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A. Vitya in the Countryside
time limit per test
memory limit per test
input
output
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.
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.
n
Input
n (1 ≤ n ≤ 92) — the number of consecutive days Vitya was watching the size of the visible part of the moon.
n integers ai (0 ≤ ai) — Vitya's records.
It's guaranteed that the input data is consistent.
Output
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.
Examples
input
5 3 4 5 6 7
output
UP
input
7 12 13 14 15 14 13 12
output
DOWN
input
1 8
output
-1
Note
8, thus the answer is "UP".
11, thus the answer is "DOWN".
7 or 9, thus the answer is -1.
题意：判断下一个值是上升还是下降，不能判断输出-1，分析最后两位，找特殊，剩下的就符合常规了。
#include<stdio.h>
#include<stdlib.h>
int main()
{
int n,s[100];
scanf("%d",&n);
for(int i=0; i<n; i++)
{
scanf("%d",&s[i]);
}
if(n==1)
{
if(s[0]==15)
printf("DOWN\n");
else if(s[0]==0)
printf("UP\n");
else
printf("-1\n");
}
else
{
if(s[n-2]==14&&s[n-1]==15)
{
printf("DOWN\n");
return 0;
}
if(s[n-2]==1&&s[n-1]==0)
{
printf("UP\n");
return 0;
}
if(s[n-2]<s[n-1])
{
printf("UP\n");
return 0;
}
else
{
printf("DOWN\n");
return 0;
}
}
return 0;
}
