Codeforces Round #562 (Div. 2) A.Circle Metro
链接:https://codeforces.com/contest/1169/problem/A
题意:
The circle line of the Roflanpolis subway has nn stations.
There are two parallel routes in the subway. The first one visits stations in order 1→2→…→n→1→2→…1→2→…→n→1→2→… (so the next stop after station xx is equal to (x+1)(x+1) if x<nx<n and 11 otherwise). The second route visits stations in order n→(n−1)→…→1→n→(n−1)→…n→(n−1)→…→1→n→(n−1)→… (so the next stop after station xx is equal to (x−1)(x−1) if x>1x>1 and nn otherwise). All trains depart their stations simultaneously, and it takes exactly 11 minute to arrive at the next station.
Two toads live in this city, their names are Daniel and Vlad.
Daniel is currently in a train of the first route at station aa and will exit the subway when his train reaches station xx.
Coincidentally, Vlad is currently in a train of the second route at station bb and he will exit the subway when his train reaches station yy.
Surprisingly, all numbers a,x,b,ya,x,b,y are distinct.
Toad Ilya asks you to check if Daniel and Vlad will ever be at the same station at the same time during their journey. In other words, check if there is a moment when their trains stop at the same station. Note that this includes the moments when Daniel or Vlad enter or leave the subway.
思路:
暴力
代码:
#include <bits/stdc++.h> using namespace std; typedef long long LL;
const int MAXN = 1e5 + 10;
const int MOD = 1e9 + 7;
int n, m, k, t;
int p, q, u, v;
int x, y, z, w;
int a, b; int main()
{
cin >> n >> a >> x >> b >> y;
bool flag = false;
int len = 2*n;
while(len--)
{
a++;
if (a > n)
a = 1;
b--;
if (b < 1)
b = n;
if (a == b)
flag = true;
if (a == x || b == y)
break;
}
if (flag)
cout << "YES" << endl;
else
cout << "NO" << endl; return 0;
}
Codeforces Round #562 (Div. 2) A.Circle Metro的更多相关文章
- [Done] Codeforces Round #562 (Div. 2) 题解
A - Circle Metro 模拟几百步就可以了. B - Pairs 爆搜一下,时间复杂度大概是 $O(4 * n)$ Code: 56306723 C - Increasing by Modu ...
- Codeforces Round #562 (Div. 2) C. Increasing by Modulo
链接:https://codeforces.com/contest/1169/problem/C 题意: Toad Zitz has an array of integers, each intege ...
- Codeforces Round #562 (Div. 2) B. Pairs
链接:https://codeforces.com/contest/1169/problem/B 题意: Toad Ivan has mm pairs of integers, each intege ...
- Codeforces Round #418 (Div. 2) D. An overnight dance in discotheque
Codeforces Round #418 (Div. 2) D. An overnight dance in discotheque 题意: 给\(n(n <= 1000)\)个圆,圆与圆之间 ...
- Codeforces Round #579 (Div. 3)
Codeforces Round #579 (Div. 3) 传送门 A. Circle of Students 这题我是直接把正序.逆序的两种放在数组里面直接判断. Code #include &l ...
- Codeforces Round #564 (Div. 1)
Codeforces Round #564 (Div. 1) A Nauuo and Cards 首先如果牌库中最后的牌是\(1,2,\cdots, k\),那么就模拟一下能不能每次打出第\(k+i\ ...
- Codeforces Round #366 (Div. 2) ABC
Codeforces Round #366 (Div. 2) A I hate that I love that I hate it水题 #I hate that I love that I hate ...
- Codeforces Round #354 (Div. 2) ABCD
Codeforces Round #354 (Div. 2) Problems # Name A Nicholas and Permutation standard input/out ...
- Codeforces Round #368 (Div. 2)
直达–>Codeforces Round #368 (Div. 2) A Brain’s Photos 给你一个NxM的矩阵,一个字母代表一种颜色,如果有”C”,”M”,”Y”三种中任意一种就输 ...
随机推荐
- 时尚创意VI矢量设计模板
时尚创意VI矢量设计模板 创意VI VI设计 企业VI 时尚背景 信封设计 封面设计 杯子 桌旗 帽子 EPS矢量素材下载 http://www.huiyi8.com/vi/
- html5--2.1新的布局元素(1)-header/footer
html5--2.1新的布局元素(1)-header/footer 学习要点 了解header/footer的语义和用法 使用header/footer进行一个简单的布局 header元素(标签) 用 ...
- UML中的6大关系详细说明
UML中的6大关系详细说明: 1.关联关系: 含义:类与类之间的连结,关联关系使一个类知道另外一个类的属性和方法:通常含有“知道”,“了解”的含义 体现:在C#中,关联关系是通过成员变量来实现的: 方 ...
- leetcode 191 Number of 1 Bits(位运算)
Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also know ...
- [acm]HDOJ 3082 Simplify The Circuit
题目地址: http://acm.hdu.edu.cn/showproblem.php?pid=3082 字符串处理+并联电阻公式 //11481261 2014-08-18 16:52:47 Acc ...
- inteliji ---idea 如何创建scala程序配置
首先创建一个新的工程: #####################################################################################
- 数据结构与算法(5)----->二叉树
1. 概念 二叉树节点的结构: class Node{ int value; // value表示二叉树的节点值 Node left; Node right; // left和right表示二叉树的 ...
- javaCV开发详解之8:转封装在rtsp转rtmp流中的应用(无须转码,更低的资源消耗)
javaCV系列文章: javacv开发详解之1:调用本机摄像头视频 javaCV开发详解之2:推流器实现,推本地摄像头视频到流媒体服务器以及摄像头录制视频功能实现(基于javaCV-FFMPEG.j ...
- WPF ValidationRule 触发ErrorTemplate 的注意事项
ValidationRule 验证时, 当验证失败后,再次验证成功, errorTemplate 还是触发, 不会被清掉. 因此需要主动调用 Validation.ClearInvalid(dtpTe ...
- 数论 最简分数 Farey序列求最简分数+POJ3374
法雷数列 定义和定理 定义一: 最简分数(也称既约分数或不可约分数).若p,q的最大公约数是1,我们称分数p/q是最简分数. 定义二: 真分数,若p,q是正整数,0<p/q<1, 我们说p ...