B. Months and Years
 
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Everybody in Russia uses Gregorian calendar. In this calendar there are 31 days in January, 28 or 29 days in February (depending on whether the year is leap or not), 31 days in March, 30 days in April, 31 days in May, 30 in June, 31 in July, 31 in August, 30 in September, 31 in October, 30 in November, 31 in December.

A year is leap in one of two cases: either its number is divisible by 4, but not divisible by 100, or is divisible by 400. For example, the following years are leap: 2000, 2004, but years 1900 and 2018 are not leap.

In this problem you are given n (1 ≤ n ≤ 24) integers a1, a2, ..., an, and you have to check if these integers could be durations in days of nconsecutive months, according to Gregorian calendar. Note that these months could belong to several consecutive years. In other words, check if there is a month in some year, such that its duration is a1 days, duration of the next month is a2 days, and so on.

Input

The first line contains single integer n (1 ≤ n ≤ 24) — the number of integers.

The second line contains n integers a1, a2, ..., an (28 ≤ ai ≤ 31) — the numbers you are to check.

Output

If there are several consecutive months that fit the sequence, print "YES" (without quotes). Otherwise, print "NO" (without quotes).

You can print each letter in arbitrary case (small or large).

Examples
input
4
31 31 30 31
output
Yes
input
2
30 30
output
No
input
5
29 31 30 31 30
output
Yes
input
3
31 28 30
output
No
input
3
31 31 28
output
Yes
Note

In the first example the integers can denote months July, August, September and October.

In the second example the answer is no, because there are no two consecutive months each having 30 days.

In the third example the months are: February (leap year) — March — April – May — June.

In the fourth example the number of days in the second month is 28, so this is February. March follows February and has 31 days, but not 30, so the answer is NO.

In the fifth example the months are: December — January — February (non-leap year).

代码:

 1 #include<iostream>
2 #include<cstdio>
3 #include<cmath>
4 using namespace std;
5 const int N=2*1e5+10;
6 int a[N];
7 int month[300]={31,28,31,30,31,30,31,31,30,31,30,31,31,28,31,30,31,30,31,31,30,31,30,31,31,28,31,30,31,30,31,31,30,31,30,31,31,29,31,30,31,30,31,31,30,
8 31,30,31,31,28,31,30,31,30,31,31,30,31,30,31,31,28,31,30,31,30,31,31,30,31,30,31,31,28,31,30,31,30,31,31,30,31,30,31,};
9 int ans[N];
10 int main(){
11 int n;
12 cin>>n;
13 for(int i=1;i<=n;i++){
14 scanf("%d",&a[i]);
15 }
16 for(int i=0;i<=150;i++){
17 if(month[i]==a[1]){
18 int flag1=0;
19 for(int j=1;j<=n;j++){
20 if(month[i+j-1]!=a[j]){
21 flag1=1;
22 break;
23 }
24 }
25 if(flag1==0){
26 cout<<"YES"<<endl;
27 return 0;
28 }
29 }
30 }
31 cout<<"NO"<<endl;
32 }

Codeforces 899 B.Months and Years的更多相关文章

  1. Codeforces 899 F. Letters Removing (二分、树状数组)

    题目链接:Letters Removing 题意: 给你一个长度为n的字符串,给出m次操作.每次操作给出一个l,r和一个字符c,要求删除字符串l到r之间所有的c. 题解: 看样例可以看出,这题最大的难 ...

  2. Codeforces 899 C.Dividing the numbers-规律

      C. Dividing the numbers   time limit per test 1 second memory limit per test 256 megabytes input s ...

  3. Codeforces 899 A.Splitting in Teams

      A. Splitting in Teams   time limit per test 1 second memory limit per test 256 megabytes input sta ...

  4. Codeforces 899 1-N两非空集合最小差 末尾最多9对数计算 pair/链表加优先队列最少次数清空

    A /*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) #define pb push_bac ...

