [Name] Southern and Volga Russia Qualifier 2019-2020 (GYM)
- [Start] Sep/17/2019 00:00UTC+8
- [Length] 04:00 (Hour)
- [Other Information]
- Final standings
- Prepared by BledDest Official ICPC Contest Northeastern Europe Region Russia, Saratov, 2019-2020 Statements: in English, in Russian
5 Recommend Documents
- 1 Codeforces游玩攻略 - About CodeForces - 知乎
- 2 在玩acm的大佬眼中codeforces这个平台怎么样? - Author:hzwer - 北京大学(计算机视觉) - About CodeForces - 知乎
- 3 在玩acm的大佬眼中codeforces这个平台怎么样? - Author:ccsu cat - About CodeForces - 知乎
- 4 洛谷
- 5 CodeForces
[C++]Yellow Cards - GYM - 102348A(Practice *) - CodeForces的更多相关文章
- Codeforces Round #585 (Div. 2) A. Yellow Cards(数学)
链接: https://codeforces.com/contest/1215/problem/A 题意: The final match of the Berland Football Cup ha ...
- A. Yellow Cards ( Codeforces Round #585 (Div. 2) 思维水题
---恢复内容开始--- output standard output The final match of the Berland Football Cup has been held recent ...
- 【Yellow Cards CodeForces - 1215A 】【贪心】
该题难点在于求最小的离开数,最大的没什么好说的,关键是求最小的. 可以这样去想,最小的离开数就是每个人获得的牌数等于他所能接受的最大牌数-1,这样就可以直接比较m=a1(k1-1)+a2(k2-1)与 ...
- Game of Cards Gym - 101128G (SG函数)
Problem G: Game of Cards \[ Time Limit: 1 s \quad Memory Limit: 256 MiB \] 题意 题意就是给出\(n\)堆扑克牌,然后给出一个 ...
- CodeForces 471C MUH and House of Cards
MUH and House of Cards Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & % ...
- Codeforces Round #384 (Div. 2) E. Vladik and cards 状压dp
E. Vladik and cards 题目链接 http://codeforces.com/contest/743/problem/E 题面 Vladik was bored on his way ...
- Gym100947E || codeforces 559c 组合数取模
E - Qwerty78 Trip Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u S ...
- Codeforces Round #281 (Div. 2)
题目链接:http://codeforces.com/contest/493 A. Vasya and Football Vasya has started watching football gam ...
- CodeForces 512B(区间dp)
D - Fox And Jumping Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64 ...
随机推荐
- PAT Advanced 1073 Scientific Notation (20 分)
Scientific notation is the way that scientists easily handle very large numbers or very small number ...
- 使用raw input 代替全局键盘钩子
//关于raw input 请查看msdn https://msdn.microsoft.com/en-us/library/windows/desktop/ms645536%28v=vs.85%29 ...
- linux服务器日志剖析
常规tomcat,apache,nginx,错误日志,还有项目log4j日志 tomcat (以tomcat7.082为例) tomcat日志配置 运行日志和访问日志结合在一起,先说下日志哪边配置,在 ...
- HttpClient获取数据
HttpClient 是Apache Jakarta Common 下的子项目,可以用来提供高效的.最新的.功能丰富的支持 HTTP 协议的客户端编程工具包,并且它支持 HTTP 协议最新的版本和建议 ...
- python_tkinter弹出对话框2
1.fledialog对话框 示例:askopenfilename(选择单个文件,获取文件路径) import tkinter # 导入消息对话框子模块 import tkinter.filedial ...
- 号外号外!WPF界面开发者福音,DevExpress支持.NET Core 3.0!
通过DevExpress WPF Controls,你能创建有着强大互动功能的XAML基础应用程序,这些应用程序专注于当代客户的需求和构建未来新一代支持触摸的解决方案. 无论是Office办公软件的衍 ...
- IDEA jetbrain Live Template
IDEA(jetbrain通用)优雅级使用教程 IDEA 强大的 Live Templates(转) 官网
- 边学边体验django--表格
在模板的末尾,我们增加一个rlt记号,为表格处理结果预留位置. 表格后面还有一个{% csrf_token %}的标签.csrf全称是Cross Site Request Forgery.这是Djan ...
- learning express step(十一)
learning express.Router() code: const express = require('express'); const app = express(); var route ...
- [crontab]修改默认编辑器
crontab默认编辑器为nano,超级不好用 想要修改成vim或者其他编辑器,方法如下. sudo select-editor 改为3或者4 再次打开就直接是vim打开了