Checking the Calendar
1 second
256 megabytes
standard input
standard output
You are given names of two days of the week.
Please, determine whether it is possible that during some non-leap year the first day of some month was equal to the first day of the week you are given, while the first day of the next month was equal to the second day of the week you are given. Both months should belong to one year.
In this problem, we consider the Gregorian calendar to be used. The number of months in this calendar is equal to 12. The number of days in months during any non-leap year is: 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31.
Names of the days of the week are given with lowercase English letters: "monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday".
The input consists of two lines, each of them containing the name of exactly one day of the week. It's guaranteed that each string in the input is from the set "monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday".
Print "YES" (without quotes) if such situation is possible during some non-leap year. Otherwise, print "NO" (without quotes).
monday
tuesday
NO
sunday
sunday
YES
saturday
tuesday
YES
In the second sample, one can consider February 1 and March 1 of year 2015. Both these days were Sundays.
In the third sample, one can consider July 1 and August 1 of year 2017. First of these two days is Saturday, while the second one is Tuesday.
分析:取差之后看能否满足即可;
代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, ls[rt]
#define Rson mid+1, R, rs[rt]
const int maxn=1e5+;
using namespace std;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}
inline ll read()
{
ll x=;int f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
int n,m,k,t,c[]={, , , , , , , , , , , };
char a[maxn],b[maxn];
map<string,int>d;
int main()
{
int i,j;
d["monday"]=;
d["tuesday"]=;
d["wednesday"]=;
d["thursday"]=;
d["friday"]=;
d["saturday"]=;
d["sunday"]=;
scanf("%s%s",a,b);
int co=(d[b]-d[a]+)%;
rep(i,,)
{
if(c[i]%==co)return *puts("YES");
}
puts("NO");
//system("Pause");
return ;
}
Checking the Calendar的更多相关文章
- Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined) A. Checking the Calendar 水题
A. Checking the Calendar 题目连接: http://codeforces.com/contest/724/problem/A Description You are given ...
- Codeforces Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined) A. Checking the Calendar(水题)
传送门 Description You are given names of two days of the week. Please, determine whether it is possibl ...
- 【63.63%】【codeforces 724A】Checking the Calendar
time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...
- codeforces724-A. Checking the Calendar 日期题
首先有这样一个显然的事实,那就是每个月的第一天可以是星期x,x可以取遍1~7 因为日期一直在往后退,总有一年能轮到分割线那天,因为本来其实压根就没有月份的划分,月份划分是人为的 而且我们也不知道开始的 ...
- Java中的Calendar日历用法详解
第一部分 Calendar介绍 public abstract class Calendar implements Serializable, Cloneable, Comparable<Cal ...
- Java Calendar,Date,DateFormat,TimeZone,Locale等时间相关内容的认知和使用(1) Calendar
Java 操作日期/时间,往往会涉及到Calendar,Date,DateFormat这些类. 最近决定把这些内容系统的整理一下,这样以后使用的时候,会更得心应手.本章的内容是主要讲解“Java时间框 ...
- codeforces 724
题目链接: http://codeforces.com/contest/724 A. Checking the Calendar time limit per test 1 second memory ...
- PMP模拟考试-1
1. A manufacturing project has a schedule performance index (SPI) of 0.89 and a cost performance ind ...
- Codeforces Gym101522 D.Distribution of Days-算日期 (La Salle-Pui Ching Programming Challenge 培正喇沙編程挑戰賽 2017)
D.Distribution of Days The Gregorian calendar is internationally the most widely used civil calendar ...
随机推荐
- discuz使用总结
使用xampp作为运行环境 xampp的初始目录. xampp中mysql root账户的密码是空
- JS学习之路,之弹性运动框架
弹性运动:顾名思义,就如同物理中的加速减速运动,当开始时速度过大,到达终点时,速度不会立刻停下,而是再前进一段距离,而后再向相反方向运动,如此往复. var timer=null; var speed ...
- iOS相关教程
Xcode Xcode 7中你一定要知道的炸裂调试神技 Xcode 6和Swift中应用程序的国际化和本地化 iOS新版本 兼容iOS 10 资料整理笔记 整理iOS9适配中出现的坑(图文) Swif ...
- linux上安装mono发布.net网站步骤
在linux上部署mono 1.自己安装好linux 2.使用桥接方式,让虚拟机和本机在一个局域网内 3.安装apache服务器 4.安装libgdiplug 5.安装mono 6.安装xsp 7.安 ...
- 仿bmfn 底部
<!DOCTYPE html> <!-- saved from url=(0019)http://www.bmfn.cn/ --> <html class="k ...
- Events and Responder Chain
事件类型(Event Type) iOS 有三种事件类型: 触控事件(UIEventTypeTouches):单点.多点触控以及各种手势操作: 传感器事件(UIEventTypeMotion):重力. ...
- ios 字符串的操作汇总
//将NSData转化为NSString NSString* str = [[NSString alloc] initWithData:response encoding:NSUTF8S ...
- 10个男孩和n个女孩共买了n2+8n+2本书,已知他们每人买的书本的数量是相同的,且女孩人数多于南海人数,问女孩人数是多少?(整除原理1.1.3)
10个男孩和n个女孩共买了n2+8n+2本书,已知他们每人买的书本的数量是相同的,且女孩人数多于南海人数,问女孩人数是多少? 解: 因为,每个人买的书本的数量是相同的, 所以,10|n2+8n+2 所 ...
- ADO.NET基础、数据增删改查
ADO.NET:数据访问技术,就是将C#和MSSQL连接起来的一个纽带.我们可以通过ADO.NET将内存中的临时数据写入到数据库中,也可以将数据库中的数据提取到内存中供程序调用. 数据库数据的增.删. ...
- Centos7 设置DNS 服务器
在CentOS 7下,手工设置 /etc/resolv.conf 里的DNS,过了一会,发现被系统重新覆盖或者清除了.和CentOS 6下的设置DNS方法不同,有几种方式: 1.使用全新的命令行工具 ...