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

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 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 ...

  7. 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/ ...

  8. 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 ...

  9. 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

  1. Word Ladder I & II

    Word Ladder I Given two words (start and end), and a dictionary, find the length of shortest transfo ...

  2. linux文件管理 -> vim编辑总结

    vi和vim命令是linux中强大的文本编辑器, 由于Linux系统一切皆文件,而配置一个服务就是在修改其配置文件的参数.vim编辑器是运维工程师必须掌握的一个工具, 没有它很多工作都无法完成.vim ...

  3. ios 个人开发者账户 给其他团队用坑爹的教程

    最新版本的 ios  支持 3个开发者证书 和 3个发布者证书  ,如果是多余3台电脑设备要真机调试,就比较麻烦 (手机支持100个设备) 解决方案就是: 在别人的电脑上要成功安装,须具备两个文件: ...

  4. Webmin试玩

    安装: # cd /opt # wget http://www.webmin.com/jcameron-key.asc # wget http://www.webmin.com/download/rp ...

  5. Mysql5.6版本内存占用过高解决方法[链接]

    传送门: http://blog.linsongzheng.com/archives/159.html

  6. MYSQL-重做系统恢复MYSQL过程

    记笔记是好习惯,记笔记是好习惯,记笔记是好习惯! 重要的事情说三遍. 说多了都是泪.第一次装MYSQL时候就遇到了很多问题,当时解决了忘记记录了.家里硬盘满了,于是买了个4T的硬盘重装系统.重装系统后 ...

  7. python面向对象(二)之封装

    封装定义: 在程序设计中,封装(Encapsulation)是对具体对象的一种抽象,即将某些部分隐藏起来,在程序外部看不到,其含义是其他程序无法调用. 即"封装"就是将抽象得到的数 ...

  8. 回归模型效果评估系列3-R平方

    决定系数(coefficient of determination,R2)是反映模型拟合优度的重要的统计量,为回归平方和与总平方和之比.R2取值在0到1之间,且无单位,其数值大小反映了回归贡献的相对程 ...

  9. DevExpress CxGrid 隐藏 Drag a column header to group by that column

  10. SQL SERVER中查询某个表或某个索引是否存在

    查询某个表是否存在: 在实际应用中可能需要删除某个表,在删除之前最好先判断一下此表是否存在,以防止返回错误信息.在SQL SERVER中可通过以下语句实现: IF OBJECT_ID(N'表名称', ...