Codeforces Round #259 (div2) B solution report,

Codeforces Round #259 (div2) B solution report,

B. Little Pony and Sort by Shifttime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

One day, Twilight Sparkle is interested in how to sort a sequence of integersA1, bytes,A2, middle..., middle ,...,ANIn non-decreasing order. Being a young unicorn, the only operation she can perform is a unit shift. That is, she can move the last element of the sequence to its beginning:

A1, bytes, A2, middle..., middle ,..., A NNavigation → navigation A N, Bytes, A1, bytes, A2, middle..., middle ,..., A NAccept-encoding 1.

Help Twilight Sparkle to calculate: what is the minimum number of operations that she needs to sort the sequence?

Input

The first line contains an integerN(2 cores ≤ CoresNLimit ≤0000105). The second line containsNInteger numbersA1, bytes,A2, middle..., middle ,...,AN(1 digit ≤ DigitAILimit ≤ limit 105 ).

Output

If it's impossible to sort the sequence output-1. Otherwise output the minimum number of operations Twilight Sparkle needs to sort it.

Sample test (s) input

22 1

Output

1

Input

31 3 2

Output

-1

Input

21 2

Output

0

Question:

N numbers are given. Each time the last number is moved to the beginning, the final state is required to be a monotonic non-decreasing sequence, and the minimum number of operations is required. If the target cannot be reached, "-1" is output ".

Solution:

It is also a very easy programming basic question to find out the monotonous non-decreasing sequence of the two teams, which is 1 ~ X and x + 1 ~ Y: determines whether the two teams overwrite the entire number and a [n] <= a [1].

A simpler approach is to convert a [1] ~ Copy a [n] to a [1] ~ A [2 * n], and then in 1 ~ In 2 * n, check whether there is a sequence with n numbers that are monotonic and not decreasing.

Code:

#include <cstdio>#define N_max 123456int n, x, y, cnt;int a[N_max];void init() {scanf("%d", &n);for (int i = 1; i <= n; i++)  scanf("%d", &a[i]);}void solve() {for (int i = 1; i <= n; i++)if (a[i] > a[i+1]) {x = i;break;}if (x == n)y = n;elsefor (int i = x+1; i <= n; i++)if (a[i] > a[i+1]) {y = i;break;}if (x == n)printf("0\n");else if (y == n && a[y] <= a[1])printf("%d\n", y-x);elseprintf("-1\n");}int main() {init();solve();}