  5. Codeforces Round #452 (Div. 2)-899A.Splitting in Teams 899B.Months and Years 899C.Dividing the numbers(规律题)

    A. Splitting in Teams time limit per test 1 second memory limit per test 256 megabytes input standar ...

  6. 【CodeForces】899 F. Letters Removing

    [题目]F. Letters Removing [题意]给定只含小写字母.大写字母和数字的字符串,每次给定一个范围要求删除[l,r]内的字符c(l和r具体位置随删除变动),求m次操作后的字符串.n&l ...

  7. 【CodeForces】899 E. Segments Removal

    [题目]E. Segments Removal [题意]给定n个数字,每次操作删除最长的连续相同数字(等长删最左),求全部删完的最少次数.n<=2*10^6,1<=ai<=10^9. ...

  8. Codeforces 899B Months and Years

    题目大意 给定 $n$($1\le n\le 24$)个正整数 $a_1,\dots, a_n$ 判断 $a_1$ 到 $a_n$ 是否可能为连续 $n$ 个月份的天数. 解法 由于 $n\le 24 ...

  9. 【Codeforces Round #452 (Div. 2) B】Months and Years

    [链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 闰,平,平 平,闰,平 平,平,闰 平,平,平 4种情况都考虑到就好. 可能有重复的情况. 但是没关系啦. [代码] #includ ...

随机推荐

  1. POJ1426-Find The Multiple(搜索)

    Find The Multiple Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 42035   Accepted: 176 ...

  2. SOA:面向服务编程——竹子整理

    .net中如webservice,wcf,webapi,均可作为服务层,单独部署,而界面UI则部署在另一台服务器上,所有的业务逻辑均在服务层的业务层中进行. 这样一来,我们的UI其实就可以不限制语言, ...

  3. JS 对于回调函数的理解,和常见的使用场景应用,使用注意点

      很经常我们会遇到这样一种情况: 例如,你需要和其他人合作,别人提供数据,而你不需要关注别人获取或者构建数据的方式方法. 你只要对这个拿到的数据进行操作. 这样,就相当于我们提供一个外在的函数,别人 ...

  4. Hyper-V:利用差异磁盘安装多个Win2008

    签于成本的原因,在学习了解一项新的技术或是产品时,在没有部署到生产环境之中前,大家都会选择在虚拟机来搭建一套实验环境.但如何快速搭建呢?如何节省磁盘空间呢? 说到此都不得不说下Hyper-V的差异磁盘 ...

  5. 成为Java高手的25个学习要点

    成为Java高手的25个学习要点 想成为Java大牛吗?不妨来学习这25个要点. 1. 你需要精通面向对象分析与设计(OOA/OOD).涉及模式(GOF,J2EEDP)以及综合模式.你应该了解UML, ...

  6. leetcode 【 Linked List Cycle II 】 python 实现

    公司和学校事情比较多,隔了好几天没刷题,今天继续刷起来. 题目: Given a linked list, return the node where the cycle begins. If the ...

  7. 【POI 2010】反对称 Antisymmetry

    题目: 对于一个 $0/1$ 字符串,如果将这个字符串 $0$ 和 $1$ 取反后,再将整个串反过来和原串一样,就称作「反对称」字符串.比如 $00001111$ 和 $010101$ 就是反对称的, ...

  8. puppet实战之master-agent

    author:JevonWei 版权声明:原创作品 blog:http://119.23.52.191/ --- master作为puppet模块的管理者,通过配置各agent节点的配置文件,使age ...

  9. Hadoop入门第三篇-MapReduce试手以及MR工作机制

    MapReduce几个小应用 上篇文章已经介绍了怎么去写一个简单的MR并且将其跑起来,学习一个东西动手还是很有必要的,接下来我们就举几个小demo来体验一下跑起来的快感. demo链接请参照附件:ht ...

  10. Java Socket实战之三 传输对象

    首先需要一个普通的对象类,由于需要序列化这个对象以便在网络上传输,所以实现java.io.Serializable接口就是必不可少的了,入下: public class User implements ...