Codeforces Round #373 (Div. 2) A. Vitya in the Countryside 水题

A. Vitya in the Countryside

题目连接：

http://codeforces.com/contest/719/problem/A

Description

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.

Input

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.

Output

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.

Sample Input

5

3 4 5 6 7

Sample Output

UP

Hint

题意

现在告诉你，有一堆数是0,1,2,3,4,5,6，。。。。。15,14,13.。。。1,0这样循环的。

给你其中的部分数字，问你下一个数是比前一个数大，还是比前一个数小。

题解：

水题，只有几种情况，讨论一下就好了。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1005;
int a[maxn];
int main()
{
    int n;scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    if(a[n]==0)return puts("UP");
    if(a[n]==15)return puts("DOWN");
    if(n==1)return puts("-1");
    if(a[n-1]<a[n])return puts("UP");
    if(a[n-1]>a[n])return puts("DOWN");
}