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.
【题解】
注意到只有1 和 15是只有一个的。
所以如果n=1,则只有单独为1或15才能确定它下一个是什么。否则都无法确定。
然后就是比较最后一个和倒数第二个。如果是递增的。除了最后一个数是15之外。都是单调递增。如果是递减的,除了最后一个数是0之外。都是单调递减的。
【代码】
#include <cstdio> #include <cstdlib> const int MAXN = 1000; int n, a[MAXN]; void input_data() { scanf("%d", &n); for (int i = 1; i <= n; i++) scanf("%d", &a[i]); } void special_judge() { if (n == 1) { if (a[1] != 0 && a[1] != 15) printf("-1 "); else if (a[1] == 0) printf("UP "); else printf("DOWN "); exit(0); } } void output_ans() { if (a[n] == 15) printf("DOWN "); else if (a[n] == 0) printf("UP "); else if (a[n] > a[n - 1]) printf("UP "); else printf("DOWN "); } int main() { // freopen("F:\rush.txt", "r", stdin); input_data(); special_judge(); output_ans(); return 0; }