Codeforces 719A 月亮
Every summer Vitya comes to visit his grandmother in the countryside. This summer, he got a huge wart. Every grandma knows that one should treat warts when the moon goes down. Thus, Vitya has to catch the moment when the moon is down.
Moon cycle lasts 30 days. The size of the visible part of the moon (in Vitya's units) for each day is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, and then cycle repeats, thus after the second 1 again goes 0.
As there is no internet in the countryside, Vitya has been watching the moon for n consecutive days and for each of these days he wrote down the size of the visible part of the moon. Help him find out whether the moon will be up or down next day, or this cannot be determined by the data he has.
The first line of the input contains a single integer n (1 ≤ n ≤ 92) — the number of consecutive days Vitya was watching the size of the visible part of the moon.
The second line contains n integers ai (0 ≤ ai ≤ 15) — Vitya's records.
It's guaranteed that the input data is consistent.
If Vitya can be sure that the size of visible part of the moon on day n + 1 will be less than the size of the visible part on day n, then print "DOWN" at the only line of the output. If he might be sure that the size of the visible part will increase, then print "UP". If it's impossible to determine what exactly will happen with the moon, print -1.
5
3 4 5 6 7
UP
7
12 13 14 15 14 13 12
DOWN
1
8
-1
In the first sample, the size of the moon on the next day will be equal to 8, thus the answer is "UP".
In the second sample, the size of the moon on the next day will be 11, thus the answer is "DOWN".
In the third sample, there is no way to determine whether the size of the moon on the next day will be 7 or 9, thus the answer is -1.
解法:
#include<stdio.h>
int main(){
int n,i;
int a[];
while(scanf("%d",&n)){
for(i=;i<n;i++)
scanf("%d",&a[i]);
if(n==&&a[n-]!=&&a[n-]!=)
printf("-1\n");
else if(a[n-]==) printf("DOWN\n");
else if(a[n-]==) printf("UP\n");
else if(a[n-]<&&a[n-]<a[n-])
printf("UP\n");
else if(a[n-]<&&a[n-]>a[n-])
printf("DOWN\n");
}
return ;
}
Codeforces 719A 月亮的更多相关文章
- codeforces 719A Vitya in the Countryside(序列判断趋势)
题目链接:http://codeforces.com/problemset/problem/719/A 题目大意: 题目给出了一个序列趋势 0 .1 .2 .3 ---14 .15 .14 ----3 ...
- CodeForces 719A Vitya in the Countryside 思维题
题目大意:月亮从0到15,15下面是0.循环往复.给出n个数字,如果下一个数字大于第n个数字输出UP,小于输出DOWN,无法确定输出-1. 题目思路:给出0则一定是UP,给出15一定是DOWN,给出其 ...
- CodeForces 719A. Vitya in the Countryside
链接:[http://codeforces.com/group/1EzrFFyOc0/contest/719/problem/A] 题意: 给你一个数列(0, 1, 2, 3, 4, 5, 6, 7, ...
- codeforces 719A:Vitya in the Countryside
Description Every summer Vitya comes to visit his grandmother in the countryside. This summer, he go ...
- CodeForces 719A Vitya in the Countryside (水题)
题意:根据题目,给定一些数字,让你判断是上升还是下降. 析:注意只有0,15时特别注意一下,然后就是14 15 1 0注意一下就可以了. 代码如下: #pragma comment(linker, & ...
- 【32.89%】【codeforces 719A】Vitya in the Countryside
time limit per test 1 second memory limit per test 256 megabytes input standard input output standar ...
- Codeforces水题集合[14/未完待续]
Codeforces Round #371 (Div. 2) A. Meeting of Old Friends |B. Filya and Homework A. Meeting of Old Fr ...
- Codeforces Round #373 (Div. 2)A B
Codeforces Round #373 (Div. 2) A. Vitya in the Countryside 这回做的好差啊,a想不到被hack的数据,b又没有想到正确的思维 = = [题目链 ...
- Codeforces 989 P循环节01构造 ABCD连通块构造 思维对云遮月参考系坐标轴转换
A 直接判存不存在连续的三个包含A,B,C就行 /*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a ...
随机推荐
- Elastic 今日在纽交所上市,股价最高暴涨122%。
10 月 6 日,Elastic 正式在纽约证券交易所上市,股票代码为"ESTC".开盘之后股价直线拉升,最高点涨幅达122%,截止到收盘涨幅回落到94%,意味着上市第一天估值接近 ...
- 四、xadmin自定义插件1
插件原理: Xadmin中每个页面都是一个AdminView对象返回的HTTPResponse结果. Xdamin插件所做的事情就是其实就是在AdminView执行过程中改变其执行逻辑或是改变其返回的 ...
- 用Flask+Redis维护Cookies池
Redis数据库:存储微博账号密码 这里需要购买账号 登录后的cookies:键值对的形式保存 GitHub:https://github.com/LXL-YAN/CookiesPool 视频讲解:h ...
- Factors of Factorial AtCoder - 2286 (N的阶乘的因子个数)(数论)
Problem Statement You are given an integer N. Find the number of the positive divisors of N!, modulo ...
- 钢琴培训班课程、课时及费用管理系统已提供ACM3.0新版下载
中小型艺术培训班课程.课时及费用管理系统. 2014新版 ACM3测试版下载:http://www.cnblogs.com/Charltsing/p/ACM3.html 您有任何功能需求,欢迎QQ 5 ...
- PS打造油画般的风景人像
- 后台管理系统之邮件开发(Java实现)
一,功能点 后台管理系统,添加用户时.对注册的新用户邮箱发送初始密码. 二,代码实现 1.Mail实体类 public class Mail { private Set<String> r ...
- ocrosoft 1015 习题1.22 求一元二次方程a*x^2 + b*x + c = 0的根
http://acm.ocrosoft.com/problem.php?id=1015 题目描述 求一元二次方程a*x2 + b*x + c = 0的根.系数a.b.c为浮点数,其值在运行时由键盘输入 ...
- Centos6.x升级内核方法支持Docker
Centos6升级内核方法_百度经验https://jingyan.baidu.com/article/7e4409531bda252fc1e2ef4c.html
- 使用ThreadLocal管理Mybatis中SqlSession对象
转自http://blog.csdn.net/qq_29227939/article/details/52029065 public class MybatisUtil { private stati ...