Codeforces Round #362 (Div. 2) A 水也挂
1 second
256 megabytes
standard input
standard output
Ted has a pineapple. This pineapple is able to bark like a bulldog! At time t (in seconds) it barks for the first time. Then every s seconds after it, it barks twice with 1 second interval. Thus it barks at times t, t + s, t + s + 1, t + 2s, t + 2s + 1, etc.

Barney woke up in the morning and wants to eat the pineapple, but he can't eat it when it's barking. Barney plans to eat it at time x (in seconds), so he asked you to tell him if it's gonna bark at that time.
The first and only line of input contains three integers t, s and x (0 ≤ t, x ≤ 109, 2 ≤ s ≤ 109) — the time the pineapple barks for the first time, the pineapple barking interval, and the time Barney wants to eat the pineapple respectively.
Print a single "YES" (without quotes) if the pineapple will bark at time x or a single "NO" (without quotes) otherwise in the only line of output.
3 10 4
NO
3 10 3
YES
3 8 51
YES
3 8 52
YES
In the first and the second sample cases pineapple will bark at moments 3, 13, 14, ..., so it won't bark at the moment 4 and will bark at the moment 3.
In the third and fourth sample cases pineapple will bark at moments 3, 11, 12, 19, 20, 27, 28, 35, 36, 43, 44, 51, 52, 59, ..., so it will bark at both moments 51 and 52.
题意: t, t + s, t + s + 1, t + 2s, t + 2s + 1, 给你t,s ,x 判断x是否为序列中的值
题解: 水题也挂终测 靠hack 上分
#include<bits/stdc++.h>
#define ll __int64
#define mod 1e9+7
#define PI acos(-1.0)
#define bug(x) printf("%%%%%%%%%%%%%",x);
using namespace std;
int t,s,x;
int main()
{
scanf("%d %d %d",&t,&s,&x);
if(t==x)
{
cout<<"YES"<<endl;
return ;
}
if(x<t||x<t+s)
{
cout<<"NO"<<endl;
return ;
}
x=x-t;
if(x%s==||x%s==)
{
cout<<"YES"<<endl;
return ;
}
cout<<"NO"<<endl;
return ;
}
Codeforces Round #362 (Div. 2) A 水也挂的更多相关文章
- Codeforces Round #365 (Div. 2) A 水
A. Mishka and Game time limit per test 1 second memory limit per test 256 megabytes input standard i ...
- Codeforces Round #362 (Div. 2) C. Lorenzo Von Matterhorn (类似LCA)
题目链接:http://codeforces.com/problemset/problem/697/D 给你一个有规则的二叉树,大概有1e18个点. 有两种操作:1操作是将u到v上的路径加上w,2操作 ...
- Codeforces Round #404 (Div. 2)(A.水,暴力,B,排序,贪心)
A. Anton and Polyhedrons time limit per test:2 seconds memory limit per test:256 megabytes input:sta ...
- Codeforces Round #408 (Div. 2)(A.水,B,模拟)
A. Buying A House time limit per test:2 seconds memory limit per test:256 megabytes input:standard i ...
- #map+LCA# Codeforces Round #362 (Div. 2)-C. Lorenzo Von Matterhorn
2018-03-16 http://codeforces.com/problemset/problem/697/C C. Lorenzo Von Matterhorn time limit per t ...
- Codeforces Round #394 (Div. 2)A水 B暴力 C暴力 D二分 E dfs
A. Dasha and Stairs time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- Codeforces Round #169 (Div. 2) A水 B C区间更新 D 思路
A. Lunch Rush time limit per test 2 seconds memory limit per test 256 megabytes input standard input ...
- Codeforces Round #337 (Div. 2) A水
A. Pasha and Stick time limit per test 1 second memory limit per test 256 megabytes input standard i ...
- Codeforces Round #316 (Div. 2) A 水
A. Elections time limit per test 1 second memory limit per test 256 megabytes input standard input o ...
随机推荐
- Maven 虐我千百遍,我待 Maven 如初恋
前言 在如今的互联网项目开发当中,特别是Java领域,可以说Maven随处可见.Maven的仓库管理.依赖管理.继承和聚合等特性为项目的构建提供了一整套完善的解决方案,可以说如果你搞不懂Maven,那 ...
- vue中created、mounted等方法整理
- java基础编程—统计二进制数中1的个数
题目描述 输入一个整数,输出该数二进制表示中1的个数.其中负数用补码表示. 题目代码 /** * 输入一个整数,输出该数二进制表示中1的个数.其中负数用补码表示. * Created by YuKai ...
- 解决Mycat对自增表不支持(第一种已测试通过)
表 INSERT INTO news_class (`class_id`,`class_name`) VALUES (next VALUE FOR MYCATSEQ_GLOBAL,'1'); sequ ...
- redhat linux6.5升级openssh
1.下载最新的openssh包 http://www.openssh.com/portable.html#http 2.升级openssh之前要先打开服务器telnet,通过telnet登录服务器,因 ...
- ActiveMQ RabbitMQ RokcetMQ Kafka实战 消息队列中间件视频教程
附上消息队列中间件百度网盘连接: 链接: https://pan.baidu.com/s/1FFZQ5w17e1TlLDSF7yhzmA 密码: hr63
- 数据结构-单链表(Linked List)
#include <stdio.h> #include <stdlib.h> #define LIST_INIT_SIZE 10 #define LISTINCREMENT 1 ...
- spark和MR比较
MapReduce: 分布式的计算框架 缺点:执行速度慢 IO瓶颈 ==> 磁盘IO 网络IO shuffle机制:数据需要输出到磁盘,而且每次shuffle都需要进行排序操作 框架的机制: 只 ...
- Python Map, Filter and Reduce
所属网站分类: python基础 > 函数 作者:慧雅 原文链接: http://www.pythonheidong.com/blog/article/21/ 来源:python黑洞网 www. ...
- hibernate的get() load() 和find()区别
如果找不到符合条件的纪录,get()方法将返回null.如果找不到符合条件的纪录,find()方法将返回null.如果找不到符合 条件的纪录,load()将会报出ObjectNotFoundEccep ...