Codeforces Round #432 (Div. 2, based on IndiaHacks Final Round 2017) A
Arpa is researching the Mexican wave.
There are n spectators in the stadium, labeled from 1 to n. They start the Mexican wave at time 0.
- At time 1, the first spectator stands.
- At time 2, the second spectator stands.
- ...
- At time k, the k-th spectator stands.
- At time k + 1, the (k + 1)-th spectator stands and the first spectator sits.
- At time k + 2, the (k + 2)-th spectator stands and the second spectator sits.
- ...
- At time n, the n-th spectator stands and the (n - k)-th spectator sits.
- At time n + 1, the (n + 1 - k)-th spectator sits.
- ...
- At time n + k, the n-th spectator sits.
Arpa wants to know how many spectators are standing at time t.
The first line contains three integers n, k, t (1 ≤ n ≤ 109, 1 ≤ k ≤ n, 1 ≤ t < n + k).
Print single integer: how many spectators are standing at time t.
10 5 3
3
10 5 7
5
10 5 12
3
In the following a sitting spectator is represented as -, a standing spectator is represented as ^.
- At t = 0 ----------
number of standing spectators = 0. - At t = 1 ^---------
number of standing spectators = 1. - At t = 2 ^^--------
number of standing spectators = 2. - At t = 3 ^^^-------
number of standing spectators = 3. - At t = 4 ^^^^------
number of standing spectators = 4. - At t = 5 ^^^^^-----
number of standing spectators = 5. - At t = 6 -^^^^^----
number of standing spectators = 5. - At t = 7 --^^^^^---
number of standing spectators = 5. - At t = 8 ---^^^^^--
number of standing spectators = 5. - At t = 9 ----^^^^^-
number of standing spectators = 5. - At t = 10 -----^^^^^
number of standing spectators = 5. - At t = 11 ------^^^^
number of standing spectators = 4. - At t = 12 -------^^^
number of standing spectators = 3. - At t = 13 --------^^
number of standing spectators = 2. - At t = 14 ---------^
number of standing spectators = 1. - At t = 15 ----------
number of standing spectators = 0.
题意:自己读一读就行
解法:暗中观察
#include<bits/stdc++.h>
using namespace std;
int main(){
int n,k,t;
cin>>n>>k>>t;
if(k>=t){
cout<<t<<endl;
}else if(t<=n){
cout<<k<<endl;
}else if(n+k<=t){
cout<<""<<endl;
}else{
cout<<k-(t-n)<<endl;
}
return ;
}
Codeforces Round #432 (Div. 2, based on IndiaHacks Final Round 2017) A的更多相关文章
- D. Arpa and a list of numbers Codeforces Round #432 (Div. 2, based on IndiaHacks Final Round 2017)
http://codeforces.com/contest/851/problem/D 分区间操作 #include <cstdio> #include <cstdlib> # ...
- Codeforces Round #432 (Div. 2, based on IndiaHacks Final Round 2017)ABCD
A. Arpa and a research in Mexican wave time limit per test 1 second memory limit per test 256 megaby ...
- Codeforces Round #432 (Div. 1, based on IndiaHacks Final Round 2017) D. Tournament Construction(dp + 构造)
题意 一个竞赛图的度数集合是由该竞赛图中每个点的出度所构成的集合. 现给定一个 \(m\) 个元素的集合,第 \(i\) 个元素是 \(a_i\) .(此处集合已经去重) 判断其是否是一个竞赛图的度数 ...
- 【前缀和】【枚举倍数】 Codeforces Round #432 (Div. 2, based on IndiaHacks Final Round 2017) D. Arpa and a list of numbers
题意:给你n个数,一次操作可以选一个数delete,代价为x:或者选一个数+1,代价y.你可以进行这两种操作任意次,让你在最小的代价下,使得所有数的GCD不为1(如果全删光也视作合法). 我们从1到m ...
- 【推导】【暴力】Codeforces Round #432 (Div. 2, based on IndiaHacks Final Round 2017) C. Five Dimensional Points
题意:给你五维空间内n个点,问你有多少个点不是坏点. 坏点定义:如果对于某个点A,存在点B,C,使得角BAC为锐角,那么A是坏点. 结论:如果n维空间内已经存在2*n+1个点,那么再往里面添加任意多个 ...
- 【推导】Codeforces Round #432 (Div. 2, based on IndiaHacks Final Round 2017) B. Arpa and an exam about geometry
题意:给你平面上3个不同的点A,B,C,问你能否通过找到一个旋转中心,使得平面绕该点旋转任意角度后,A到原先B的位置,B到原先C的位置. 只要A,B,C构成等腰三角形,且B为上顶点.那么其外接圆圆心即 ...
- Codeforces Round #432 (Div. 2, based on IndiaHacks Final Round 2017) D
Arpa has found a list containing n numbers. He calls a list bad if and only if it is not empty and g ...
- Codeforces Round #432 (Div. 2, based on IndiaHacks Final Round 2017) C
You are given set of n points in 5-dimensional space. The points are labeled from 1 to n. No two poi ...
- Codeforces Round #432 (Div. 2, based on IndiaHacks Final Round 2017) B
Arpa is taking a geometry exam. Here is the last problem of the exam. You are given three points a, ...
随机推荐
- LoadRunner中的函数
函数是LoadRunner提供给性能测试工程师的利器,有了它,性能测试工程师可以对脚本进行更为自由的开发,更适应实际测试的需求,进一步扩展脚本的功能. LoadRunner函数的格式: 返回值 函数 ...
- animation-delay负值
<!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8&quo ...
- 【Lintcode】120.Word Ladder
题目: Given two words (start and end), and a dictionary, find the length of shortest transformation se ...
- bzoj 3232 圈地游戏——0/1分数规划(或网络流)
题目:https://www.lydsy.com/JudgeOnline/problem.php?id=3232 当然是0/1分数规划.但加的东西和减的东西不在一起,怎么办? 考虑把它们合在一起.因为 ...
- 五 python 发送邮件
需求: 系统中使用一个邮箱(这里用QQ)给其他邮箱发消息,用python完成,步骤如下: 1: 开启QQ邮箱的SMTP服务.设置-> 账号 2: 开启邮箱服务:这个需要手机验证,最后会给你一个授 ...
- u-boot.lds 链接脚本分析(hi3515)
目录:/u-boot_hi3515/board/hi3515v100 OUTPUT_FORMAT("elf32-littlearm", "elf32-littlearm& ...
- stm32之入门知识
一.stm32最小系统 stm32最小系统组成如下(除了stm32芯片外): 1.电源模块,3.3V电源,需要用稳压器件,有时要用感容网络产生stm32所使用的模拟电源. 2.时钟模块,有源或者无源晶 ...
- 【249】◀▶IEW-Unit14
Unit 14 Money and Finance 线图写作技巧 1.Model1对应图片分析 The graph contains information about the price in US ...
- MSSQl分布式查询(转)
MSSQlServer所谓的分布式查询(Distributed Query)是能够访问存放在同一部计算机或不同计算机上的SQL Server或不同种类的数据源, 从概念上来说分布式查询与普通查询区别 ...
- [51nod1065]最小正子段和
题意:求一个序列中大于0的最小子段和. 解题关键: 先求出前缀和和,对于每个位置求某个位置到当前位置和大于1的和的最小值.然而这是复杂度是O(n^2)的.其实可以通过排序优化到O(nlogn).对前缀 ...