Codeforces Round #353 (Div. 2) A. Infinite Sequence 水题
A. Infinite Sequence
题目连接:
http://www.codeforces.com/contest/675/problem/A
Description
Vasya likes everything infinite. Now he is studying the properties of a sequence s, such that its first element is equal to a (s1 = a), and the difference between any two neighbouring elements is equal to c (si - si - 1 = c). In particular, Vasya wonders if his favourite integer b appears in this sequence, that is, there exists a positive integer i, such that si = b. Of course, you are the person he asks for a help.
Input
The first line of the input contain three integers a, b and c ( - 109 ≤ a, b, c ≤ 109) — the first element of the sequence, Vasya's favorite number and the difference between any two neighbouring elements of the sequence, respectively.
Output
If b appears in the sequence s print "YES" (without quotes), otherwise print "NO" (without quotes).
Sample Input
1 7 3
Sample Output
YES
Hint
题意
A0=a,Ai=Ai-1+b,问你c可不可能出现在这个无限长的数列里面
题解:
水题,用数学方法去做就好了
代码
#include<bits/stdc++.h>
using namespace std;
int main()
{
long long a,b,c;
cin>>a>>b>>c;
b-=a;
if(c==0&&b!=0)return puts("NO"),0;
if(c==0&&b==0)return puts("YES"),0;
if(b%c!=0)return puts("NO"),0;
else if(b/c<0)return puts("NO"),0;
return puts("YES"),0;
}
Codeforces Round #353 (Div. 2) A. Infinite Sequence 水题的更多相关文章
- Codeforces Round #353 (Div. 2) A. Infinite Sequence
Vasya likes everything infinite. Now he is studying the properties of a sequence s, such that its fi ...
- Codeforces Round #353 (Div. 2) B. Restoring Painting 水题
B. Restoring Painting 题目连接: http://www.codeforces.com/contest/675/problem/B Description Vasya works ...
- Codeforces Round #367 (Div. 2) A. Beru-taxi (水题)
Beru-taxi 题目链接: http://codeforces.com/contest/706/problem/A Description Vasiliy lives at point (a, b ...
- Codeforces Round #334 (Div. 2) A. Uncowed Forces 水题
A. Uncowed Forces Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/604/pro ...
- Codeforces Round #353 (Div. 2) C. Money Transfers (思维题)
题目链接:http://codeforces.com/contest/675/problem/C 给你n个bank,1~n形成一个环,每个bank有一个值,但是保证所有值的和为0.有一个操作是每个相邻 ...
- Codeforces Round #335 (Div. 2) B. Testing Robots 水题
B. Testing Robots Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://www.codeforces.com/contest/606 ...
- Codeforces Round #188 (Div. 2) A. Even Odds 水题
A. Even Odds Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/318/problem/ ...
- Codeforces Round #327 (Div. 2) A. Wizards' Duel 水题
A. Wizards' Duel Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/591/prob ...
- Codeforces Round #Pi (Div. 2) A. Lineland Mail 水题
A. Lineland MailTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/567/probl ...
随机推荐
- 【驱动】USB驱动·入门【转】
转自:http://www.cnblogs.com/lcw/p/3159371.html Preface USB是目前最流行的系统总线之一.随着计算机周围硬件的不断扩展,各种设备使用不同的总线接口,导 ...
- Add custom daemon on Linux System
Ubuntu add custom service(daemon) Task 需要在系统启动的时候自动启动一个服务(后台程序),在系统关闭的时候关闭服务. 比如在部署某个应用之前,需要将某个任务设置成 ...
- js点击页面其他地方如何隐藏div元素菜单
web页面常用的一个需求,写下拉菜单是我们往往不是用select_option,而是自定义一个元素列出选项来满足需求,当我们点击按钮出现菜单, 点击按钮或菜单以外页面空白地方隐藏该菜单,这里提供一种简 ...
- 用monit监控系统关键进程
原地址: https://feilong.me/2011/02/monitor-core-processes-with-monit monit是一款功能强大的系统状态.进程.文件.目录和设备的监控软件 ...
- 数据库-mysql安装
MySQL 安装 所有平台的Mysql下载地址为: MySQL 下载. 挑选你需要的 MySQL Community Server 版本及对应的平台. Linux/UNIX上安装Mysql Linux ...
- webpack3学习笔记
地址:https://segmentfault.com/a/1190000006843916 地址:https://www.chungold.com/my/course/32 地址:http://js ...
- day06作业
一.方法 1.方法是完成特定功能的代码块. 修饰符 返回值类型 方法类型(参数类型 参数名1,参数类型 参数名2,...){ 方法体语句: return返回值: } 修饰符:目前就用publi ...
- 蛮力法解决0_1背包问题新思路-——利用C语言位域类型
废话不说了,直接上代码 #include<stdio.h> #include<math.h> #define N 5 //物品种类数目 #define CAPACITY 6 / ...
- java 多线程总结篇3之——生命周期和线程同步
一.生命周期 线程的生命周期全在一张图中,理解此图是基本: 线程状态图 一.新建和就绪状态 当程序使用new关键字创建了一个线程之后,该线程就处于新建状态,此时它和其他的Java对象一样,仅仅由Jav ...
- 【51nod】1531 树上的博弈
题解 我们发现每次决策的时候,我们可以判断某个点的决策,至少小于等于几个点或者至少大于等于几个点 我们求最大值 dp[u][1 / 0] dp[u][1]表示u这个点先手,至少大于等于几个点 dp[u ...