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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.
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.
题意：0....15 14 ...0，这个规律，然后给你一串数字，问你下一位是下降还是上升
题解：很坑，首先必须想清楚，只有一个数的情况，0和15应该输出上升，15输出下降，其他输出-1，接下来讨论一般情况，末尾数字为0和15是上升和下降，其他有a[n-2]<a[n-1]和a[n-2]>a[n-1]，讨论完毕
#include<bits/stdc++.h>
using namespace std;
int n, a[10000];
int main()
{
scanf("%d",&n);
for(int i=0; i<n; i++)
{
scanf("%d",&a[i]);
}
if(n==1)
{
if(a[0]==0)
{
puts("UP");
}
else if(a[0]==15)
{
puts("DOWN");
}
else
{
puts("-1");
}
}
else
{
if(a[n-1]==0)
{
puts("UP");
}
else if(a[n-1]==15)
{
puts("DOWN");
}
else if(a[n-1]>a[n-2]&&a[n-1]!=15)
{
puts("UP");
}
else if(a[n-1]<a[n-2]&&a[n-1]!=0)
{
puts("DOWN");
}
}
return 0;
}
Codeforces Round #373 (Div. 2) A
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原文地址：http://www.cnblogs.com/yinghualuowu/p/5902688.html
